maths study material - elementary definite integral
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1. If f & g are functions of x such that g′(x) = f(x) then,
∫ f(x) dx = g(x) + c ⇔ d
dx{g(x)+c} = f(x), where c is called the constant of integration.
2 .2 .2 .2 . Standard Formula:Standard Formula:Standard Formula:Standard Formula:
(i) ∫ (ax + b)n dx =( )
( )ax b
a n
n+
+
+1
1 + c, n ≠ −1
(ii) ∫ dx
ax b+ =
1
a ln (ax + b) + c
(iii) ∫ eax+b dx =1
a eax+b + c
(iv) ∫ apx+q dx =1
p
a
na
px q+
� + c; a > 0
(v) ∫ sin (ax + b) dx = −1
a cos (ax + b) + c
(vi) ∫ cos (ax + b) dx =1
a sin (ax + b) + c
(vii) ∫ tan(ax + b) dx =1
a ln sec (ax + b) + c
(viii) ∫ cot(ax + b) dx =1
a ln sin(ax + b)+ c
(ix) ∫ sec² (ax + b) dx =1
a tan(ax + b) + c
(x) ∫ cosec²(ax + b) dx = −1
acot(ax + b)+ c
(xi) ∫ sec (ax + b). tan (ax + b) dx =1
a sec (ax + b) + c
(xii) ∫ cosec (ax + b). cot (ax + b) dx = −1
a cosec (ax + b) + c
(xiii) ∫ secx dx = ln (secx + tanx) + c OR ln tan π
4 2+
x+ c
(xiv) ∫ cosec x dx = ln (cosecx − cotx) + c OR ln tanx
2
(xv) ∫d x
a x2 2− = sin−1
x
a + c
(xvi) ∫d x
a x2 2+
=1
a tan−1
x
a + c
(xvii) ∫d x
x x a2 2− =
1
a sec−1
x
a + c
(xviii) ∫d x
x a2 2+
= ln [ ]x x a+ +2 2 OR sinh−1
x
a + c
(xix) ∫d x
x a2 2− = ln [ ]x x a+ −2 2
OR cosh−1x
a + c
(xx) ∫d x
a x2 2−
=1
2a ln
xa
xa
−
+ + c
(xxi) ∫d x
x a2 2− =
1
2a ln
ax
ax
+
− + c
(xxii) ∫ a x2 2− dx =x
2a x2 2− +
a2
2 sin−1
x
a + c
(xxiii) ∫ x a2 2+ dx =x
2x a2 2+ +
a2
2 �n
++
a
axx 22
+ c
(xxiv) ∫ x a2 2− dx =x
2x a2 2− −
a2
2 �n
−+
a
axx 22
+ c
(xxv) ∫ eax. sin bx dx =e
a b
ax
2 2+ (a sin bx − b cos bx) + c
(xxvi) ∫ eax. cos bx dx =e
a b
ax
2 2+ (a cos bx + b sin bx) + c
3 .3 .3 .3 . Theorems on integrat ionTheorems on integrat ionTheorems on integrat ionTheorems on integrat ion
(i) ∫ dx).x(fc = c ∫ dx).x(f
(ii) ∫ ± dx))x(g)x(f( = ∫ ± dx)x(gdx)x(f
(iii) ∫ += c)x(gdx)x(f ⇒ ∫ + dx)bax(f = a
)bax(g + + c
Note : (i) every contineous function is integrable
(ii) the integral of a function reffered only by a constant.
∫ dx).x(f = g(x) + c
= h(x) + c
g′(x) = f(x) & h′(x) = f(x)
g′(x) – h′(x) = 0
means, g(x) – h(x) = c
Example : Evaluate : ∫ dxx4 5
Solution. ∫ dxx4 5 =
6
4 x6 + C =
3
2 x6 + C.
Example : Evaluate : ∫
++−+ dx
x
2
x
74x5x 23
Solution. ∫
++−+ dx
x
2
x
74x5x 23
= ∫ dxx3 + ∫ dxx5 2
– ∫ dx4 + ∫ dxx
7 + ∫ dx
x
2
= ∫ dxx3 + 5 . ∫ dxx2
– 4 . ∫ dx.1 + 7 . ∫ dxx
1 + 2 . ∫
− dxx 2/1
= 4
x4
+ 5 . 3
x3
– 4x + 7 log | x | + 2
2/1
x 2/1
+ C
= 4
x4
+ 3
5x3 – 4x + 7 log | x | + 4 x + C
Example : Evaluate : ∫ ++ alogaxlogaalogx eee dx
Solution. We have,
∫ ++ alogaxlogaalogx eee dx
= ∫ ++aax alogxlogalog eee dx
= ∫ ++ )axa( aax dx
= ∫ dxax + ∫ dxxa
+ ∫ dxaa
= alog
ax
+ 1a
x 1a
+
+
+ aa . x + C.
Example : Evaluate : ∫+x
xx
5
32 dx
Solution. ∫+x
xx
5
32 dx
= ∫
+
x
x
x
x
5
3
5
2 dx
= ∫
+
xx
5
3
5
2 dx =
5/2log
)5/2(
e
x
+ 5/3log
)5/3(
e
x
+ C
Example: Evaluate : ∫ xcosxsin 33 dx
Solution. = 8
1
3)xcosxsin2(∫ dx
= 8
1 ∫ x2sin3 dx
= 8
1 ∫
−
4
x6sinx2sin3 dx
= 32
1 ∫ − )x6sinx2sin3( dx
= 32
1
+− x6cos
6
1x2cos
2
3 + C
Example : Evaluate : ∫ +1x
x2
4
dx
Solution. ∫ +1x
x2
4
dx
= ∫ +
+−
1x
11x2
4
dx = ∫ +
−
1x
1x2
4
+ 1x
12 +
dx
= )1x( 2 −∫ dx + ∫ +1x
12 dx =
3
x3
– x + tan–1 x + C
Example: Evaluate : ∫ + 2x94
1 dx
Solution. We have
∫ + 2x94
1
= 9
1 ∫
+ 2x9
4
1 dx
= 9
1 ∫ + 22 x)3/2(
1 dx
= 9
1 .
)3/2(
1 tan–1
3/2
x + C =
6
1 tan–1
2
x3 + C
Example : ∫ dxx2cosxcos
Solution. ∫ dxx2cosxcos
= 2
1∫ dxx2cosxcos2
= 2
1∫ + )xcosx3(cos dx
= 2
1
+
1
xsin
3
x3sin + c
Self Practice Problems
1. Evaluate : ∫ xtan2 dx Ans. tanx – x + C
2. Evaluate : ∫ + xsin1
1 dx Ans. tanx – sec x + C
4.4 .4 .4 . Integration by SubsitutionsIntegration by SubsitutionsIntegration by SubsitutionsIntegration by Subsitutions
If we subsitute x = φ(t) in a integral then(i) everywhere x will be replaced in terms of t.(ii) dx also gets converted in terms of dt.(iii) φ(t) should be able to take all possible value that x can take.
Example : Evaluate : ∫43 xsinx dx
Solution. We have
Ι = ∫43 xsinx dx
L e t xL e t xL e t xL e t x
4 = t ⇒ d(x4) = dt ⇒ 4x3 dx = dt ⇒ dx = 3x4
1 dt
Example : ∫ x
)xn( 2�
dx
Solution. ∫ x
)xn( 2�
dx
Put �nx = t
⇒x
1 dx = dt
= ∫
x
dx.t2
= ∫ dtt2
= 3
t3
+ c
)xn( 3
Example : Evaluate ∫ + dxxcos)xsin1( 2
Solution. Put sinx = tcosx dx = dt
∫ + dt)t1( 2 = t +
3
t3
+ c
� + c=
3
= sin x + 3
xsin3
+ c
Example : Evaluate : ∫ ++ 1xx
x24 dx
Solution. We have,
Ι = ∫ ++ 1xx
x24 dx = ∫ ++ 1x)x(
x222 dx
Let x2 = t, then, d (x2) = dt ⇒ 2x dx = dt ⇒ dx = x2
dt
Ι = ∫ ++ 1tt
x2 .
x2
dt
= 2
1 ∫ ++ 1tt
12 dt
= 2
1 ∫
+
+
22
2
3
2
1t
1 dt
= 2
1 .
2
3
1 tan–1
+
2
3
2
1t
+ C
= 3
1 tan–1
+
3
1t2 + C =
3
1 tan–1
+
3
1x2 2
+ C.
Note: (i) ∫ [ f(x)]n f ′(x) dx = 1n
))x(f( 1n
+
+
(ii) ∫[ ]
′f x
f xn
( )
( ) dx =
n1
))x(f( n1
−
−
(iii) ∫ d x
x xn
( )+1 n ∈ N Take xn common & put 1 + x−n = t.
(iv) ∫( )
dx
x xnn
n21
1+−( ) n ∈ N, take xn common & put 1+x−n = tn
(v)
( )dx
x xn n
n
11
+∫ / take xn common as x and put 1 + x −n = t.
Self Practice Problems
1. ∫ +dx
xtan1
xsec2
Ans. �n |1 + tan x| + C
2. ∫ dxx
)nxsin(�Ans. – cos (�n x) + C
5 .5 .5 .5 . Integrat ion by Par t :Integrat ion by Par t :Integrat ion by Par t :Integrat ion by Par t :
( )∫ )x(g)x(f dx = f(x) ( )∫ )x(g dx – ( ) ( ) dxdx)x(g)x(fdx
d∫ ∫
(i) when you find integral ∫ dx)x(g then it will not contain arbitarary constant.
(ii) ∫ dx)x(g should be taken as same both terms.
(iii) the choice of f(x) and g(x) is decided by ILATE rule.
the function will come later is taken an integral function.
Ι → Inverse function
L → Logrithimic function
A → Algeberic function
T → Trigonometric function
E → Exponential function
Example : Evaluate : ∫− dxxtanx 1
Solution. ∫− dxxtanx 1
= (tan–1 x) 2
x2
– ∫ + 2x1
1 .
2
x2
dx
= 2
x2
tan–1 x – 2
1 ∫ +
−+
1x
11x2
2
dx = 2
x2
tan–1 x – 2
1 ∫ +
−1x
11
2 dx
= 2
x2
tan–1 x – 2
1 [x – tan–1 x] + C.
Example : Evaluate : ∫ + dx)x1log(x
Solution. ∫ + dx)x1log(x
= log (x + 1) . 2
x2
– ∫ +1x
1 .
2
x2
dx
= 2
x2
log (x + 1) – 2
1 ∫ +1x
x2
dx = 2
x2
log (x + 1) – 2
1 ∫ +
+−
1x
11x2
dx
= 2
x2
log (x + 1) – 2
1 ∫ +
−
1x
1x2
+ 1x
1
+ dx
= 2
x2
log (x + 1) – 2
1
++−∫ dx
1x
1)1x(
= 2
x2
log (x + 1) – 2
1
++− |1x|logx
2
x2
+ C
Example : Evaluate : ∫ x3sine x2 dx
Solution. Let Ι = ∫ x3sine x2dx. Then,
Ι = ∫ x3sine x2dx
⇒ Ι = e2x
−
3
x3cos – ∫
x2e2
−
3
x3cos dx
⇒ Ι = – 3
1 e2x cos 3x +
3
2 ∫ x3cose x2
dx
⇒ Ι = – 3
1 e2x cos 3x +
3
2
− ∫ dx
3
x3sine2
3
x3sine x2x2
⇒ Ι = – 3
1 e2x cos 3x +
9
2e2x sin 3x –
9
4 ∫ x3sine x2
dx
⇒ Ι = – 3
1 e2x cos 3x +
9
2e2x sin 3x –
9
4 Ι
⇒ Ι + 9
4 Ι =
9
e x2
(2 sin 3x – 3 cos 3x)
⇒9
13 Ι =
9
e x2
(2 sin 3x – 3 cos 3x)
⇒ Ι = 13
e x2
(2 sin 3x – 3 cos 3x) + C
Note :
(i) ∫ ex [f(x) + f ′(x)] dx = ex. f(x) + c
(ii) ∫ [f(x) + xf ′(x)] dx = x f(x) + c
Example : ∫xe 2)1x(
x
+ dx
Solution. ∫xe 2)1x(
11x
+
−+ dx
⇒ ∫ xe
+−
+ 2)1x(
1
)1x(
1 dx =
)1x(
ex
+ + c
Example : ∫ xe
−
−
xcos1
xsin1 dx
Solution. ∫xe
−
2
xsin2
2
xcos
2
xsin21
2 dx
⇒ ∫xe
−
2
xcoteccos
2
1 2 dx = – ex cot
2
x + c
Example : ∫
+
2)nx(
1)nx(n�
�� dx
Solution. put x = et
⇒ ∫te
+
2t
1nt� dt
⇒ ∫te
++−
2t
1
t
1
t
1nt� dt = et
−
t
1nt� + c
⇒ x
−
nx
1)nx(n�
�� + c
Self Practice Problems
1. ∫ dxxsinx Ans. – x cosx + sin x + C
2. ∫ dxex x2Ans. x2 ex – 2xex + 2ex + C
6.6 .6 .6 . Integration of Rational Algebraic Functions by using Partial Fractions:Integration of Rational Algebraic Functions by using Partial Fractions:Integration of Rational Algebraic Functions by using Partial Fractions:Integration of Rational Algebraic Functions by using Partial Fractions:
PARTIAL FRACTIONS :
If f(x) and g(x) are two polynomials, then )x(g
)x(f defines a rational algebraic function of a rational function
of x.
If degree of f(x) < degree of g(x), then )x(g
)x(f is called a proper rational function.
If degree of f(x) ≥ degree of g(x) then )x(g
)x(f is called an improper rational function
If )x(g
)x(f is an improper rational function, we divide f(x) by g(x) so that the rational function
)x(g
)x(f is
expressed in the form φ(x) + )x(g
)x(Ψ where φ(x) and ψ(x) are polynomials such that the degree of ψ(x) is
less than that of g(x). Thus, )x(g
)x(f is expressible as the sum of a polynomial and a proper rational
function.
Any proper rational function )x(g
)x(f can be expressed as the sum of rational functions, each having a
simple factor of g(x). Each such fraction is called a partial fraction and the process of obtained them is
called the resolutions or decomposition of )x(g
)x(f into partial fractions.
The resolution of )x(g
)x(f into partial fractions depends mainly upon the nature of the factors of g(x) as
discussed below.
CASE I When denominator is expressible as the product of non-repeating linear factors.
Let g(x) = (x – a1) (x – a
2) .....(x – a
n). Then, we assume that
)x(g
)x(f =
1
1
ax
A
− +
2
2
ax
A
− + ..... +
n
n
ax
A
−
where A1, A
2, ...... A
n are constants and can be determined by equating the numerator on R.H.S. to the
numerator on L.H.S. and then substituting x = a1, a
2, ........,a
n.
Example : Resolve 6x11x6x
2x323 −+−
+ into partial fractions.
Solution. We have, 6x11x6x
2x323 −+−
+ =
)3x)(2x)(1x(
2x3
−−−
+
Let)3x)(2x)(1x(
2x3
−−−
+ =
1x
A
− +
2x
B
− +
3x
B
−. Then,
⇒)3x)(2x)(1x(
2x3
−−−
+ =
)3x)(2x)(1x(
)2x)(1x(C)3x)(1x(B)3x)(2x(A
−−−
−−+−−+−−
⇒ 3x + 2 = A(x – 2) (x – 3) + B (x – 1) (x – 3) + C(x – 1) (x – 2) ...........(i)
Putting x – 1 = 0 or x = 1 in (i), we get
5 = A(1 – 2) (1 – 3) ⇒ A = 2
5,
Putting x – 2 = 0 or, x = 2 in (i), we obtain
8 = B (2 – 1) (2 – 3) ⇒ B = –8.
Putting x – 3 = 0 or, x = 3 in (i), we obtain
11 = C (3 – 1) (3 – 2) ⇒ C = 2
11.
∴6x11x6x
2x323 −+−
+ =
)3x)(2x)(1x(
2x3
−−−
+ =
)1x(2
5
− –
2x
8
− +
)3x(2
11
−
Note : In order to determine the value of constants in the numerator of the partial fraction corresponding to the
non-repeated linear factor px + q in the denominator of a rational expression, we may proceed as
follows :
Replace x = – p
q (obtained by putting px + q = 0) everywhere in the given rational expression except in
the factor px + q itself. For example, in the above illustration the value of A is obtained by replacing x
by 1 in all factors of )3x)(2x)(1x(
2x3
−−−
+ except (x – 1) i.e.
A = )31)(21(
213
−−
+× =
2
5
Similarly, we have
B = )32)(21(
123
−−
+× = –8 and, C =
)23)(13(
233
−−
+× =
2
11
Example : Resolve 6x5x
2x10x6x2
23
+−
−+− into partial fractions.
Solution. Here the given function is an improper rational function. On dividing we get
6x5x
2x10x6x2
23
+−
−+− = x – 1 +
)6x5x(
)4x(2 +−
+−...........(i)
we have, 6x5x
4x2 +−
+− =
)3x)(2x(
4x
−−
+−
So, let )3x)(2x(
4x
−−
+− =
2x
A
− +
3x
B
− – x + 4 = A(x – 3) + B(x – 2) ...........(ii)
Putting x – 3 = 0 or, x = 3 in (ii), we get
1 = B(1) ⇒ B = 1.
Putting x – 2 = 0 or, x = 2 in (ii), we get
2 = A (2 – 3) ⇒ A = – 2
∴)3x)(2x(
4x
−−
+− =
2x
2
−
− +
3x
1
−
Hence6x5x
2x10x6x2
23
+−
−+− = x – 1 –
2x
2
− +
3x
2
−
CASE II When the denominator g(x) is expressible as the product of the linear factors such that some
of them are repeating.
Example )x(g
1 =
)ax).......(ax)(ax()ax(
1
r21k −−−−
this can be expressed as
ax
A1
− + 2
2
)ax(
A
− + 3
3
)ax(
A
− + ....+ k
k
)ax(
A
− +
)ax(
B
1
1
− +
)ax(
B
2
2
− + ...... +
)ax(
B
r
r
−
Now to determine constants we equate numerators on both sides. Some of the constants are determined
by substitution as in case I and remaining are obtained by
The following example illustrate the procedure.
Example : Resolve )2x)(1x()1x(
2x32 ++−
− into partial fractions, and evaluate ∫ ++−
−
)2x)(1x()1x(
dx)2x3(2
Solution. Let )2x)(1x()1x(
2x32 ++−
− =
1x
A1
− + 2
2
)1x(
A
− +
1x
A3
+ +
2x
A4
+
⇒ 3x – 2 = A1 (x – 1) (x + 1) (x + 2) + A
2 (x + 1) (x + 2)
+ A3 (x – 1)2 (x + 2) + A
4 (x – 1)2 (x + 1) .......(i)
Putting x – 1 = 0 or, x = 1 in (i) we get
1 = A2 (1 + 1) (1 + 2) ⇒ A
2 =
6
1
Putting x + 1 = 0 or, x = –1 in (i) we get
– 5 = A3 (–2)2 (–1 + 2) ⇒ A
3 = –
4
5
Putting x + 2 = 0 or, x = –2 in (i) we get
– 8 = A4 (–3)2 (–1) ⇒ A
4 =
9
8
Now equating coefficient of x3 on both sides, we get 0 = A1 + A
3 + A
4
⇒ A1 = –A
3 – A
4 =
4
5 –
9
8 =
36
13
∴)2x)(1x()1x(
2x32 ++−
− =
)1x(36
13
− + 2)1x(6
1
− –
)1x(4
5
+ +
)2x(9
8
+
and hence ∫ ++−
−
)2x)(1x()1x(
dx)2x3(2
=36
13 �n |x – 1| –
)1x(6
1
− –
4
5 �n |x + 1| +
9
8 �n |x + 2| + c
CASE III When some of the factors of denominator g(x) are quadratic but non-repeating. Corresponding
to each quadratic factor ax2 + bx + c, we assume partial fraction of the type cbxax
BAx2 ++
+, where A and
B are constants to be determined by comparing coefficients of similar powers of x in the numerator of
both sides. In practice it is advisable to assume partial fractions of the type cbxax
)bax2(A2 ++
+ +
cbxax
B2 ++
The following example illustrates the procedure
Example : Resolve )2x)(1x(
1x22 ++
− into partial fractions and evaluate ∫ ++
−
)2x)(1x(
1x22 dx
Solution. Let)2x)(1x(
1x22 ++
− =
1x
A
+ +
2x
CBx2 +
+. Then,
)2x)(1x(
1x22 ++
− =
)2x)(1x(
)1x)(CBx()2x(A2
2
++
++++
⇒ 2x – 1 = A (x2 + 2) + (Bx + C) (x + 1) ...(i)
Putting x + 1 = 0 or, x = –1 in (i), we get – 3 = A(3) ⇒ A = –1.
Comparing coefficients of the like powers of x on both sides of (i), we get
A + B = 0, C + 2A = –1 and C + B = 2
∴ –1 + B = 0, C – 2 = –1 (Putting A = –1)
⇒ B = 1, C = 1
∴)2x)(1x(
1x22 ++
− = –
1x
1
+ +
2x
1x2 +
+
Hence ∫ ++
−
)2x)(1x(
1x22 dx
= – �n |x + 1| + 2
1 �n |x2 + 1| +
2
1 tan–1
2
x + c
CASE IV When some of the factors of the denominator g(x) are quadratic and repeating fractions of the
form
+++
++
+
cbxax
A
cbxax
)bax2(A2
1
2
0 + ( ) ( )
+++
++
+22
2
22
1
cbxax
A
cbxax
)bax2(A
+ .......+ ( ) ( )
+++
++
+−
k2
k2
k2
1k2
cbxax
A
cbxax
)bax2(A
The following example illustrates the procedure.
Example: Resolve 22 )1x)(1x(
3x2
+−
− into partial fractions.
Solution. Let 22 )1x)(1x(
3x2
+−
− =
1x
A
− +
1x
CBx2 +
+ + 22 )1x(
EDx
+
+. Then,
2x – 3 = A(x2 + 1)2 + (Bx + C) (x – 1) (x2 + 1) + (Dx + E) (x – 1) ......(i)
Putting x = 1 in (i), we get – 1 = A (1 + 1)2 ⇒ A = – 4
1
Equation coefficients of like powers of x, we have
A + B = 0, C – B = 0, 2A + B – C + D = 0, C + E – B – D = 2 and A – C – E = –3.
Putting A = – 4
1 and solving these equations, we get
B = 4
1 = C, D =
2
1 and E =
2
5
∴ 22 )1x)(1x(
3x2
+−
− =
)1x(4
1
−
− +
)1x(4
1x2 +
+ + 22 )1x(2
5x
+
+
Example : Resolve 1x
x23 −
into partial fractions.
Solution. We have, 1x
x23 −
= )1xx)(1x(
x22 ++−
So, let )1xx)(1x(
x22 ++−
= 1x
A
− +
1xx
CBx2 ++
+. Then,
2x = A (x2 + x + 1) + (Bx + C) (x – 1) .......(i)
Putting x – 1 = 0 or, x = 1 in (i), we get 2 = 3 A ⇒ A = 3
2
Putting x = 0 in (i), we get A – C = 0 ⇒ C = A = 3
2
Putting x = – 1 in (i), we get –2 = A + 2B – 2 C.
⇒ – 2 = 3
2 + 2B –
3
4 ⇒ B = –
3
2
∴1x
x23 −
= 3
2 .
1x
1
− +
1xx
3/2x3/22 ++
+− or, ,
1x
x23 −
= 3
2
1x
1
− +
3
2
1xx
x12 ++
−
Self Practice Problems
1. (i) ∫ ++dx
)3x)(2x(
1Ans. �n
3x
2x
+
+ + C
(ii) ∫ ++ )1x)(1x(
dx2 Ans.
2
1 �n |x + 1| –
4
1 �n (x2 + 1) +
2
1 tan–1 (x) + C2222 , ∫
++ ccccbxbxbxbxaxaxaxax dxdxdxdx2222 , ∫ ++ ccccbxbxbxbxaxaxaxax 2222 dx
Express ax2 + bx + c in the form of perfect square & then apply the standard results.
Example : Evaluate : ∫ ++ 5x2x2 dx
Solution. We have,
2
2 dx= ∫ x + 2x + 1+ 4
∫ x + 2x + 5
7777 .... IIIInnnntttteeeeggggrrrraaaatttt iiiioooonnnn ooooffff ttttyyyyppppeeee ∫ axaxaxax +bxbxbxbx+ccccdxdxdxdx
= 2
1 (x + 1) 22 2)1x( +− +
2
1 . (2)2 log |(x + 1) + 22 2)1x( ++ | + C
= 2
1 (x + 1) 5x2x2 +− + 2 log |(x + 1) + 5x2x2 ++ | + C
Example : Evaluate : ∫ +− 1xx
12 dx
Solution. ∫ +− 1xx
12 dx
= ∫+−+− 1
4
1
4
1xx
1
2 dx
= ∫ +− 4/3)2/1x(
12 dx
= ( )∫+−
22 2/3)2/1x(
1
dx = 2/3
1 tan–1
−
2/3
2/1x + C
= 3
2 tan–1
−
3
1x2 + C.
Example : Evaluate : ∫−+ 2xx89
1 dx
Solution. ∫−+ 2xx89
1 dx
= ∫−−− }9x8x{
1
2 dx
= ∫−+−− }2516x8x{
1
2 dx
= ∫ −−− }5)4x{(
122 dx = ∫
−− 22 )4x(5
1 dx = sin–1
−
5
4x + C
Self Practice Problems
1. ∫ −+ 1xx2
12 dx Ans.
3
1 �n
2x2
1x2
+
− + C
2. ∫−+ 2x3x2
1
2 dx Ans.2
1 log 1x
2
3x
4
3x 2 −++
+ + C
8 .8 .8 .8 . Integrat ion of typeIntegrat ion of typeIntegrat ion of typeIntegrat ion of type
∫ ++
+ ccccbxbxbxbxaxaxaxax qqqqpxpxpxpx2222 dx, ∫++
+ ccccbxbxbxbxaxaxaxax qqqqpxpxpxpx2222 dx, ∫∫∫∫ ++++++++++++ cbxax)qpx( 2 dx
Express px + q = A (differential co−efficient of denominator) + B.
Example : Evaluate : ∫++
+
1x4x
3x2
2 dx
Solution. ∫++
+
1x4x
3x2
2 dx
= ∫++
−+
1x4x
1)4x2(
2 dx
= ∫++
+
1x4x
4x2
2 dx – ∫++ 1x4x
1
2 dx
= ∫ t
dt –
( )∫
−+22 3)2x(
1 dx, where t = x2 + 4x + 1
= 2 t – log | (x + 2) + 1x4x2 ++ | + C
= 2 1x4x2 ++ – log | x + 2 + 1x4x2 ++ | + C
Example : Evaluate : ∫ +− xx)5x( 2 dx
Solution. Let (x – 5) = λ . dx
d (x2 + x) + µ. Then,
x – 5 = λ (2x + 1) + µ.
Comparing coefficients of like powers of x, we get
1 = 2λ and λ + µ = – 5 ⇒ λ = 2
1 and µ = –
2
11
∫ +− xx)5x( 2 dx
= ∫
−+
2
11)1x2(
2
1 xx2 + dx
= ∫ + )1x2(2
1 xx2 + dx –
2
11 ∫ + xx2
dx
= 2
1∫ + )1x2( xx2 + dx –
2
11 ∫ + xx2
dx
= 2
1
∫ t dt – 2
11 ∫
−
+
22
2
1
2
1x dx where t = x2 + x
= 2
1 .
2/3
t 2/3
– 2
11
−
+
+
22
2
1
2
1x
2
1x
2
1
– 2
1 .
2
2
1
log
−
++
+
22
2
1
2
1x
2
1x + C
= 3
1 t3/2 –
2
11
++
+−+
+xx
2
1xn
8
1xx
4
1x2 22� + C
= 3
1(x2 + x)3/2 –
2
11
++
+−+
+xx
2
1xn
8
1xx
4
1x2 22� + C
Self Practice Problems
1. ∫ ++
+
3xx
1x2 dx Ans.
2
1 log |x2 + x + 3| +
11
1 tan–1
+
11
1x2 + C
2. ∫+−
−
1x5x3
5x6
2 dx Ans. 2 1x5x3 2 +− + C
3. ∫ ++− 2xx1)1x( dx
Ans.3
1 (x2 + x + 1)3/2 –
8
3 (2x + 1) 2xx1 ++ –
16
9 log (2x +1 + 2 1xx2 ++ ) + C
9.9 .9 .9 . Integrat ion of trigonometric functionsIntegrat ion of trigonometric functionsIntegrat ion of trigonometric functionsIntegrat ion of trigonometric functions
(i) ∫ xxxxsinsinsinsinbbbbaaaa xxxxdddd 2222+ OR ∫ xxxxcoscoscoscosbbbbaaaa xxxxdddd 2222+
OR ∫ xxxxcoscoscoscosccccxxxxcoscoscoscosxxxxsinsinsinsinbbbbxxxxsinsinsinsinaaaa xxxxdddd 22222222 ++
Multiply Nr & Dr by sec² x & put tan x = t.
(ii) ∫ sinxsinxsinxsinxbbbbaaaa xxxxdddd+
OR ∫ cosxcosxcosxcosxbbbbaaaa xxxxdddd+
OR ∫ xxxxcoscoscoscosccccxxxxsinsinsinsinbbbbaaaa xxxxdddd++
Hint:
Convert sines & cosines into their respective tangents of half the angles and then,
put tan2
x = t
(iii) ∫ nnnnxxxxsinsinsinsin....mmmmxxxxcoscoscoscos.... ccccxxxxsinsinsinsin....bbbbxxxxcoscoscoscos....aaaa++
++
� dx. Express Nr ≡≡≡≡ A(Dr) + B
d
d x (Dr) + c & proceed.
Example : Evaluate : ∫ ++ xcosxsin1
1 dx
Solution. Ι = ∫ ++ xcosxsin1
1 dx
= ∫
+
−+
++
2/xtan1
2/xtan1
2/xtan1
2/xtan21
1
2
2
2 dx
= ∫ −+++
+
2/xtan12/xtan22/xtan1
2/xtan122
2
dx = ∫ + 2/xtan22
2/xsec2
dx
Putting tan2
x = t and
2
1 sec2
2
x dx = dt, we get
Ι = ∫ + 1t
1 dt = log | t + 1| + C = log 1
2
xtan + + C
Example : Evaluate : ∫ +
+
xsin2xcos3
xcos2xsin3 dx
Solution. Ι = ∫ +
+
xsin2xcos3
xcos2xsin3 dx
Let 3 sin x + 2 cos x = λ. dx
d (3 cos x + 2 sin x ) + µ (3 cos x + 2 sin x)
⇒ 3 sin x + 2 cos x = λ (–3 sin x + 2 cos x) + µ (3 cos x + 2 sin x )
Comparing the coefficients of sin x and cos x on both sides, we get
– 3λ + 2µ = 3 and 2λ + 3µ = 2 ⇒ µ = 13
12 and λ = –
13
5
∴ Ι = ∫ +
+λ++−µ
xsin2xcos3
)xsin2xcos3()xcos2xsin3( dx
= λ ∫ dx.1 + µ ∫ +
+−
xsin2xcos3
xcos2xsin3 dx
= λ x + µ ∫ t
dt, where t = 3 cos x + 2 sin x
= λ x + µ �n | t | + C = 13
5− x +
13
12 �n | 3 cos x + 2 sin x | + C
Example : Evaluate : ∫ ++
+
3xcos2xsin
2xcos3 dx
Solution. We have,
Ι = ∫ ++
+
3xcos2xsin
2xcos3 dx
Let 3 cos x + 2 = λ (sin x + 2 cos x + 3) + µ (cos x – 2 sin x) + ν
Comparing the coefficients of sin x, cos x and constant term on both sides, we get
λ – 2µ = 0, 2λ + µ = 3, 3λ + ν = 2
⇒ λ = 5
6, µ
5
3 and ν = –
5
8
∴ Ι = ∫ ++
ν+−µ+++λ
3xcos2xsin
)xsin2x(cos)3xcos2x(sin dx
⇒ Ι = λ ∫ µ+dx ∫ ++
−
3xcos2xsin
xsin2xcos dx + ν ∫ ++ 3xcos2xsin
1 dx
⇒ Ι = λ x + µ log | sin x + 2 cos x + 3 | + ν Ι1, where
Ι1 = ∫ ++ 3xcos2xsin
1 dx
Putting, sin x = 2/xtan1
2/xtan22+
, cos x = 2/xtan1
2/xtan12
2
+
− we get
Ι1 = ∫
++
−+
+3
2/xtan1
)2/xtan1(2
2/xtan1
2/xtan2
1
2
2
2
dx
= ∫ ++−+
+
)2/xtan1(32/xtan222/xtan2
2/xtan122
2
dx
= ∫ ++ 52/xtan22/xtan
2/xsec2
2
dx
Putting tan 2
x = t and
2
1 sec2
2
x = dt or sec2
2
x dx = 2 dt, we get
Ι1 = ∫ ++ 5t2t
dt22
= 2 ∫ ++ 22 2)1t(
dt =
2
2 tan–1
+
2
1t = tan–1
+
2
12
xtan
Hence, Ι = λx + µ log | sin x + 2 cos x + 3 | + ν tan–1
+
2
12
xtan
+ C
where λ = 5
6, µ =
5
3 and ν = –
5
8
Example : ∫ + xcos31
dx2
Solution. = ∫ + 4xtan
dxxsec2
2
= 2
1 tan–1
2
xtan + C
Self Practice Problems
1. ∫ +
+
xcos4xsin5
xcos5xsin4 dx Ans.
41
40x +
41
9 log |5sinx + 4cosx| + C
10 .10 .10 .10 . Integration of type Integration of type Integration of type Integration of type ∫ dxdxdxdxxxxxcoscoscoscosx.x.x.x.sinsinsinsin nnnnmmmmCase - ΙΙΙΙ
If m and n are even natural number then converts higher power into higher angles.
Case - ΙΙΙΙΙΙΙΙ
If at least m or n is odd natural number then if m is odd put cosx = t and vice-versa.
Case - ΙΙΙ ΙΙΙ ΙΙΙ ΙΙΙ
When m + n is a negative even integer then put tan x = t.
Example: ∫ dxxcosxsin 45
Solution. put cos x = t ⇒ – sinx dx = dt
= – ∫ − 22 )t1( . t4 . dt
= – ∫ +− )1t2t( 24t4 dt
= – ∫ +− )tt2t( 468 dt
= – 9
t9
+ 7
t2 7
– 5
t5
+ c
= – 9
xcos9
+ 2 7
xcos7
– 5
xcos5
+ c Ans.
Example : ∫− dx)x(cos)x(sin 3/73/1
Solution. ∫− dx)x(cos)x(sin 3/73/1
= ∫3/1)x(tan
xcos
12 dx
put tanx = t ⇒ sec2x dx = dt
= dtt 3/1
∫
= 4
3 t4/3 + c
= 4
3 (tanx)4/3 + c Ans.
Example : ∫ dxxcosxsin 42
Solution.8
1 ∫ + dx)x2cos1(x2sin2
= 8
1 ∫ dxx2sin2
+ 8
1 ∫ dxx2cosx2sin2
= 16
1 ∫ +−
16
1dx)x4cos1(
3
x2sin3
= 16
1 –
64
x4sin +
48
x2sin3
+ c
11 .11 .11 .11 . Integration of typeIntegration of typeIntegration of typeIntegration of type
∫ 1111xxxxKKKKxxxx 1111xxxx 22224444 2222++
± dx where K is any constant.
Divide Nr & Dr by x² & put x ∓ x
1 = t.
Example : ∫ 42
2
xx1
x1
++
−dx
Solution. ∫ 1x
1x
dxx
11
2
2
2
++
−
x + x
1 = t
⇒ – ∫ 1t
dt2 −
– 2
1 �n
1t
1t
+
− + C
– 2
1 �n
1x
1x
1x
1x
++
−+
+ C
Example : Evaluate : ∫ + 1x
14 dx
Solution. We have,
Ι = ∫ + 1x
14 dx
⇒ Ι = ∫+
2
2
2
x
1x
x
1
dx
⇒ Ι = 2
1 ∫
+2
2
2
x
1x
x
2
dx
⇒ Ι = 2
1 ∫
+
−
−
+
+
2
2
2
2
2
2
x
1x
x
11
x
1x
x
11
dx
⇒ Ι = 2
1 ∫
+
+
2
2
2
x
1x
x
11
dx – 2
1 ∫
+
−
2
2
2
x
1x
x
11
dx
⇒ Ι = 2
1 ∫
+
−
+
2x
1x
x
11
2
2
dx – 2
1 ∫
−
+
−
2x
1x
x
11
2
2
dx
Putting x – x
1 = u in 1st integral and x +
x
1 = ν in 2nd integral, we get
Ι = 2
1
( )∫+
22 2u
du
– 2
1 ( )∫
−ν
ν22 2
d
= 22
1 tan–1
2
u –
2
1
22
1 log
2
2
+ν
−ν + C
= 22
1 tan–1
−
2
x/1x –
24
1 log
2x/1x
2x/1x
++
−+ + C
= 22
1 tan–1
−
x2
1x2
– 24
1 log
12xx
1x2x2
2
++
+− + C
Self Practice Problem :
1. ∫ 1x7x
1x24
2
+−
− dx Ans.
6
1 �n
3x
1x
3x
1x
++
−+
+ C
2. ∫ xtan dx Ans.2
1 tan–1
2
y +
22
1 �n
2y
2y
+
− + C where y = tan x –
xtan
1
1 2 .1 2 .1 2 .1 2 . Integrat ion of typeIntegrat ion of typeIntegrat ion of typeIntegrat ion of type
∫ ++ qqqqpxpxpxpxb)b)b)b)(ax(ax(ax(ax dxdxdxdx OR ( )∫ +++ qqqqpxpxpxpxccccbxbxbxbxaxaxaxax dxdxdxdx2222 ; put px + q = t2.
Example: Evaluate : ∫ +− 1x)3x(
1 dx
Solution. Let Ι = ∫ +− 1x)3x(
1 dx
Here, P and Q both are linear, so we put Q = t2 i.e. x + 1 = t2 and dx = 2t dt
∴ Ι = 22
t
t2
)31t(
1∫ −−
dt
⇒ Ι = 2 ∫ − 22 2t
dt = 2 .
)2(2
1 log
2t
2t
+
− + C
⇒ Ι = 2
1 log
21x
21x
++
−+ + C.
Example : Evaluate : ∫ +++
+
1x)3x3x(
2x2 dx
Solution. Let Ι = ∫ +++
+
1x)3x3x(
2x2 dx
Putting x + 1 = t2, and dx = 2t dt, we get Ι = ∫+−+−
+
2222
2
t}3)1t(3)1t{(
dtt2)1t(
⇒ Ι = 2 ∫ ++
+
1tt
)1t(24
2
dt = 2 ∫++
+
1t
1t
t
11
2
2
2
dt
⇒ Ι = 2 ( )∫ ′+
22 3u
du
where t – t
1 = u.
⇒ Ι = 3
2 tan–1
3
u + C =
3
2 tan–1
−
3
t
1t
+ C
⇒ Ι = 3
2 tan–1
−
3t
1t2
+ C = 3
2 tan–1
+ )1x(3
x + C
13 .13 .13 .13 . Integration of typeIntegration of typeIntegration of typeIntegration of type
∫+++ rrrrqxqxqxqxpxpxpxpxb)b)b)b)(ax(ax(ax(ax dxdxdxdx 2222 , put ax + b = tttt1111 ; ∫
++ qqqqpxpxpxpxb)b)b)b)(ax(ax(ax(ax dxdxdxdx 22222222111121
Solution = ∫+
−
−
t
11
t
1
t
22
dt= ∫
+−
−
1t
1
t
1t
2
dt= ∫
+−
−
1tt2
+
−
−
4
3
2
1t
dtdt= ∫
1t
Example : ∫(x + ) x + x + 1dx , put x = tttt
= – �n
+
−+−
4
3
2
1t
2
1t
2
+ C
Example : ∫−+ 22 x1)x1(
dxSolution. Put x =
t
1 ⇒ Ι = ∫
−+ 1t1t 22 )(
dtput t2 – 1 = y2
⇒ Ι = – ∫ + y)2y( 2
dyy= –
2
1 tan–1
2
y + C
= – 2
1 tan–1
−
x2
x1 2
+ C
Self Practice Problems :
1. ∫ ++ 1x)2x(
dxAns. 2 tan–1 ( )1x + + C
2. ∫ +++ 1x)6x5x(
dx2 Ans. 2 tan–1 ( )1x + 2 tan–1
+
2
1x + C
3. ∫−++ 2xx1)1x(
dxAns. sin–1
+−
2
5
1x
1
2
3
+ C
4. ∫−+ 22 x1)1x2(
dxAns. –
3
1 tan–1
−2
2
x3
x1 + C
5. ∫−+++ 4x2x)2x2x(
dx
22 Ans. – 62
1�n
++−+
+−−+
)1x(64x2x
)1x(64x2x
2
2
+ C
1 4 .1 4 .1 4 .1 4 . Integrat ion of typeIntegrat ion of typeIntegrat ion of typeIntegrat ion of type
∫ −
− xxxxββββ ααααxxxx dx or ( )( )∫ −− xxxxββββααααxxxx ; put x = αααα cos2 θθθθ + ββββ sin2 θθθθ
∫ −
−ββββxxxx ααααxxxx dx or ( )( )∫ −− ββββxxxxααααxxxx ; put x = αααα sec2 θ −θ −θ −θ − ββββ tan2 θθθθ
( )( )∫ −− ββββxxxxααααxxxx dxdxdxdx; put x −−−− αααα = t2 or x −−−− ββββ = t2.
1 5 .1 5 .1 5 .1 5 . Reduct ion formula of Reduct ion formula of Reduct ion formula of Reduct ion formula of ∫ dxxntan , , , , ∫ dxxncot , , , , ∫ dxxnsec ,,,,
∫ dxxecncos
1. Ιn = ∫ dxxtann
= ∫ − dxxtanxtan 2n2 = ∫ − )1x(sec2
tann – 2x dx
⇒ Ιn = ∫
−2n2 tanxsec + dx – Ιn – 2
⇒ Ιn =
1n
xtan 1n
−
−
– Ιn – 2
2. Ιn = ∫ dxxcotn
= ∫− dxxcot.cot 2n2
= ∫−− dxxcot)1xec(cos 2n2
⇒ Ιn = ∫
− dxxcotxeccos 2n2 – Ι
n – 2
⇒ Ιn = –
1n
xcot 1n
−
−
– Ιn – 2
3. Ιn = ∫ dxxsecn
= ∫− dxxsecxsec 2n2
⇒ Ιn = tanx secn – 2x – ∫ − )2n)(x(tan secn – 3 x. secx tanx dx.
⇒ Ιn = tanx secn – 2 x dx – (n – 2) (sec2 x – 1) secn – 2x dx
⇒ (n – 1) Ιn = tanx secn – 2x + (n – 2) Ι
n – 2
Ιn =
1n
xsecxtan 2n
−
−
+ 1n
2n
−
− Ι
n – 2
4. Ιn = ∫ dxeccos n
= ∫ xeccos 2 cosecn – 2 x dx
⇒ Ιn = – cotx cosecn – 2x + ∫ − )2n)(x(cot (– cosecn – 3x cosec x cot x) dx
⇒ – cotx cosecn – 2x – (n – 2) ∫− dxxeccosxcot 2n2
⇒ Ιn = – cotx cosecn – 2x – (n – 2) ∫ − )1xec(cos 2
cosecn – 2 x dx
⇒ (n – 1) Ιn = – cotx cosecn – 2 x + (n – 2) 2
n – 2
Ιn =
1n
xeccosxcot 2n
−
−
+ 1n
2n
−
− Ι
n – 2
Example : Obtain reducation formula for Ιn = ∫ sinnx dx. Hence evaluate ∫ sin4x dx
Solution. Ιn = ∫ (sin x) (sin x)n –1 dx
ΙΙ Ι
= – cos x (sin x)n–1 + (n – 1) ∫ (sin x)n–2 cos2x dx
= – cos x (sin x)n–1 + (n – 1) ∫ (sin x)n–2 (1 – sin2x) dx
Ιn = – cos x (sin x)n–1 + (n – 1) Ι
n–2 – (n – 1) Ι
n
⇒ Ιn = –
n
)x(sinxcos 1n−
+ n
)1n( − Ι
n–2(n ≥ 2)
Hence Ι4 = –
4
)x(sinxcos 3
+ 4
3
+− x
2
1
2
)x(sinxcos + C
Self Practice Problems :
1. ∫ 4x
3x
−
− dx Ans. )4x)(3x( −− + �n ( )4x3x −+− + C
2. ∫ 2/3)]x2)(1x[(
dx
−−Ans. 8
−
−−
−
−
1x
x2
x2
1x + C
3. ∫ 7/168 ])1x()2x[(
dx
−+Ans. 7
7/1
2x
1x
+
− + C
4. Deduce the reduction formula for Ιn = ∫ n4 )x1(
dx
+ and Hence evaluate Ι
2 = ∫ 24 )x1(
dx
+
Ans. Ιn = 1n4 )x1)(1n(4
x−+−
+ )1n(4
5n4
−
− Ι
n–1
Ι2 =
)x1(4
x4+
+ 4
3
++
−+−
−
−
2x
1x
2x
1x
n24
1
2
x
1x
tan22
1 1� + C
5. If Ιm,n
= ∫ (sin x)m (cos x)n dx then prove that
Ιm,n
= nm
)x(cos)x(sin 1n1m
+
−+
+ nm
1n
+
− . Ι
m,n–2
Definite IntegralsPART A :
A Let f(x) be a continuous function defined on [a, b],
∫ dx)x(f = F(x) + c. Then ∫b
a
dx)x(f = F(b) – F(a) is called definite integral. This formula is known as Newton-
Leibnitz formula.
Note :
1. The indefinite integral ∫ dx)x(f is a function of x, where as definite integral ∫b
a
dx)x(f is a number..
2. Given ∫ dx)x(f we can find ∫b
a
dx)x(f , but given ∫b
a
dx)x(f we cannot find ∫ dx)x(f
Illustration. 1 Evaluate ∫ ++
2
1)2x)(1x(
dx
Sol. ∵)2x)(1x(
1
++ =
1x
1
+ –
2x
1
+(by partial fractions)
∫ ++
2
1)2x)(1x(
dx = [ ]2
1ee )2x(log)1x(log +−+
= loge
3 – loge
4 – loge
2 + loge
3 =
8
9
elog
Self Practice Problems
Evaluate the following
1. ∫ ++
2
1
2
2
3x4x
x5 dx Ans. 5 –
2
5
− 2
3
e4
5
e loglog9
2. ( )∫
π
++
2
0
32 2xxsec2 dx Ans.1024
4π +
2
π + 2
3. ∫
π
+
3
0xsec1
x dx Ans.
18
2π –
33
π + 2 log
e
3
2
PART B :Properties of definite integral
P – 1 ∫b
a
)x(f dx = ∫b
a
)t(f dt
i.e. definite integral is independent of variable of integration.
P – 2 ∫b
a
)x(f dx = – ∫a
b
)x(f dx
P – 3 ∫b
a
)x(f dx = ∫c
a
)x(f dx + ∫b
c
)x(f dx, where c may lie inside or outside the interval [a, b].
Illustration 2 If f(x) =
≥+
<+
3x:1x3
3x:3x2 , then find ∫
5
2
)x(f dx
Sol. ∫5
2
)x(f dx = ∫3
2
)x(f dx + ∫5
3
)x(f dx
= ∫ +
3
2
)3x( dx + ∫ +
5
3
2 )1x3( dx
= 2
49 − + 3 (3 – 2) + 53 – 33 + 5 – 3 =
2
211
Illustration 3 Evaluate ∫ −
8
2
|5x| dx
Sol. ∫ −
8
2
|5x| dx = ∫ +−
5
2
)5x( dx + ∫ +
8
5
)5x( dx = 9
Illustration 4 Show that ∫ +
2
0
)1x2( dx = ∫ +
5
0
)1x2( + ∫ +
2
5
)1x2(
Sol. L.H.S. = x2 + x ]20 = 4 + 2 = 6
R.H.S. = 25 + 5 – 0 + (4 + 2) – (25 + 5) = 6∴ L.H.S. R.H.S
Self Practice Problems
Evaluate the following
1. ∫ −+
2
0
2 |3x2x| dx Ans. 4
2. ∫3
0
]x[ dx , where [x] is integral part of x. Ans. 3
3. [ ]∫9
0
t dt Ans. 13
PART C :
P – 4 ∫−
a
a
)x(f dx = ∫ −+
a
0
))x(f)x(f( dx
= 2 ∫a
0
)x(f dx if f(–x) = f(x) i.e. f(x) is even
= 0 if f(–x) = –f (x) i.e. f(x) is odd
Illustration 5 Evaluate ∫−
−
+
+1
1
x
xx
e1
ee dx
Sol. ∫−
−
+
+1
1
x
xx
e1
ee dx = ∫
+
++
+
+−
−−1
0
x
xx
x
xx
e1
ee
e1
ee dx
= ∫
+
++
+
+ −−1
0
x
xxx
x
xx
1e
)ee(e
e1
ee dx = ∫ −+
1
0
xx )ee( dx = e – 1 + 1
)1e( 1
−
−−
= e
1e2 −
Illustration 6 Evaluate ∫
π
π−
2
2
xcos dx
Sol. ∫
π
π−
2
2
xcos dx = 2 ∫
π
2
0
xcos dx = 2 (∵ cos x is even function)
Illustration 7 Evaluate ∫−
+
−1
1
ex2
x2log dx
Sol. Let f(x) = loge
+
−
x2
x2
⇒ f(–x) = loge
−
+
x2
x2 = – log
e
+
−
x2
x2 = – f(x)
i.e. f(x) is odd function
∴ ∫−
+
−1
1
ex2
x2log dx = 0
Self Practice Problems
Evaluate the following
1. ∫−
1
1
|x| dx Ans. 1
2. ∫
π
π−
2
2
7 xsin dx Ans. 0
3. ∫
π
π−
+
2
2
xe1
xcos dx Ans. 1
PART D :
P – 5 ∫b
a
)x(f dx = ∫ −+
b
a
)xba(f dx
Further ∫a
0
)x(f dx = ∫ −
a
0
)xa(f dx
Illustration 8 Prove that ∫
π
+
2
0)x(cosg)x(sing
)x(sing dx = ∫
π
+
2
0)x(cosg)x(sing
)x(cosg dx =
4
π
Sol. Let Ι = ∫
π
+
2
0)x(cosg)x(sing
)x(sing dx
⇒ Ι = ∫
π
−
π+
−
π
−
π2
0 x2
cosgx2
sing
x2
sing
= ∫
π
+
2
0)x(sing)x(cosg
)x(cosg dx
on adding, we obtain
2Ι = ∫
π
++
+
2
0)x(sing)x(cosg
)x(cosg
)x(cosg)x(sing
)x(sing dx = ∫
π
2
0
dx ⇒ Ι = 4
π
Note : 1. The above illustration can be remembered as a formula
2. Other similar formulae are
∫
π
+
2
0)x(cotg)x(tang
)x(tang dx = ∫
π
+
2
0)x(cotg)x(tang
)x(cotg dx =
4
π
∫
π
+
2
0)x(secg)ecx(cosg
)ecx(cosg dx = ∫
π
+
2
0)x(secg)ecx(cosg
)x(secg dx =
4
π
∫ −+
a
0)xa(g)x(g
)x(g dx =
2
a
Self Practice Problems
Evaluate the following
1. ∫π
+0
xsin1
x dx Ans. π
2. ∫
π
+
2
0xcosxsin
x dx Ans.
22
π log
e ( )21+
3. ∫
π
+
2
0
44 xcosxsin
xcosxsinx dx Ans.
16
2π
4. ∫
π
π+
3
6
xtan1
dx
Ans.12
π
PART E :
P – 6 ∫a2
0
)x(f dx = ∫ −+
a
0
))xa2(f)x(f( dx
= 2 ∫a
0
)x(f dx if f (2a – x) = f(x)
= 0 if f (2a – x) = –f(x)
Illustration 9 Evaluate ∫π
0
33 xcosxsin dx
Sol. Let f(x) = sin3x cos3x ⇒ f(π – x) = – f(x)
∴ ∫π
0
33 xcosxsin dx = 0
Illustration 10 Evaluate ∫π
+0
2 xsin21
dx dx
Sol. Let f(x) = xsin21
12+
⇒ f(π – x) = f(x)
⇒ ∫π
+0
2 xsin21
dx = 2 ∫
π
+
2
0
2 xsin21
dx = 2 ∫
π
++
2
0
22
2
xtan2xtan1
dxxsec
= 2 ∫
π
+
2
0
2
2
xtan31
dxxsec =
3
2 ( )[ ]2
01 xtan3tan
π
−
∵ tan 2
π is undefined, we take limit
= 3
2
( ) ( )
− −−
π→
−0tan3tanxtan3tanLt 11
2x
= 3
2
2
π =
3
π
Note : We can evaluate the integral without using this property
Alternatively : ∫π
+0
2 xsin21
dx = ∫
π
+0
2
2
2xeccos
xeccos dx = ∫
π
+0
2
2
3xcot
dxxeccos
Observe that we are not converting in terms of tan x as it is not continuous in (0, π)
= – 3
1
π
−
0
1
3
xcottan = –
3
1
−
−
→
−
π→ +− 3
xcottanLt
3
xcottanLt 1
0x
1
x
= – 3
1
π−
π−
22 =
3
π
Note : If we convert in terms of tan x, then we have to break integral using property P – 3.
Illustration 11 Prove that ∫
π
2
0
e xsinlog dx = ∫
π
2
0
e xcoslog dx = ∫
π
2
0
e )x2(sinlog dx = – 2
π log
e
2 .
Sol. Let Ι = ∫
π
2
0
e xsinlog dx ..........(i)
⇒ Ι = ∫
π
−
π2
0
e x2
sinlog dx (by property P – 5)
Ι = ∫
π
2
0
e )x(coslog dx ..........(ii)
Adding (i) and (ii)
2 Ι = ∫
π
2
0
e )xcos.x(sinlog dx = ∫
π
2
0
e2
x2sinlog dx
2 Ι = ∫
π
2
0
e )x2(sinlog dx – ∫
π
2
0
2elog dx
2 Ι = Ι1 –
2
π log2
e..........(iii)
where Ι1 = ∫
π
2
0
e )x2(sinlog dx
put 2x = t ⇒ dx = 2
1 dt
L . L : x = 0 ⇒ t = 0
U . L : x = 2
π⇒ t = π
⇒ Ι1 = ∫
π
0
e )t(sinlog 2
1 dt
= 2
1 × 2 ∫
π
2
0
e )t(sinlog dt (by using property P – 6)
⇒ Ι1 = Ι ∴ (iii) gives Ι = –
2
π 2
elog
Self Practice Problems
Evaluate the following
1. ∫∞
+
+
0
2
e
x1
x
1xlog
dx : Ans : π loge2
2. ∫−1
0
1
x
xsin dx : Ans :
2
π log
e2
3. ∫π
0
e xsinlogx dx Ans : –2
2π log
e2
PART F :
P – 7 If f(x) is a periodic function with period T, then
(i) ∫nT
0
)x(f dx = n ∫T
0
)x(f dx, n ∈ z
(ii) ∫+nTa
a
)x(f dx = n ∫T
0
)x(f dx, n ∈ z, a ∈ R
(iii) ∫nT
mT
)x(f dx = (n – m) ∫T
0
)x(f dx, m, n ∈ z
(iv) ∫+nTa
nT
)x(f dx = ∫a
0
)x(f dx, n ∈ z, a ∈ R
(v) ∫+
+
nTb
nTa
)x(f dx = ∫a
a
)x(f dx, n ∈ z, a, b ∈ R
Illustration 12 Evaluate ∫−
2
1
}x{e dx
Sol. ∫−
2
1
}x{e dx = ∫+−
−
31
1
}x{e dx = 3 ∫1
0
}x{e dx = 3 ∫1
0
}x{e dx = 3(e – 1)
Illustration 13 Evaluate ∫+π vn
0
|xcos| dx , 2
π < v < π and n ∈ z
Sol. ∫+π vn
0
|xcos| dx = ∫v
0
|xcos| dx + ∫+π vn
v
|xcos| dx
= ∫
π
2
0
xcos – ∫π
v
xcos dx + n ∫π
0
|xcos| dx
= (1 – 0) – (sin v – 1) + 2n ∫
π
2
0
xcos dx
= 2 – sin v + 2n (1 – 0) = 2n + 2 – sin v
Self Practice Problem
Evaluate the following
1. ∫−
2
1
}x3{e dx Ans. 3 (e – 1)
2. ∫π
+
2000
0
xsine1
dx dx Ans. 1000 π
3. ∫
π
π+
4
5
44 xcosxsin
x2sin dx Ans.
4
π
PART G :
P – 8 If ψ(x) ≤ f(x) ≤ φ (x) for a ≤ x ≤ b, then
∫ψ
b
a
)x( dx ≤ ∫b
a
)x(f dx ≤ ∫φ
b
a
)x( dx
P – 9 If m ≤ f(x) ≤ M for a ≤ x ≤ b, then m (b – a) ≤ ∫b
a
)x(f dx ≤ M (b – a)
Further if f(x) is monotonically decreasing in (a, b) then f(b) (b – a) < ∫b
a
)x(f dx < f(a) (b – a) and if f(x)
is monotonically increasing in (a, b) then f(a) (b – a) < ∫b
a
)x(f dx < f(b) (b – a)
P – 10 ∫b
a
dx)x(f ≤ ∫b
a
dx)x(f
P – 11 If f(x) ≥ 0 on [a, b] then ∫b
a
dx)x(f ≥ 0
Illustration 14 For x ∈ (0, 1) arrange f1(x) =
2x4
1
−, f
2(x) =
2x24
1
− and f
3(x) =
32 xx4
1
−− in ascending
order and hence prove that 6
π < ∫
−−
1
032 xx4
dx <
24
π
Sol. ∵ 0 < x3 < x2 ⇒ x2 < x2 + x3 < 2x2 ⇒ –2x2 < – x2 – x3 < –x2
⇒ 4 – 2x2 < 4 –x2 – x3 < 4 – x2
⇒ 2x24 − < 32 xx4 −− < 2x4 −
⇒ f1(x) < f
3(x) < f
2(x) for x ∈ (0, 1)
⇒ ∫1
0
1 )x(f dx < ∫1
0
3 )x(f dx < ∫1
0
2 )x(f dx
sin–1
1
02
x
< ∫
−−
1
032 xx4
dx <
2
1 sin–1
1
02
x
6
π < ∫
−−
1
032 xx4
dx <
24
π
Illustration 15 Estimate the value of ∫
π
2
0x
xsin dx
Sol. Let f(x) = x
xsin
f′(x) = 2x
xsinxcosx − = 2x
)xtanx)(x(cos − < 0
⇒ f(x) is monotonically decreasing function.f(0) is not defined, so we evaluate
+→0xLt f(x) = +→0x
Lt x
xsin = 1. Take f(0) = +→0x
Lt f(x) = 1
f
π
2 =
π
2
π
2 .
−
π0
2 < ∫
π
2
0x
xsin dx < 1 .
−
π0
2
1 < ∫
π
2
0x
xsin dx <
2
π
Note : Here by making the use of graph we can make more appropriate approximation as in next illustration.
Illustration 16 Estimate the value of ∫1
0
x dxe2
using (i) rectangle, (ii) triangle
Sol. (i) By using rectangle
Area OAED < ∫1
0
x dxe2
< Area OABC
1 < ∫1
0
x dxe2
< 1 . e
1 < ∫1
0
x dxe2
< e
(ii) By using triangle
Area OAED < ∫1
0
x dxe2
< Area OAED + Area of triangle DEB
1 < ∫1
0
x dxe2
< 1 + 2
1 . 1. (e – 1) 1 < ∫
1
0
x dxe2
< 2
1e +
Illustration 17 Estimate the value of ∫1
0
x dxe2
by using ∫1
0
xdxe
Sol. For x ∈ (0, 1), 2xe < ex
⇒ 1 × 1 < ∫1
0
x dxe2
< ∫1
0
xdxe
1 < ∫1
0
x dxe2
< e – 1
Exercise : Prove the following :
1. ∫−
1
0
2x xcose dx < ∫−
1
0
2x xcose2
dx
2. 0 < ∫
π
+2
0
1n xsin dx < ∫
π
2
0
2 xsin dx
3. 4
1
e−
< ∫−
1
0
xx2
e dx < 1
4. – 2
1 ≤ ∫ +
1
0
2
3
x2
xcosx dx <
2
1
5. 1 < ∫
π
2
0
xsin dx < 2
π
6. 0 < ∫ +
2
0
3x16
dxx <
6
1
PART - H
Leibnitz Theorem : If F(x) = ∫)x(h
)x(g
dt)t(f , then
dx
)x(dF = h′(x) f(h(x)) – g′(x) f(g(x))
Proof : Let P(t) = ∫ dt)t(f
⇒ F(x) = ∫)x(h
)x(g
dt)t(f = P(h(x)) – P(g(x))
⇒dx
)x(dF= P′(h(x)) h′(x) – P′(g(x)) g′(x)
= f(h(x)) h′(x) – f (g(x)) g′(x)
Illustration 18 If F(x) = dttsin
2x
x
∫ , then find F′(x)
Sol. F′(x) = 2x . 2xsin – 1 . xsin
Illustration 19 If F(x) = dtlog
tx3
x2
e
e
te
∫ , then find first and second derivative of F(x) with respect to logxe
at x = 2elog
Sol. ( )xelogd
)x(dF =
dx
)x(dF
)x(logd
dx =
−
x2x3 ee
x2x2
ee
x3x3
log
ee.2
log
e.e.3 x = e6x – e4x.
( )2xe
2
logd
)x(Fd = ( )x
elogd
d (e6x – e4x) =
dx
d (e6x – e4x) ×
dx
logd
1xe
= (6 e6x – 4 e4x) x
First derivative of F(x) at x = log2e (i.e. ex = 2) is 26 – 24 = 48
Second derivative of F(x) at x = log2e (i.e. ex = 2) is (6 . 26 – 4 . 24) . log
e2 = 5 . 26 . log2
e.
Illustration 20 Evaluate ∞→x
Lt
∫
∫
x
0
t2
2x
0
t
dte
dte
2
2
Sol.∞→x
Lt
∫
∫
x
0
t2
2x
0
t
dte
dte
2
2
∞
∞form
Applying L′ Hospital rule
=∞→x
Lt2
22
x2
x
0
xt
e.1
e.dte.2 ∫
=∞→x
Lt2
2
x
x
0
t
e
dte.2 ∫=
∞→xLt 2
2
x
x
e.x2
e.2 = 0
Modified Leibnitz Theorem :
If F(x) = ∫)x(h
)x(g
)t,x(f dt, then
FFFF ′(x) = ∫ ∂
∂)x(h
)x(gx
)t,x(f dt + f(x, h(x))h′(x) – f(x, g(x)) . g′(x)
Illustration 21 If f(x) = ∫ +
x
logxe
tx
dt
, then find f′(x)
Sol. f′(x) = ( )∫ +
−x
log
2xe
dttx
1 + 1 .
x2
1 –
x
1 ( )x
elogx
1
+ =
x
logxe
)tx(
1
+ + x2
1 – ( )x
elogxx
1
+
= x2
1 – x
elogx
1
+ +
x2
1 – ( )x
elogxx
1
+=
x
1 – ( )x
elogxx
1x
+
+ = ( )x
e
xe
logxx
1log
+
−
Alternatively : f(x) =
x
log
e
x
log xe
xe
)tx(logtx
dt
+=
+∫ (treating ‘t’ as constant)
f(x) = loge2x – log
e (x + log x
e)
f′(x) = x
1 – ( )x
elogx
1
+
+
x
11 = ( )x
e
xe
logxx
1log
+
−
Definite Integrals dependent on parameters :
Illustration 23 Evaluate ∫−
1
0
xe
b
log
1x, ‘b’ being parameter
Sol. Let Ι(b) = ∫−
1
0
xe
b
log
1x dx
db
)b(dΙ = ∫
1
0
xe
xe
b
log
logx dx + 0 – 0
(using modified Leibnitz Theorem)
= ∫1
0
b dxx =
1
0
1b
1b
x
+
+
= 1b
1
+
Ι(b) = loge (b + 1) + c
b = 0 ⇒ Ι(0) = 0∴ c = 0 ∴ Ι(b) = log
e (b+1)
Illustration 24 Evaluate ∫−
−1
02
1
x1x
)ax(tan dx , ‘a’ being parameter
Sol. Let Ι(a) = ∫−
−1
02
1
x1x
)ax(tan dx
da
)a(dΙ = ∫ +
1
0
22 )xa1(
x 2x1x
1
− dx = ∫
−+
1
0222 x1)xa1(
dx
Put x = sin t ⇒ dx = cos t dtL.L. : x = 0 ⇒ t = 0
U.L. : x = 1 ⇒ t = 2
π
da
)a(dΙ = ∫
π
+
2
0
22 tsina1
1
tcos
1 cos t dt = ∫
π
+
2
0
22 tsina1
dt
= ∫
π
++
2
0
22
2
ttan)a1(1
dttsec =
2a1
1
+ tan–1
2
0
2 ttana1
π
+
= 2a1
1
+ .
2
π
⇒ Ι(a) = 2
π log
e
++ 2a1a + c
But Ι(0) = 0 ⇒ c = 0
⇒ Ι(a) = 2
π log
e
++ 2a1a
Self Practice Problems :
1. If f(x) = ∫3x
0
tcos dt, find f′(x). Ans. 3x2 3xcos
2. If f(x) = eg(x) and g(x) = ∫ +
x
2
4t1
t dt then find the value of f′(2). Ans.
17
2
3. If x = ∫+
y
02t41
dt and 2
2
dx
yd = Ry then find R Ans. 4
4. If f(x) = ∫2x
x
2 dttsinx then find f′(x). Ans. x2 (2x sin x2 – sin x) + (cos x – cos x2) x
5. If φ(x) = cos x – ∫ φ−
x
0
)t()tx( dt, then find the value of φ′′(x) + φ(x). Ans. – cos x
6. Find the value of the function f(x) = 1 + x + ( )∫ +
x
1
te
2te log2)(log dt where f′(x) vanishes. Ans. 1 +
e
2
7. Evaluate 0x
Lt→
xsinx
dttcos
2x
0
2∫. Ans. 1
8. Evaluate ∫π
+
0
e )xcosb1(log dx, ‘b’ being parameter.. Ans. π loge
−+
2
b11 2
PART - ΙΙΙΙ
Definite Integral as a Limit of Sum.
Let f(x) be a continuous real valued function defined on the closed interval [a, b] which is divided into n parts
as shown in figure.
y = f(x)
a a+h a+2h ................ a+(n-1)h a+nh=bx
The point of division on x-axis are a, a + h, a + 2h ..........a + (n – 1)h, a + nh, where n
ab − = h.
Let Sn denotes the area of these n rectangles.
Then, Sn = hf(a) + hf(a + h) + hf(a + 2h) + ........+hf(a + (n – 1)h)
Clearly, Sn is area very close to the area of the region bounded by curve y = f(x), x–axis and the ordinates
x = a, x = b.
Hence ∫b
a
dx)x(f = ∞→n
Lt Sn
∫b
a
dx)x(f = ∞→n
Lt ∑−
=
+
1n
0r
)rha(fh = ∞→n
Lt ∑−
=
−1n
0rn
ab f
−+
n
r)ab(a
Note :
1. We can also write
Sn = hf(a + h) + hf (a + 2h) + .........+ hf(a + nh) and ∫
b
a
dx)x(f = ∞→n
Lt ∑=
−n
1rn
ab f
−+ r
n
aba
2. If a = 0, b = 1, ∫1
0
dx)x(f = ∞→nLt ∑
−
=
1n
0r n
rf
n
1
Steps to express the limit of sum as definte integral
Step 1. Replace n
r by x,
n
1 by dx and ∞→n
Lt ∑ by ∫
Step 2. Evaluate ∞→nLt
n
r by putting least and greatest values of r as lower and upper limits respectively..
For example ∞→nLt ∑
=
pn
1rn
rf
n
1 = ∫
p
0
dx)x(f (∵ 1r
n n
rLt
=∞→
= 0,
nprn n
rLt
=∞→
= p)
Illustration 25 : Evaluate
∞→nLt
++
++
++
+ n2
1.........
n3
1
n2
1
n1
1
Sol. ∞→nLt
++
++
++
+ n2
1.........
n3
1
n2
1
n1
1
= ∞→nLt ∑
=+
n
1rnr
1
= ∞→nLt ∑
=
n
1rn
1
1n
r
1
+
= ∫ +
1
01x
dx = [ ]10e )1x(log + = log
e2.
Illustration 26 : Evaluate ∞→nLt
++
+
++
+
++
+
+
n5
3.........
3n
3n
2n
2n
1n
1n222222
Sol. ∞→nLt ∑
= +
+n2
1r22 rn
rn = ∞→n
Lt ∑=
n2
1rn
1 2
n
r1
n
r1
+
+
∵ ∞→nLt
n
r = 0, when r = 1, lower limit = 0
and ∞→nLt
n
r = ∞→n
Lt
n
n2 = 2, when r = 2n, upper limit = 2
∫ +
+2
0
2x1
x1 dx = ∫ +
2
0
2x1
1 dx +
2
1 ∫ +
2
0
2x1
x2 dx
= tan–1x]20 +
2
0
2e )x1(log
2
1
+
= tan–1 2 + 2
1 log
e5
Illustration 27 : Evaluate
∞→nLt
n
1
nn
!n
Sol. Let y = ∞→nLt
n
1
nn
!n
loge y = ∞→n
Lt n
1 log
e
nn
!n
= ∞→nLt
n
1log
e
nn
n........3.2.1
= ∞→nLt
n
1
++
+
+
n
nlog.....
n
3log
n
2log
n
1log eeee
= ∞→nLt
n
1 ∑=
n
1r
en
rlog
=
1
0
e
1
0
e xxlogxdxxlog
−=∫
= (0 – 1) – +→0x
Lt x logex + 0
= – 1 – 0 = –1
⇒ y = e
1
Self Practice Problems :
Evaluate the following limits
1. ∞→nLt
+++
++
++
22222 nn
1....
n2n
1
nn
1
n
1Ans. 2 ( )12 −
2. ∞→nLt
++
++
++
+ n5
1.......
n3
1
n2
1
n1
1Ans. log
e5
3. ∞→nLt 2n
1
π++
π+
π+
π
n4
nsinn........
n4
3sin3
n4
2sin2
n4sin 3333
Ans. 29
2
π (52 – 15π)
4. ∞→nLt ∑
−
=
1n
0r22 rn
1
−Ans.
2
π
5. ∞→nLt
n
3
−+++
++
++
++
)1n(3n
n......
9n
n
6n
n
3n
n1 Ans. 2
PART – J
Reduction Formulae in Definite Integrals
1. If Ιn = ∫
π
2
0
n xsin dx , then show that Ιn =
−
n
1n Ι
n–2
Proof : Ιn = ∫
π
2
0
n xsin dx
Ιn = [ ]201n xcosxsin
π
−− + ∫
π
−−
2
0
22n dxxcos.xsin)1n(
= (n – 1) ∫
π
− −
2
0
22n dx)xsin1(.xsin
= (n – 1) ∫
π
− −−
2
0
2n )1n(dxxsin ∫
π
2
0
n dxxsin
Ιn + (n – 1) Ι
n = (n – 1) Ι
n–2
Ιn =
−
n
1n Ι
n–2
Note : 1. ∫
π
2
0
n dxxsin = ∫
π
2
0
n dxxcos
2. Ιn =
−
n
1n
−
−
2n
3n
−
−
4n
5n ..... Ι
0 or Ι
1
according as n is even or odd. Ι0 =
2
π, Ι
1 = 1
Hence Ιn =
−
−
−
−
−
π
−
−
−
−
−
oddisnif1.3
2........
4n
5n
2n
3n
n
1n
evenisnif2
.2
1........
4n
5n
2n
3n
n
1n
2. If Ιn = ∫
π
4
0
n dxxtan , then show that Ιn + Ι
n–2 =
1n
1
−
Sol. Ιn = ∫
π
−4
0
2n)x(tan . tan2x dx
= ∫
π
−4
0
2n)x(tan (sec2x – 1) dx
= ∫
π
−4
0
2n)x(tan sec2x dx – ∫
π
−4
0
2n)x(tan dx
= 4
0
1n
1n
)x(tan
π−
− – Ι
n–2
Ιn =
1n
1
− – Ι
n–2
∴ Ιn + Ι
n–2 =
1n
1
−
3. If Ιm,n
= ∫
π
2
0
nm dxxcos.xsin , then show that Ιm,n
= nm
1m
+
− Ι
m–2 , n
Sol. Ιm,n
= ∫
π
−2
0
n1m )xcosx(sinxsin dx
= 2
0
1n1m
1n
xcos.xsin
π+−
+− + ∫
π
+
+
2
0
1n
1n
xcos (m – 1) sinm–2 x cos x dx
=
+
−
1n
1m ∫
π
−2
0
2n2m xcos.xcos.xsin dx
=
+
−
1n
1m ( )∫
π
− −
2
0
nmn2m xcos.xsinxcos.xsin dx
=
+
−
1n
1m Ι
m–2,n –
+
−
1n
1m Ι
m,n
⇒
+
−+
1n
1m1 Ι
m,n =
+
−
1n
1m Ι
m–2,n
Ιm,n
=
+
−
nm
1m Ι
m–2,n
Note : 1. Ιm,n
=
+
−
nm
1m
−+
−
2nm
3m
−+
−
4nm
5m ........ Ι
0,n or Ι
1,n according as m is even or odd.
Ι0,n
= ∫
π
2
0
n dxxcos and Ι1,n
= ∫
π
2
0
n dxxcos.xsin = 1n
1
+
2. Walli’s Formula
Ιm,n
=
−+−++
−−−−−−
π
−+−++
−−−−−−
otherwise)........4nm()2nm()nm(
).......5n()3n()1n.........()5m()3m()1m(
evenaren,mbothwhen2)........4nm()2nm()nm(
).......5n()3n()1n.........()5m()3n()1n(
Illustration 28 : Evaluate ∫
π
π−
+2
2
22 dx)xcosx(sinxcosxsin
Sol. Given integral = ∫
π
π−
2
2
23 dxxcosxsin + ∫
π
π−
2
2
32 dxxcosxsin
= 0 + 2 ∫
π
2
0
32 dxxcosxsin (∵ sin3x cos2x is odd and sin2x cos3x is even)
= 2. 1.3.5
2.1 =
15
4
Illustration 29 : Evaluate ∫π
0
65 dxxcosxsinx
Sol. Let Ι = ∫π
0
65 dxxcosxsinx
Ι = ∫π
−π−π−π
0
65 dx)x(cos)x(sin)x(
= π ∫π
0
65 dxxcos.sin – ∫π
0
65 dxxcos.xsinx
⇒ 2Ι = π . 2 ∫
π
2
0
65 dxxcos.xsin
Ι = π 1.3.5.7.9.11
1.3.5.2.4
Ι = 693
8π
Illustration 30 : Evaluate ∫ −
1
0
53 dx)x1(x
Sol. Put x = sin2θ ⇒ dx = 2 sin θ cos θ dθL . L : x = 0 ⇒ θ = 0
U.L. : x = 1 ⇒ θ = 2
π
∴ ∫ −
1
0
53 dx)x1(x = ∫
π
θθ
2
0
526 )(cossin 2 . sin θ . cos θ dθ
= 2 . ∫
π
θθθ
2
0
117 dcossin
= 2 . 2.4.6.8.10.12.14.16.18
2.4.6.8.10.2.4.6=
504
1
Self Practice Problems:
Evaluate the following
1. ∫
π
2
0
5 dxxsin Ans.15
18
2. ∫
π
2
0
45 dxxcosxsin Ans.315
8
3. ∫ −
1
0
16 dxxsinx Ans.14
π –
245
16
4. ( )∫ −
a
0
2
722 dxxax Ans.
9
a9
5. ∫ −
2
0
2/3 dxx2x Ans.2
π
KEY CONCEPTS1. DEFINITION :
If f & g are functions of x such that g′(x) = f(x) then the function g is called a PRIMITIVE OR
ANTIDERIVATIVE OR INTEGRAL of f(x) w.r.t. x and is written symbolically as
∫ f(x) dx = g(x) + c ⇔ d
dx{g(x) + c} = f(x), where c is called the constant of integration.
2. STANDARD RESULTS :
(i) ∫ (ax + b)n dx = ( )
( )ax b
a n
n++
+1
1 + c n ≠ −1 (ii) ∫ dx
ax b+ =
1
a ln (ax + b) + c
(iii) ∫ eax+b dx = 1
a eax+b + c (iv) ∫ apx+q dx =
1
p
a
na
px q+
� (a > 0) + c
(v) ∫ sin (ax + b) dx = − 1
a cos (ax + b) + c (vi) ∫ cos (ax + b) dx =
1
a sin (ax + b) + c
(vii) ∫ tan(ax + b) dx = 1
a
ln sec (ax + b) + c (viii) ∫ cot(ax + b) dx = 1
a ln sin(ax + b)+ c
(ix) ∫ sec² (ax + b) dx = 1
a tan(ax + b) + c (x) ∫ cosec²(ax + b) dx = −
1
acot(ax + b)+ c
(xi) ∫ sec (ax + b) . tan (ax + b) dx = a
1 sec (ax + b) + c
(xii) ∫ cosec (ax + b) . cot (ax + b) dx = a
1− cosec (ax + b) + c
(xiii) ∫ secx dx = ln (secx + tanx) + c OR ln tan π4 2
+
x+ c
(xiv) ∫ cosec x dx = ln (cosecx − cotx) + c OR ln tan 2
x + c OR − ln (cosecx + cotx)
(xv) ∫ sinh x dx = cosh x + c (xvi) ∫ cosh x dx = sinh x + c (xvii) ∫ sech²x dx = tanh x + c
(xviii) ∫ cosech²x dx = − coth x + c (xix) ∫ sech x . tanh x dx = − sech x + c
(xx) ∫ cosech x . coth x dx = − cosech x + c (xxi) ∫22 xa
xd
− = sin−1
a
x + c
(xxii) ∫ 22 xa
xd
+ =
a
1 tan−1
a
x + c (xxiii) ∫ 22 axx
xd
− =
a
1 sec−1
a
x + c
(xxiv) ∫ d x
x a2 2+ = ln
++ 22 axx OR sinh−1
a
x + c
(xxv) ∫22 ax
xd
− = ln
−+ 22 axx OR cosh−1
a
x + c
(xxvi) ∫ 22 xa
xd
− =
a2
1 ln
xa
xa
−+
+ c (xxvii) ∫ 22ax
xd
− =
a2
1 ln
ax
ax
+−
+ c
(xxviii) ∫ 22xa − dx =
2
x 22
xa − + 2
a2
sin−1
a
x + c
(xxix) ∫ 22 ax + dx = 2
x 22 ax + +
2
a2
sinh−1
a
x + c
(xxx) ∫ 22 ax − dx = 2
x 22 ax − −
2
a2
cosh−1
a
x + c
(xxxi) ∫ eax. sin bx dx = 22
ax
ba
e
+ (a sin bx − b cos bx) + c
(xxxii) ∫ eax . cos bx dx = 22
ax
ba
e
+ (a cos bx + b sin bx) + c
3. TECHNIQUES OF INTEGRATION :(i) Substitution or change of independent variable .
Integral I = ∫ f(x) dx is changed to ∫ f(φ (t)) f ′ (t) dt , by a suitable substitution
x = φ (t) provided the later integral is easier to integrate .
(ii) Integration by part : ∫ u.v dx = u ∫ v dx − ∫
∫ xdv.
xd
ud dx where u & v are differentiable
function . Note : While using integration by parts, choose u & v such that
(a) ∫ v dx is simple & (b) ∫
∫ xdv.
xd
ud dx is simple to integrate.
This is generally obtained, by keeping the order of u & v as per the order of the letters in ILATE,where ; I − Inverse function, L − Logarithmic function ,A − Algebraic function, T − Trigonometric function & E − Exponential function
(iii) Partial fraction , spiliting a bigger fraction into smaller fraction by known methods .
4. INTEGRALS OF THE TYPE :
(i) ∫ [ f(x)]n f ′(x) dx OR ∫ [ ]n)x(f
)x(f ′ dx put f(x) = t & proceed .
(ii)dx
ax bx c2 + +∫ , dx
ax bx c2 + +∫ , ax bx c2 + +∫ dx
Express ax2 + bx + c in the form of perfect square & then apply the standard results .
(iii)px q
ax bx c
++ +∫ 2 dx ,
px q
ax bx c
+
+ +∫ 2
dx .
Express px + q = A (differential co-efficient of denominator) + B .
(iv) ∫ ex [f(x) + f ′(x)] dx = ex . f(x) + c (v) ∫ [f(x) + xf ′(x)] dx = x f(x) + c
(vi) ∫)1x(x
xdn +
n ∈ N Take xn common & put 1 + x−n = t .
(vii) ∫( ) n
)1n(n2 1xx
xd−
+ n ∈ N , take xn common & put 1+x−n = tn
(viii)( )
dx
x xn nn
11
+∫ /
take xn common as x and put 1 + x −n = t .
(ix) ∫xsinba
xd2+
OR ∫xcosba
xd2+
OR ∫xcoscxcosxsinbxsina
xd22 ++
Multiply ..rN & ..rD by sec² x & put tan x = t .
(x) ∫xsinba
xd
+ OR ∫
xcosba
xd
+ OR ∫
xcoscxsinba
xd
++
Hint :Convert sines & cosines into their respective tangents of half the angles , put tan 2
x = t
(xi) ∫nxsin.mxcos.
cxsin.bxcos.a
++++
� dx . Express Nr ≡ A(Dr) + B
xd
d (Dr) + c & proceed .
(xii) ∫1xKx
1x24
2
+++
dx OR ∫1xKx
1x24
2
++−
dx where K is any constant .
Hint : Divide Nr & Dr by x² & proceed .
(xiii)dx
ax b px q( )+ +∫ & ( )dx
ax bx c px q2 + + +∫ ; put px + q = t2 .
(xiv) dx
ax b px qx r( )+ + +∫ 2
, put ax + b = 1
t ;
( )dx
ax bx c px qx r2 2+ + + +∫ , put x =
1
t
(xv)x
x
−−∫
αβ
dx or ( ) ( )x x− −∫ α β ; put x = α cos2 θ + β sin2 θ
x
x
−−∫
αβ
dx or ( ) ( )x x− −∫ α β ; put x = α sec2 θ − β tan2 θ
( ) ( )dx
x x− −∫
α β ; put x − α = t2 or x − β = t2 .
DEFINITE INTEGRAL
1. ∫b
a
f(x) dx = F(b) − F(a) where ∫ f(x) dx = F(x) + c
VERY IMPORTANT NOTE : If ∫b
a
f(x) dx = 0 ⇒ then the equation f(x) = 0 has atleast one
root lying in (a , b) provided f is a continuous function in (a , b) .
2. PROPERTIES OF DEFINITE INTEGRAL :
P−−−−1 ∫b
a
f(x) dx = ∫b
a
f(t) dt provided f is same P −−−− 2 ∫b
a
f(x) dx = − ∫a
b
f(x) dx
P−−−−3 ∫b
a
f(x) dx = ∫c
a
f(x) dx + ∫b
c
f(x) dx , where c may lie inside or outside the interval [a, b] . This property
to be used when f is piecewise continuous in (a, b) .
P−−−−4 ∫−
a
af(x) dx = 0 if f(x) is an odd function i.e. f(x) = − f(−x) .
= 2 ∫a
0
f(x) dx if f(x) is an even function i.e. f(x) = f(−x) .
P−−−−5 ∫b
a
f(x) dx = ∫b
a
f(a + b − x) dx , In particular ∫a
0
f(x) dx = ∫a
0
f(a − x)dx
P−−−−6 ∫a2
0
f(x) dx = ∫a
0
f(x) dx + ∫a
0
f(2a − x) dx = 2 ∫a
0
f(x) dx if f(2a − x) = f(x)
= 0 if f(2a − x) = − f(x)
P−−−−7 ∫an
0
f(x) dx = n ∫a
0
f(x) dx ; where‘a’is the period of the function i.e. f(a + x) = f(x)
P−−−−8a nT
b nT
+
+
∫ f(x) dx = a
b
∫ f(x) dx where f(x) is periodic with period T & n ∈ I .
P−−−−9ma
na
∫ f(x) dx = (n − m) 0
a
∫ f(x) dx if f(x) is periodic with period 'a' .
P−−−−10 If f(x) ≤ φ(x) for a ≤ x ≤ b then ∫b
a
f(x) dx ≤ ∫b
a
φ (x) dx
P−−−−11 xd)x(fb
a
∫ ≤ ∫b
a
f(x)dx .
P−−−−12 If f(x) ≥ 0 on the interval [a, b] , then a
b
∫ f(x) dx ≥ 0.
3. WALLI’S FORMULA :
∫π 2/
0
sinnx . cosmx dx = [ ][ ]
2or1)....4nm()2nm()nm(
2or1)....3m()1m(2or1)....5n()3n()1n(
−+−++−−−−−
K
Where K = 2
π if both m and n are even (m, n ∈ N) ;
= 1 otherwise4. DERIVATIVE OF ANTIDERIVATIVE FUNCTION :
If h(x) & g(x) are differentiable functions of x then ,
xd
d ∫)x(h
)x(g
f(t) dt = f [h (x)] . h′(x) − f [g (x)] . g′(x)
5. DEFINITE INTEGRAL AS LIMIT OF A SUM :
∫b
a
f(x) dx = ∞→nLimit h [f (a) + f (a + h) + f (a + 2h) + ..... + f ( )a n h+ −1 ]
= 0hLimit
→ h ∑=
−
r
n
0
1
f (a + rh) where b − a = nh
If a = 0 & b = 1 then , ∞→nLimit h ∑
=
−
r
n
0
1
f (rh) = ∫1
0
f(x) dx ; where nh = 1 OR
∞→nLimit
n
1
1n
1r
−
=∑ f
n
r = ∫
1
0
f(x) dx .
6. ESTIMATION OF DEFINITE INTEGRAL :
(i) For a monotonic decreasing function in (a , b) ; f(b).(b − a) < ∫b
a
f(x) dx < f(a).(b − a) &
(ii) For a monotonic increasing function in (a , b) ; f(a).(b − a) < ∫b
a
f(x) dx < f(b).(b − a)
7. SOME IMPORTANT EXPANSIONS :
(i) 1 − 2ln.....5
1
4
1
3
1
2
1=∞++−+ (ii)
6.....
4
1
3
1
2
1
1
1 2
2222
π=∞++++
(iii)12
.....4
1
3
1
2
1
1
1 2
2222
π=∞+−+− (iv)
8.....
7
1
5
1
3
1
1
1 2
2222
π=∞++++
(v)24
.....8
1
6
1
4
1
2
1 2
2222
π=∞++++
EXERCISE–1
Q.1 ∫ θθ+θ
θd
sincos
2tan
66Q.2
( )5 4
1
4 5
52
x x
x x
+
+ +∫ dx Q.3 ∫ +
dxxtan1
xcos2
Q.4
( )dx
x42
1−∫ Q.5 Integrate ∫
dx
x x x2 2 1+ − by the substitution z = x + x x
22 1+ −
Q.6 x
e
e
xnx dx
x x +
∫ � Q.7 ∫ cos 2θ . lnθ−θθ+θ
sincos
sincosdθ Q.8 ∫ x2sinxsin
xd2 +
Q.9 ∫ xcosbxsina
xcosbxsina2424
2222
++
dx Q.10 ( )∫++
2)x1(xx
dxQ.11 ∫ 2xx 2++ dx
Q.12 ∫ sin( )
sin( )
x a
x a
−+
dx Q.13 ∫ (sin x)−11/3 (cos x)−1/3dx Q.14 ∫ )1x(sec)xsin1(
dxxcot
+−
Q.15 ∫ 1
1
−+
x
x dx Q.16 ∫ sin−1
xa
x
+ dx
Q.17 ∫( )[ ]
−++4
22
x
xln21xnl1x dx Q.18 ∫
+
2)x(ln
1)x(lnnl dx Q.19 ∫ ( )2xex1x
1x
+
+ dx
Q.20 Integrate 1
2 f ′ (x) w.r.t. x4 , where f (x) = tan −1x + ln 1+ x − ln 1− x
Q.21 ∫ ++
)1x(x
dx)1x(3 Q.22
dx
x xsin cos2 2
3∫ Q.23 ∫ ++
+dx
)1xe(
xx2x
2
Q.24 ∫ xsec21
xsec.
xcotxeccos
xcotxeccos
++−
dx Q.25 ∫ x2sin97
xsinxcos
−−
dx Q.26 ∫xeccosxsec
xd
+ dx
Q.27 ∫ xsecxsin
xd
+ Q.28 ∫ tan x.tan 2x.tan 3x dx Q.29 ( )
dx
x xsin sin 2 +∫
α
Q.30 dx)xcosxsinx)(xsinxcosx(
x2
∫ +−Q.31 ∫ xcosxsin23
xcos2xsin43
++++
dx
Q.32 ∫( )�n x x
x
cos cos
sin
+ 2
2dx
Q.33 sin
sin cos
x
x x+∫ dx Q.34 ∫ xtanxsin
xd
+Q.35 ∫ −
+dx
)1x(
1x332
2
Q.36 ∫+
dxxsin
)xcosxsinx(e2
3xcos
Q.37 ∫( )
( )
ax b dx
x c x ax b
2
2 2 2 2
−
− +Q.38 ∫
( )2
2x
x1)x1(
x2e
−−
−dx
Q.39 ∫ 2/3)x10x7(
x
2−− dx Q.40 ∫ ( ) 2/32 1x
xlnx
− dx Q.41 ∫
1
13
−+
x
x
dx
x
Q.42 2 3
2 3
1
1
−+
+−∫
x
x
x
x dx Q.43
cot tan
sin
x x
x
−+∫
1 3 2 dx Q.44 ∫ +
−+−+−dx
)1x(x
7x4x2x8x7x4222
2345
Q.45 ∫ ( ) 1x
xd
3x3x
2x2 +++
+Q.46 ∫
2 2
2
− −x x
x dx Q.47 ∫ )x()x()x(
xd
β−α−α−
Q.48 ∫++−+ 1x4x6x4x
dxx
234Q.49 ∫ xsin
x2cos dx Q.50 ∫ +α−
+42
2
xcosx21
dx)x1(α∈(0, π)
EXERCISE–2
Q.1 ∫π
0
x dx
x x9 2 2cos sin+ Q.2 1 2
1 20
2 −+z sin
sin
x
xdx
π
Q.3 Evaluate In = ∫
e
1
(lnn x) dx hence find I3.
Q.4 ∫π 2/
0
sin2x · arc tan(sinx) dx Q.5 ∫π 2/
0
cos4 3x . sin2 6x dx Q.6 ∫π 4/
0
x dx
x x xcos (cos sin )+
Q.7 Let h (x) = (fog) (x) + K where K is any constant. If ( ))x(hdx
d = –
)x(coscos
xsin2
then compute the
value of j (0) where j (x) = ∫)x(f
)x(g
dt)t(g
)t(f, where f and g are trigonometric functions.
Q.8 Find the value of the definite integral ∫π
+0
dxxcos2xsin2 .
Q.9 Evaluate the integral : x x x x+ − + − −
∫ 2 2 4 2 2 4
3
5
dx
Q.10 If P = ∫∞
0
x
xdx
2
41+; Q = ∫
∞
0
xdx
x1 4+ and R = ∫
∞
0
dx
x1 4+ then prove that
(a) Q = π4
, (b) P = R, (c) P – 2 Q + R = π
2 2
Q.11 Prove that x n x n a b x nab
x a x bdx
b a
a b
n
a
b n n− − −− + − + +
+ +=
−+z
1 2
2 2
1 12 1
2
b ge j( )( )
( ) ( ) ( )
Q.12 x x
xdx
4 4
2
0
11
1
( )−+∫ Q.13 ∫
1
02
2
x1
xln.x
− dx Q.14 Evaluate: ∫
− +
−2
22
2
dx4x
xx
Q.15 ∫ +−
3
0
2
1 dxx1
x2sin Q.16
0
2π/
∫ ( )xsin
xcosbxsina
4+
+π dx Q.17
0
2π
∫dx
x2 2+ sin
Q.18 ∫π
+π2
0
x dx2
x
4cose Q.19 ∫
−
2
2 2x
1xx12x7x10x3x22
23567
+++−−−+
dx
Q.20 Let α, β be the distinct positive roots of the equation tan x = 2x then evaluate ∫ βα1
0
dx)xsin·x(sin ,
independent of α and β.
Q.21 ∫π 4/
0 x2sin10
xsinxcos
+−
dx Q.22 ∫π
0
( )sec tan
tan
ax b x x
x
++4 2
dx (a,b>0) Q.23 Evaluate: ∫π
++
02
dx)xcos1(
xsin)3x2(
Q.24 If a1, a
2 and a
3 are the three values of a which satisfy the equation
∫ +1
0
3dx)xcosax(sin – ∫−π
1
0
dxxcosx2
a4 = 2
then find the value of 1000( 23
22
21 aaa ++ ).
Q.25 Show that 0
p q+
∫π
| cos x| dx = 2q + sinp where q ε N & − < <π π2 2
p
Q.26 Show that the sum of the two integrals ∫−
−
5
4
e(x+5)² dx + 3 ∫3/2
3/1
e9(x− 2/3)² dx is zero.
Q.27 ∫ +−
−1
02
1
dx1xx
xsinQ.28 ∫
π 2/
0xcosbxsina
xsin2222
2
+dx (a>0, b>0) Q.29 �n
x
x
x
x
1
1 11
1 3
2
+− −−
∫ dx
Q.30 0
2π/
∫ tan−1
−−+
−++
xsin1xsin1
xsin1xsin1dx Q.31 ∫
+
+−−
2
ba
2
ba3
2222
22
22)xb()ax(
xd.x
Q.32 Comment upon the nature of roots of the quadratic equation x2 + 2x = k + ∫ +1
0
dt|kt| depending on the
value of k + R.
Q.33 ∫a2
0
x sin−1
−a
xa2
2
1dx Q.34 Prove that
( )dx
x
dx
xn
nn
1 11
0
1
0 + −= ∫∫
∞
/ (n > 1)
Q.35 Show that ∫∫−− =
x
0
4z4xx
0
zxz dzeedze·e222
Q.36 ∫π
0
( )π−
π
x2
xcos.sin.x2sinx2
2
dx
Q.37 (a) ∫1
0 32 xxx
xd.
x1
x1
+++−
, (b) dxx
1x1n
1xx
1x2
51
124
2
−++−
+∫
+
l
Q.38 Show that dx
x x20 2 1+ +
∞
∫cosθ
= 2 dx
x x20
1
2 1+ +∫cosθ
=
θθ
θ π
θ πθ
θ π π
sin( , )
sin( , )
if
if
∈
−∈
L
N
MMMM
0
22
Q.39 ∫π
+
2
02
2
dxxsin8
xsinx
Q.40 ∫π
+−4
02
2
xdxcos)x2sin1(
)x2cosx2(sinxQ.41 Prove that ∫
x
0
f t dt
u
( )
0
∫
du = ∫x
0
f (u).(x − u) du.
Q.42 ∫π
0
dx
x( cos )5 4 2+Q.43 Evaluate ∫
1
0
ln ( )x1x1 ++− dx Q.44 ∫16
11xtan 1 −− dx
Q.45 ∫
−∞→
+n1
n1
2
ndx|x|)xcos2007xsin2006(nLim . Q.46 Show that f
a
x
x
a
x
xdx a f
a
x
x
a
dx
x( ).
lnln . ( ).
0 0
∞ ∞
∫ ∫+ = +
Q.47 Evaluate the definite integral, ∫− +
++1
1666
6911668998332
dxx1
)xsin·x4xx2(
Q.48 Prove that
(a) α
β
∫ )x()x( −βα− dx = ( )
8
2 πα−β(b)
α
β
∫x
x
−βα− dx = ( )
2
πα−β
(c) α
β
∫)x()x(x
xd
−βα− =
πα β
where α , β > 0 (d) α
β
∫ )x()x(
xd.x
−βα− = ( )2
πβ+α where α < β
Q.49 If f(x) =
4 1 1
1 1 1
1 1
2
2 2 2
2 2 2
cos
(cos ) (cos ) (cos )
(cos ) (cos ) cos
x
x x x
x x x
− + −+ +
, find f x dx( )
−
∫π
π
2
2
Q.50 Evaluate : ∫−−
1
0
1xtann dx)x(cossin·e1l
.
EXERCISE–3Q.1 If the derivative of f(x) wrt x is
)x(f
xcos then show that f(x) is a periodic function .
Q.2 Find the range of the function, f(x) = sin
cos
x dt
t x t1 2 21
1
− +−∫ .
Q.3 A function f is defined in [−1 , 1] as f′(x) = 2 x sin x
1 − cos
x
1; x ≠ 0 ; f(0) = 0;
f (1/π) = 0. Discuss the continuity and derivability of f at x = 0.
Q.4 Let f(x) = [ − − ≤ ≤− < ≤
1 2 0
1 0 2
if x
x if x and g(x) = ∫
−
x
2
f(t) dt. Define g (x) as a function of x and test the
continuity and differentiability of g(x) in (−2, 2).
Q.5 Prove the inequalities:
(a) π6
< dx
x x4
2
82 30
1
− −<∫
π(b) 2 e−1/4 < ∫
−2
0
xx2
e dx < 2e².
dx2
0
<+∫
π
then find a & b. (d) 2
1 ≤ ∫ +
2
02x2
dx ≤
6
5
Q.6 Determine a positive integer n ≤ 5, such that ∫1
0
ex (x − 1)n dx = 16 − 6e.
Q.7 Using calculus
(a) If x < 1 then find the sum of the series ∞++
++
++
++
......x1
x8
x1
x4
x1
x2
x1
18
7
4
3
2.
(b) If x < 1 prove that 284
73
42
3
2 xx1
x21......
xx1
x8x4
xx1
x4x2
xx1
x21
+++
=∞++−
−+
+−−
++−
−.
(c) a < b10 3cos x
(c) Prove the identity f (x)= tanx + 1
2tan
x
2 +
1
22tan
x
22 + .... + 1
2 1n− tanxn2 1− =
1
2 1n− cot xn2 1− – 2cot 2x
Q.8 If φ(x) = cos x − ∫x
0
(x − t) φ(t) dt. Then find the value of φ′′ (x) + φ(x).
Q.9 If y = dt)tx(asin·)t(fa
1x
0
−∫ then prove that yadx
yd 2
2
2
+ = f (x).
Q.10 If y = ∫x
1
tdtnl
x , find xd
yd at x = e.
Q.11 If f(x) = x + ∫1
0
Q.12 A curve C1 is defined by:
dx
dy = ex cos x for x ∈ [0, 2π] and passes through the origin. Prove that the
roots of the function (other than zero) occurs in the ranges 2
π < x < π and
2
3π < x < 2π.
Q.13(a) Let g(x) = xc . e2x & let f(x) = 0
x
∫ e2t . (3 t2 + 1)1/2 dt . For a certain value of 'c', the limit of ′′
f x
g x
( )
( )
as x → ∞ is finite and non zero. Determine the value of 'c' and the limit.
(b) Find the constants 'a' (a > 0) and 'b' such that, 0x
Lim→
t d ta t
bx x
x 2
0+
−
∫
sin = 1.
Q.14 Evaluate: ∫ +−+
+∞→
x3
x
1sin2
2
4
xdt
)3t)(3t(
1t3
dx
dLim
Q.15 Given that Un = {x(1 − x)}n & n ≥ 2 prove that
2n
2
xd
Ud = n (n − 1) U
n−2 − 2 n(2n − 1)U
n−1,
further if Vn = ∫
1
0
ex . Un dx, prove that when n ≥ 2, V
n + 2n (2n − 1).V
n−1− n (n − 1) V
n−2 = 0
Q.16 If ∫∞
0
dttx
tn22 +
�=
π �n 2
4 (x > 0) then show that there can be two integral values of ‘x’ satisfying this
equation.
Q.17 Let f(x) =
1 0 1
0 1 2
2 2 32
− ≤ ≤< ≤
− < ≤
x if x
if x
x if x( )
. Define the function F(x) = ∫x
0
f(t) dt and show that F is
continuous in [0, 3] and differentiable in (0, 3).
Q.18 Let f be an injective function such that f(x) f(y) + 2 = f(x) + f(y) + f(xy) for all non negative real
′ (0) = 0 & f ′ (1) = 2 ≠ f(0) . Find f(x) & show that, 3 ∫ f(x) dx − x (f(x) + 2) is a constant.
Q.19 Evaluate: (a) ∞→nLim
n/1
2
2
2
2
2
2
2 n
n1.....
n
31
n
21
n
11
+
+
+
+ ;
(b) ∞→nLim 1 1
1
2
2
3
4n n n
n
n++
++ +
..... ; (c) ∞→n
Limn/1
nn
!n
;
(d) Given
n1
nn2
nn3
n C
CLim
∞→=
b
a where a and b are relatively prime, find the value of (a + b).
[xy² + x²y] f(y) dy where x and y are independent variable. Find f(x).
x &
y with f
Q.20 Prove that sin x + sin 3x + sin 5x + .... + sin (2k − 1) x = xsin
xksin 2
, k ∈ N and hence
prove that , ∫π 2/
0 xsin
xksin2
dx = 1k2
1......
7
1
5
1
3
11
−+++++ .
Q.21 If Un= ∫
π 2/
0 xsin
xnsin2
2
dx , then show that U1 , U
2 , U
3 , ..... , U
n constitute an AP .
Hence or otherwise find the value of Un.
Q.22 Solve the equation for y as a function of x, satisfying
x · ∫∫ +=x
0
x
0
dt)t(y·t)1x(dt)t(y , where x > 0, given y (1) = 1.
Q.23 Prove that : (a) Im , n
= ∫1
0
xm . (1 − x)n dx = !)1nm(
!n!m
++ m , n ∈ N.
(b) Im , n
= ∫1
0
xm . (ln x)n dx = (−1)n 1n)1m(
!n++
m , n ∈ N.
Q.24 Find a positive real valued continuously differentiable functions f on the real line such that for all x
f 2(x) = ( ) ( )( )∫ +
x
0
22dt)t(')t( ff + e2
Q.25 Let f(x) be a continuously differentiable function then prove that, ∫x
1
[t] f ′ (t) dt = [x]. f(x) −∑=
]x[
1k
)k(f
where [. ] denotes the greatest integer function and x > 1.
Q.26 Let f be a function such that f(u) − f(v) ≤ u − v for all real u & v in an interval [a, b] . Then:(i) Prove that f is continuous at each point of [a, b] .
(ii) Assume that f is integrable on [a, b]. Prove that, f x dx b a f ca
b
( ) ( ) ( )− −∫ ≤ ( )b a− 2
2, where a ≤ c ≤ b
Q.27 Let F (x) = ∫−
+x
1
2 dtt4 and G (x) = ∫ +1
x
2 dtt4 then compute the value of (FG)' (0) where dash
denotes the derivative.
Q.28 Show that for a continuously thrice differentiable function f(x)
f(x) − f(0) = xf′ (0) + ′′f x( ).0
2
2
+ 1
2
2
0
′′′ −∫ f t x t dtx
( )( )
Q.29 Prove that k
n
=∑
0
(− 1)k ( )kn
1
1k m+ + =
k
m
=∑
0
(− 1)k ( )km
1
1k n+ +
(a) Prove that f (x) + g (x) = 6 for all x. (b) Find f (x) and g (x).
EXERCISE–4
Q.1 Find Limitn → ∞ S
n , if : S
n =
1
2
1
42
1
1
42
4
1
32
2 1n
n n n n
+−
+−
+ ++ −
.......... . [REE ’97, 6]
Q.30 Let f and g be function that are differentiable for all real numbers x and that have the followingproperties:(i) f ' (x) = f (x) – g (x) (ii) g ' (x) = g (x) – f (x)(iii) f (0) = 5 (iv) g (0) = 1
Q.2 (a) If g (x) = cos4
0
t dtx
∫ , then g (x + π) equals :
(A) g (x) + g (π) (B) g (x) − g (π) (C) g (x) g (π) (D) [ g (x)/g (π) ]
(b) Limitn → ∞
1
2 21
2
n
r
n rr
n
+=∑ equals :
(A) 1 5+ (B) − +1 5 (C) − +1 2 (D) 1 2+
(c) The value of π πsin ( ln )x
x
e
1
37
∫ dx is _______ .
(d) Let d
dxF x
e
x
x
( )sin
= , x > 0 . If 2
2
1
4e
x
xsin
∫ dx = F (k) − F (1) then one of the possible
values of k is ______.
(e) Determine the value of 2 1
12
x x
xdx
( sin )
cos
+
+−∫π
π. [JEE ’97, 2 + 2 + 2 + 2 + 5]
Q.3 (a) If f t dt x t f t dtx
x
( ) ( )= + ∫∫1
0
, then the value of f (1) is
(A) 1/ 2 (B) 0 (C) 1 (D) − 1/ 2
(b) Prove that tan tan− −
− +
=∫ ∫
1
20
11
0
11
12
x xdx x dx . Hence or otherwise, evaluate the integral
( )tan− − +∫1 2
0
1
1 x x dx [JEE’98, 2 + 8]
Q.4 Evaluate 1
5 2 2 12 2 40
1
( )( )( )+ − + −∫x x e
dxx
[REE ’98, 6 ]
Q.5 (a) If for al real number y, [y] is the greatest integer less than or equal to y, then the value of the
integral π
π
/
/
2
3 2
∫ [2 sin x] dx is :
(A) − π (B) 0 (C) − π2
(D) π2
(b) dx
x14
3 4
+∫cos
/
/
π
π is equal to :
(A) 2 (B) − 2 (C) 1
2(D) −
1
2
(c) Integrate :
( )x x
x x
3
22
3 2
1 1
+ +
+ +∫
( ) dx
(d) Integrate: ∫π
−+0
xcosxcos
xcos
dxee
e[JEE '99, 2 + 2 + 7 + 3 (out of 200)]
Q.6 Evaluate the integral 3 2 1
0
6 cos
cos
/ x
x
−∫
π
dx. [ REE '99, 6]
Q.7 (a) The value of the integral e
e
−∫
1
2
loge x
x d x is :
(A) 3/2 (B) 5/2 (C) 3 (D) 5
(b) Let g (x) = 0
x
∫ f (t) d t , where f is such that 1
2 ≤ f (t) ≤ 1 for t ∈ (0, 1] and 0 ≤ f (t) ≤
1
2for t ∈ (1, 2]. Then g (2) satisfies the inequality :
(A) − 3
2 ≤ g (2) <
1
2(B) 0 ≤ g (2) < 2 (C)
3
2 < g (2) ≤
5
2(D) 2 < g (2) < 4
(c) If f (x) = { e x for x
otherwise
xcos. sin | | ≤ 2
2. Then
−∫2
3
f (x)d x :
(A) 0 (B) 1 (C) 2 (D) 3
(d) For x > 0, let f (x) = 1
x
∫�n t
t1 + dt. Find the function f (x) + f (1/x) and show that,
f (e) + f (1/e) = 1/2 . [JEE 2000, 1 + 1 + 1 + 5]
Q.8 (a) Sn =
1
1 + n +
1
2 2+ n + ........ +
1
2n n+ . Find Limit
n → ∞ Sn .
(b) Given 0
1
∫sin t
t1 + d t = α , find the value of
4 2
4
π
π
−∫
sin t
t
2
4 2π + − d t in terms of α .
[ REE 2000, Mains, 3 + 3 out of 100]
Q.9 Evaluate sin− +
+ +
∫
1
2
2 2
4 8 13
x
x xdx .
Q.10 (a) Evaluate cos
cos sin
/ 9
3 30
2x
x xdx
+∫π
. (b) Evaluate xdx
x1+∫cos sinα
π
0
[ REE 2001, 3 + 5]
Q.11 (a) Let f(x) = 2 2
1
−∫ t dt
x
. Then the real roots of the equation x2 – f ′ (x) = 0 are
(A) +1 (B) + 1
2(C) +
1
2(D) 0 and 1
(b) Let T > 0 be a fixed real number. Suppose f is a continuous function such that for all x ∈ R
f (x + T) = f (x). If I = f( )x
T
0
∫ dx then the value of f(2 )x
T
3
3 3+
∫ dx is
(A) 3
2 I (B) 2 I (C) 3 I (D) 6 I
(c) The integral [ ]x nx
x+
+−
−
∫ l1
11
2
1
2
dx equals
(A) – 1
2(B) 0 (C) 1 (D) 2ln
1
2
[JEE 2002(Scr.), 3+3+3](d) For any natural number m, evaluate
x x x x x dxm m m m m m3 2 2
1
2 3 6+ + + +z c h c h , where x > 0 [JEE 2002 (Mains),4]
Q.12 If f is an even function then prove that ∫π 2
0
f (cos2x) cosx dx = ∫π 4
0
2 f (sin2x) cosx dx
[JEE 2003,(Mains) 2 out of 60]
Q.13 (a) ∫ +−1
0
dxx1
x1 =
(A) 12
+π
(B) 12
−π
(C) π (D) 1
(b) If 5t
0
t5
2dx)x(fx
2
=∫ , t > 0, then
25
4f =
(A) 5
2(B)
2
5(C) –
5
2(D) 1
[JEE 2004, (Scr.)]
(c) If ( ) θθ+θ
= ∫π
d.sin1
cos.xcosxy
2
16/2
x
2 then find
dx
dy at x = π. [JEE 2004 (Mains), 2]
(d) Evaluate dx
3|x|cos2
x43/
3/
3
∫π
π−
π+−
+π. [JEE 2004 (Mains), 4]
Q.14 (a) If ( )∫1
xsin
2 dt)t(ft = (1 – sin x), then f
3
1 is [JEE 2005 (Scr.)]
(A) 1/3 (B) 31 (C) 3 (D) 3
(b) ( )∫−
++++++0
2
23 dx)1xcos()1x(3x3x3x is equal to [JEE 2005 (Scr.)]
(A) – 4 (B) 0 (C) 4 (D) 6
(c) Evaluate: ∫π
+
0
|xcos| dxxsinxcos2
1cos3xcos
2
1sin2e . [JEE 2005, Mains,2]
Q.15 dx1x2x2x
1x
243
2
∫+−
− is equal to
(A) 2
24
x
1x2x2 +− + C (B)
3
24
x
1x2x2 +− + C
(C) x
1x2x2 24 +− + C (D)
2
24
x2
1x2x2 +− + C [JEE 2006, 3]
Comprehension
Q.16 Suppose we define the definite integral using the following formula ( ) ( ))b(f)a(f2
abdxxf
b
a
+−
=∫ , for
more accurate result for c ∈ (a, b) F(c) = ( ) ( ))c(f)b(f2
cb)c(f)a(f
2
ac+
−++
−. When c =
2
ba +,
( ) ))c(f2)b(f)a(f(4
abdxxf
b
a
++−
=∫
(a) ∫π 2/
0
dxxsin is equal to
(A) ( )218
+π
(B) ( )214
+π
(C) 28
π(D)
24
π
(b) If f (x) is a polynomial and if
( )
( )0
at
)a(f)t(f2
atdx)x(f
Lim3
t
a
at=
−
+−
−∫
→ for all a then the degree of f (x) can
atmost be(A) 1 (B) 2 (C) 3 (D) 4
(c) If f ''(x) < 0, ∀ x ∈ (a, b) and c is a point such that a < c < b, and (c, f (c) ) is the point lying on the curvefor which F(c) is maximum, then f '(c) is equal to
(A) ( ) ( )
ab
afbf
−−
(B) ( ) ( )( )
ab
afbf2
−−
(C) ( ) ( )
ab2
afbf2
−−
(D) 0
[JEE 2006, 5 marks each]
Q.17 Find the value of
( )
( )∫
∫
−
−
1
0
10150
1
0
10050
dxx1
dxx15050
[JEE 2006, 6]
ANSWER
EXERCISE–1
Q.1 ln
θθ++
2cos
2cos311 2
+ C Q.2 − x
x x
++ +
1
15 + c
Q.3 1
4ln(cos x + sin x) +
x
2 +
8
1 (sin 2x + cos 2x) + c Q.4
3
8 tan−1 x − ( )
x
x4 14 −−
3
16 ln
x
x
−+
1
1 + c
Q.5 2 tan−1 x x x+ + −
2 2 1 + c Q.6 x
e
e
xc
x x −
+
Q.7 (c) 2
1 (sin 2 θ) ln cos sin
cos sin
θ θθ θ
+−
−
2
1 ln (sec 2 θ) + c Q.8
2
1 ln
2xtan
xtan
+ + c
Q.9
+
+−
2
21
22 b
xtanatanx
ba
1+ c Q.10 2ln
1t2
t
+ +
1t2
1
+ + C when t = x + xx2 +
Q.11 3
1
2/32
2xx
++ − 2/1
2 2xx
2
++
+ c
Q.12 cos a . arc cos
acos
xcos − sin a . ln
−+ asinxsinxsin 22
+ c Q.13 ( )
3/8
2
)x(tan8
xtan413 +− + c
Q.14 2
1 ln
2
xtan +
4
1 sec²
2
x + tan
2
x + c Q.15 cxcosarcx12x1x ++−−−
Q.16 (a + x) arc tan a
x − xa + c Q.17
( )
+−++
23
22
x
11ln32.
x9
1x1x
Q.18 xln (lnx) − xnl
x + c Q.19 xx
x
ex1
1
ex1
exln
++
+ + c
Q.20 − ln (1 − x4) + c Q.21
−+++− − ttan)t1(n
2
1t
2
t
4
t6 12
24
l + C where t = x1/6
Q.22 4
2cos x
+ 2 tan−1 cos x2 − ln
1
1
2
2
+
−
cos
cos
x
x+ c Q.23 C – ln(1 + (x + 1)e–x) – xe)1x(1
1−++
Q24. sin−1
2
xsec
2
1 2 + c
Q.25 c)xcos3xsin34(
)xcos3xsin34(nl
24
1+
−−++
Q.26 2
1
π+−−
82
xtannl
2
1xcosxsin + c
Q.27 c)xcosx(sintanarcxcosxsin3
xcosxsin3nl
32
1+++
+−−+ Q.28 − − +
� � �n x n x n x(sec ) (sec ) (sec )
1
22
1
33 + c
Q.29 − 1
sinα ln [ ]cot cot cot cot cotx x x+ + + −α α2 2 1 + c Q.30 ln
xsinxcosx
xcosxsinx
−+
Q.31 c12
xtantanarc3x2 +
+− Q.32 cos
sin
2x
x − x − cot x . ln ( )( )e x xcos cos+ 2 + c
Q.33 ln (1 + t) − 1
4 ln (1 + t4) +
1
2 2 ln
t t
t t
2
2
2 1
2 1
− ++ +
− 1
2 tan−1 t2 + c where t = cotx
Q.34 c2
xtan
4
1
2
xtannl
2
1 2 +− Q.35 22 )1x(
xc
−−
Q.36 c – ecos x (x + cosec x) Q.37 sin− +
+1
2ax b
cxk Q.38 ex
x1
x1
−+
+ c Q.39 cx10x79
)20x7(2
2+
−−
−
Q.40 c1x
xnlxsecarc
2+
−− Q.41 �n
u
u u
uc where u
x
x
| |tan
2
4 2
12
31
13
1 2
3
1
1
−
+ ++
++ =
−+
−
Q.42 ( )8
3
1
2 5
5 1
5 11
1 1 2tan sin
− −+−+
− − −t nt
tx x� + c where t =
1
1
+−
x
x
Q.43 tan−1 2 2sin
sin cos
x
x x+
+ c Q.44 4 ln x +
x
7 + 6 tan–1(x) + 2x1
x6
+ + C
Q.45 c)1x(3
xtanarc
3
2+
+Q.46 −
− −+
− + − −
−+
−2 2
4
4 2 2 2 2 1
3
2 21x x
xn
x x x
x
x� sin + c
Q.47 cx
x.
2+
α−β−
β−α−
Q.48 2
1 ln
−
+++++ 122x
1x2
x
1x
2
+ C
Q.49
+−
−
−+
t1
t1n
2
1
t2
t2n
2
1ll where t = cosθ and θ = cosec–1(cotx)
Q.50
α
−
α −
2eccos
x2
1xtan·
2eccos
2
1 21
EXERCISE–2
Q.1 6
2π Q.2 2nl Q.3 6 − 2e Q.4
π2
1− Q.5 5
64
πQ.6
8
π ln 2
Q.7 1 – sec(1) Q.8 62 Q.9 2 2 + 4
3 ( )3 3 2 2− Q.12
22
7−
π Q.13
8
π(1 − ln 4)
Q.14 )12(n424 +− l Q.15 3
3π Q.16
22
)ba( +π Q.17
2
3
πQ.18 – )1e(
5
23 2 +π
Q.19 5
216
22−
πQ.20 0 Q.21
−
3
1tanarc
3
2tanarc
3
1
Q.22 ( )a bπ π+ 2
3 3Q.23
2
)3( +ππQ.24 5250 Q.27
36
2π
Q.28 )ba(a2 +
πQ.29
5
3
πQ.30
16
3 2πQ.31
12
πQ.32 real & distinct ∀ k ∈ R
Q.33 4
a2πQ.36
π8
Q.37 (a) 3
π; (b) 2n
8l
π Q.39 –
3
2 2π ln 2 Q.40
π2
16 −
π4
ln2
Q.42 27
5π Q.43
1
22
21ln + −
πQ.44 32
3
16−
πQ.45 2007 Q.47
666
4+π
Q.49 –2π – 32
15Q.50
2
1)2n1(
48
2
++π
−π
l
EXERCISE–3
Q.2 −
π π2 2
, Q.3 cont. & der. at x = 0
Q.4 g(x) is cont. in (−2 , 2); g(x) is der. at x = 1 & not der. at x = 0 . Note that ;
g(x) =
− + − ≤ ≤− + − < <
− − ≤ ≤
( )x for x
x for x
x for x
x
x
2 2 0
2 0 1
1 1 2
2
22
2
Q.5 (c) a = 13
2π & b =
7
2π Q.6 n = 3
Q.7 (a) x1
1
−Q.8 – cos x Q.10 1 + e Q.11 f(x) = x +
119
61 x +
119
80 x²
Q.13 (a) c = 1 and Limitx → ∞ will be
3
2 (b) a = 4 and b =1 Q.14 13.5
Q.16 x = 2 or 4 Q.17 F(x) =
( )3x2if
2x1if
1x0ifx
21
3
2x
21
2x
3
2
≤<+
≤<≤≤−
−
Q.18 f (x) = 1 + x2 Q.19 (a) 2 e(1/2) (π − 4); (b) 3 − ln 4; (c) e
1; (d) 43 Q.21 U
n =
2
nπ
Q.22 y = x1
3e
x
e −Q.24 f (x) = ex + 1 Q.27 0
Q.30 f (x) = 3 + 2e2x; g (x) = 3 – 2e2x
EXERCISE–4Q.1 π/6 Q.2 (a) A (b) B (c) 2 (d) 16 (e) π² Q.3 (a) A (b) ln2
Q.4 1
2 11
11 1
11 1�n
+−
Q.5 ( a )( a )( a )( a ) C, (b) A ; (c) 3
2 tan-1 x −
1
2 ln (1 + x) +
1
4 ln (1 + x2) +
x
x1 2+ + c, (d)
2
π
Q.6 2
3 π − 2 tan−1 2 Q.7 (a) B, (b) B, (c) C, (d)
1
2 ln2 x
Q.8 (a) 2 ln 2, (b) – α Q.9 (x + 1) tan–1 )13x8x4(n4
3
3
)1x(2 2 ++−+
� + C
Q.10 (a) 1
8
5
4
1
3
π−
, (b) I =
ππ∈απ−αα
π
π∈αα
πα
)2,(if)2(sin
),0(ifsin
Q.11 (a) A, (b) C, (c) B, (d) 1
6 12 3 63 2
1
( )mx x x Cm m m
m
m
++ + +
+
d i
Q.13 (a) B, (b) A, (c) 2π, (d)
π −
2
1tan
3
4 1
Q.14 (a) C, (b) C, (c)
−
+
1
2
1sin
2
e
2
1cose
5
24
Q.15 D Q.16 (a) A, (b) A, (c) A Q.17 5051
ELEMENTARY DEFINITE INTEGRAL
(SELF PRACTICE)Evaluate the following definite integrals.
Q.1 sin
( )
−
−∫1
0
1
1
x
x x
dx Q.2 0
2�n
∫ x e−x dx Q.3 sin
cos
/x dx
x1 20
3 4
+∫π
Q 4. 0
2π /
∫ e2x . cos x dx Q 5. x dx
x5 41
1
−−∫ Q 6.
1 12
2� �n x n x
e
−
∫ dx
Q 7. sin
sin cos
/2
4 40
4x
x x+∫π
dx Q 8. cos
( sin ) ( sin )
/x dx
x x1 20
2
+ +∫π
Q 9. ( )
sin . cos
sin cos
/ 2 2
3 32
0
4x x
x x+∫
π dx
Q 10. ( ) ( )x x− −∫ 1 21
2
dx Q 11. dx
x x( ) ( )− −∫1 52
3
Q 12. dx
x x( )
/
+ +∫
1 1 20
3 4
Q 13. ( )sin cos sin cos/
φ φ φ φπ
a b2 2 2 2
0
2
+∫ dφ a≠b Q14. x x2 2
0
1
4. −∫ dx
Q 15. 0
4π /
∫ x cos x cos 3x dx Q16. dx
x5 40
2
+∫sin
/π
Q17. dx
x x x( )− −∫
1 222
3
Q 18. dx
x10
2
+∫cos . cos
/
θ
π
θ ∈ (0, π) Q 19. x dx
x x+ + +∫1 5 10
3
Q 20. ( )
dx
x12
3 2
1
3
+∫ /
Q 21. 0
2π /
∫ sin4x dx Q 22. 0
4π /
∫ cos 2x 1 2− sin x dx Q23. x
x30
3
−∫ dx
Q.24 ( )dx
x x1 2 12 20
1 2
− −∫/
Q 25. ( )dx
x x41
2
1+∫ Q 26. xa x
a x
a 2 2
2 20
−+∫ dx
Q27. sin cos
cos cos
/x x
x x2
0
2
2 2+ +∫π
dx Q 28. 0
1
∫ x (tan−1 x)2 dx Q 29.
( )sin
/
−
−∫
1
23 2
0
12
1
x
x
dx
Q 30. dx
x x20
1
2 1+ +∫cosα
where −π < α < π Q 31. x
x
2
40 1+
∞
∫ dx
Q 32. dx
xa
b
1 2+∫ where a =
e e− −1
2 & b =
e e2 2
2
− −
Q33. 1
1
2
2 4
0
1 −+ +∫
x
x xdx
Q 34. xx
x
52
20
11
1∫
dx Q 35. dx
x x3 20
+ +∫sin cos
π
Q 36. sin cos
sin
/ θ θθ
π ++∫
9 16 20
4
dθ
Q 37. 0
π
∫ θ sin2 θ cos θ dθ Q 38. 1 2
22
0
2 ++∫
cos
( cos )
/x
x
π
dx Q 39. x x
x
++∫
sin
cos
/
10
2π
dx
Q 40. 0
2π /
∫ cos3x sin 3x dx Q 41. 2
1 1
2
20
1 −
+ −∫
x
x x( )
dx Q 42. dxe1
1
dx
d1
1
x/1∫−
+
+−
Q 43. ∫e
0xx )exln(
dxQ 44.
−∫1
1
x2 d (ln x)
Q 45. If f(π) = 2 & 0
π
∫ (f(x) + f ′′ (x)) sin x dx = 5, then find f(0)
Q.46 ∫b
a
dxx
|x|Q.47 cos cos
2 2
0
3
8 4
11
8 4
π ππ
−
− +
∫
x x dx
Q.48sec tan
sec tan
cos
cos
/x x
x x
ecx
ecx
−+ +∫
1 20
2π
dx Q.49 dx)x(''fx
1
0
∫ , where f (x) =cos(tan–1x)
Q.50�
�
n
n
2
3
∫ f (x)dx, where f(x) = e −x + 2 e −2x + 3 e −3x + ......∞
ANSWER KEY
Q 1. π2
4Q 2.
1
2 2�n
e Q3. π
4
1
2
1+ −tan Q 4. e
π − 2
5
Q 5. 1
6Q6. e −
2
2�nQ 7.
π4
Q 8. ln 4
3
Q 9. 1
6Q 10.
π8
Q 11. π6
Q 12. 1
2
9 4 2
7�n
+
Q 13. 1
3
a b
a b
3 3
2 2
−−
Q 14. π3
3
4− Q 15.
π − 3
16Q 16.
2
3 tan−1
1
3
Q 17. π3
Q 18 θ
θsinQ 19.
14
15Q 20.
3 2
2
−
Q 21. 3
16
πQ 22.
1
3Q 23.
3
2
πQ 24. ( )1
22 3�n +
Q 25. 1
4 ln
32
17Q 26.
a2
4 (π − 2) Q 27.
π4
− tan−1 2 + 1
2 ln
5
2
Q 28. π π4 4
11
22−
+ �n Q 29.
π4
1
2− ln 2 Q 30.
αα
α α2
01
20
sin;if if≠ =
Q 31. π
2 2Q 32. 1 Q 33.
1
2 ln 3 Q 34.
3 8
24
π +
Q 35. π4
Q 36. 1
20 ln 3 Q 37. −
4
9Q 38.
1
2
Q 39. π2
Q 40. 5
12Q 41.
π2
Q 42. e1
2
+
Q 43. ln 2 Q 44. e e2 2
2
− −
Q 45. 3 Q.46 | b | – | a |
Q.47 2 Q.48 π/3 Q.49 22
31− Q.50
2
1
EXERCISE–5Part : (A) Only one correct option
1. ∫ ′′−′′ )]x(g)x(f)x(g)x(f[ dx is equal to
(A) )x(g
)x(f
′ (B) f′(x) g(x) – f(x) g′(x)
(C) f(x) g′(x) – f′(x) g(x) (D) f(x) g′(x) + f′(x) g′(x)
2. ∫xcosxsin
1
3 dx is equal to
(A) xtan
2− + c (B) xtan2 + c (C)
xtan
2 + c (D) – xtan2 – c
3. ∫�
�
n x
x n x
| |
| |1 + dx equals :
(A)2
31 + �n x (�nx − 2) + c (B)
2
31 + �n x (�nx + 2) + c
(C)1
31 + �n x (�nx − 2) + c (D) 2 1 + �n x (3 �nx − 2) + c
4. If ∫+
−
2
1
x1
xtanx dx = 2x1+ f(x) + A A �n (x + 1x2 + ) + C, then
(A) f(x) = tan–1 x, A = –1 (B) f(x) = tan–1 x, A = 1(C) f(x) = 2 tan–1 x , A = –1 (D) f(x) = 2 tan–1 x, A = 1
5. ∫ −
−
xcosxsin21
xcosxsin22
88
dx =
(A) 2
1 sin 2x + c (B) –
2
1 sin 2x + c (C) –
2
1 sin x + c (D) – sin2x + c
6. ∫ +−
−−+
xa
xa
xa
xa dx is equal to
(A) – 2 22 xa − + C (B) 22 xa − + C (C) – 22 ax − + C (D) none of these
7. ∫ α+α− )xtan()xtan( tan 2x dx is equal to
(A) �n )xsec(
)xsec(.x2sec
α−α+
+ C (B) �n )xsec()xsec(
x2sec
α+α− + C
(C) �n )xsec(
)xsec(.x2sec
α+α+
+ C (D) none of these
8. ∫ −1xsec dx is equal to
(A) 2 �n
−+
2
1
2
xcos
2
xcos 2
+ C (B) �n
−+
2
1
2
xcos
2
xcos 2
+ C
(C) – 2 �n
−+
2
1
2
xcos
2
xcos 2
+ C (D) none of these
9. ∫ x2sinxcos
dx3 is equal to
(A) 2
+ xtan5
1xcos 2/5
+ C (B) 2
+ xtan5
1xtan 2/5
+ C
(C) 2
− xtan5
1xtan 2/5
+ C (D) none of these
10. Primitive of
( )3 1
1
4
4 2
x
x x
−
+ + w.r.t. x is:
(A)x
x x4 1+ + + c (B) −
x
x x4 1+ + + c
(C)x
x x
++ +
1
14 + c (D) −x
x x
++ +
1
14 + c
11. If( )x
x x
4
2 2
1
1
+
+∫ dx = A �n x +
B
x1 2+ + c, where c is the constant of integration then:
(A) A = 1; B = − 1 (B) A = − 1; B = 1(C) A = 1; B = 1 (D) A = − 1; B = − 1
12. ∫1
1
−+
x
x dx equals :
(A) x 1 − x − 2 1 − x + cos −1 ( )x + c (B) x 1 − x + 2 1 − x + cos −1 ( )x + c
(C) x 1 − x − 2 1 − x − cos −1 ( )x + c (D) x 1 − x + 2 1 − x − cos −1 ( )x + c
13. ∫ sin x. cos x. cos 2x. cos 4x. cos 8x. cos 16 x dx equals:
(A)sin 16
1024
x + c (B) −
cos 32
1024
x + c (C)
cos 32
1096
x + c (D) −
cos 32
1096
x + c
14. ∫1
6 6cos sinx x+ d
x equals :
(A) tan −1 (tan x + cot x) + c (B) − tan −1 (tan x + cot x) + c(C) tan −1 (tan x − cot x) + c (D) − tan −1 (tan x − cot x) + c
15. ∫
−π
++ dx2
x
4tanx)xsin1ln( is equal to:
(A) x �n (1 + sinx) + c (B) �n (1 + sin x) + c (C) – x �n (1 + sin x) + c (D) �n (1 – sin) + c
16.dx
x xcos . sin3
2∫ equals:
(A)2
5 (tan x)5/2 + 2 tanx + c (B)
2
5 (tan2 x + 5) tanx + c
(C)2
5 (tan2 x + 5) 2tanx + c (D) none
17. Ifdx
x xsin cos3 5∫ = a cot x + b tan3 x + c where c is an arbitrary constant of integration then the
values of ‘a’ and ‘b’ are respectively:
(A) − 2 & 2
3(B) 2 & −
2
3(C) 2 &
2
3(D) none
18. ∫+−
−
1x2x2x
1x
243
2
dx is equal to [IIT - 2006, (3, –1)]
(A) 2
24
x
1x2x2 +− + c (B)
3
24
x
1x2x2 +− + c
(C) x
1x2x2 24 +− + c (D)
2
24
x2
1x2x2 +− + c
Part : (B) May have more than one options correct
19. If( )x dx
x x x
−
− +∫
1
2 2 12 2 is equal to
f x
g x
( )
( ) + c then
(A) f(x) = 2x2 – 2x + 1 (B) g(x) = x + 1
(C) g(x) = x (D) f(x) = x2x2 2 −
20.dx
x5 4+∫cos
= Ι tan−1 mx
tan2
+ C then:
(A) l = 2/3 (B) m = 1/3 (C) l = 1/3 (D) m = 2/3
21. If3 3
3 3
cot cot
tan tan
x x
x x
−−∫ dx = p f(x) + q g(x) + c where 'c' is a constant of integration, then
(A) p = 1; q =1
3; f(x) = x; g(x) = �n
3
3
−+
tan
tan
x
x
(B) p = 1; q = −1
3; f(x) = x; g(x) = �n
3
3
−+
tan
tan
x
x
(C) p = 1; q = −2
3; f(x) = x; g(x) = �n
3
3
+−
tan
tan
x
x
(D) p = 1; q = −1
3; f(x) = x; g(x) = �n
3
3
+−
tan
tan
x
x
22.sin
sin cos
24 4
x
x x+∫ dx is equal to:
(A) cot −1 (cot2 x) + c (B) − cot −1 (tan2 x) + c(C) tan −1 (tan2 x) + c (D) − tan −1 (cos 2 x) + c
23.
�n
x
xx
−+
−∫
11
21
dx equal:
(A)1
2 �n2
x
x
−+
1
1 + c (B)
1
4 �n2
x
x
−+
1
1 + c (C)
1
2 �n2
x
x
+−
1
1 + c (D)
1
4 �n2
x
x
+−
1
1 + c
24.�n x
x x
(tan )
sin cos∫ dx equal:
(A)1
2 �n2 (cot x) + c (B)
1
2 �n2 (sec x) + c
(C)1
2 �n2 (sin x sec x) + c (D)
1
2 �n2 (cos x cosec x) + c
EXERCISE–6
1. Integrate with respect ∫ − )xcosx(sin
xsin.xeccos 2
. dx
2. Integrate with respect to x 42
2
xx1
x1
+−−
3. Integrate with respect to x 2x)1x(
1
++ 2
4. ∫( )x
x x
−
+ +
1
1
2
4 2 dx 5. ∫
2 2
6 42
sin cos
cos sin
φ φ
φ φ
−
− − dφ
6. ∫tan tan
tan
θ θ
θ
+
+
3
31 dθ 7. ∫
cos cos
cos
5 4
1 2 3
x x
xdx
+−
8. ∫3 4 2
3 2
+ ++ +
sin cos
sin cos
x x
x x dx 9. ∫
1 ++
cos cos
cos cos
αα
x
x dx
10. ∫d x
x x x( ) ( ) ( )− − −α α β11. ( )
dx
x x x x x3 2 23 3 1 2 3+ + + + −∫
12. ∫ ex
( )x x
x
3
22
2
1
− +
+ dx 13. ∫
−
−dx
xcot4xcos
)3x2(cos
24
14. ∫( ){ }
−++4
22
x
xln21xnl1x dx 15. ∫ ( )
d x
a b x+ cos2
, (a > b)
16. ∫cos cot
cos cot.
sec
sec
ec x x
ec x x
x
x
−+ +1 2
dx 17. ∫ ( )x
x x7 102
3 2
− −/ dx
18. ∫2 2
2
− −x x
x dx 19. ∫ tan −1 x. �n (1 + x2) dx.
20.( )
a b x
b a x
+
+∫
sin
sin2 dx 21.
( )dx
x x4 32
1+∫
22. ( )1
1 2 2
+−∫
x x
x x e x
cossin
dx
23.
( )x
x x
cos
cos/
α
α
+
+ +∫
1
2 123 2
dx =f x
g x
( )
( ) + c then find f(x) and g(x)
24. Evaluate ∫+
xcos
)xsin1(n2
2�
dx.
25. Integrate,
( )x x
x x
3
22
3 2
1 1
+ +
+ +∫
( ) dx. [IIT - 1999, 7]
26. For any natural number m, evaluate,
∫ ( )x x xm m m3 2+ + ( )2 3 62
1x x
m mm
+ +/
d x, x > 0. [IIT - 2002, 5]
ANSWER
EXERCISE–51. C 2. A 3. A 4. A 5. B 6. A
7. B 8. C 9. B 10. B 11. C 12. A
13. B 14. C 15. A 16. B 17. A 18. D
19. AC 20. AB 21. AD 22. ABCD 23. BD
24. ACD
EXERCISE–6
1. �n 2
xtan21+ + c 2. –
32
1 �n
xx
1x
3x
1x
++
−+ + c
3. – 3
1 �n 9
2
3
1t
3
1t
2
+
−+
− + c
where t= 1x
1
+
4.1
3 tan −1
−
3x
1x2
−3
2 tan −1
+
3
1x2 2
+ c
5. 2 �n 5sin4sin2 +φ−φ + 7 tan −1(sinφ− 2) + c
6. − 1
3�n1+tanθ+
1
6�ntan2 θ − tan θ + 1+
1
3
tan− 12 1
3
tan θ −
+ c
7. − +(sinsin
)xx2
2+ c
8. 2 32
1x arcx
c− +
+tan tan
9. x cos α + sin α �ncos ( )
cos ( )
1212
α
α
−
+
x
x + c
10. cx
x.
2+
α−β−
β−α−
11.x x
x
2
2
2 3
8 1
+ −+( )
+1
16. cos−1
2
1x +
+ c
12. exx
x
+
+
1
12 + c
13. c – 3
1 tanx.(2 + tan2x). xcot4 2−
14.( )x x
x x
2 2
3 2
1 1
92 3 1
1+ +− +
. ln
15. – ( ) )xcosba(ba
xsinb22 +−
+ ( ) 2/322 ba
a2
−
arctanba
ba
+−
.tan2
x + c
16. sin−11
2 2
2sec
x
+ c
17.2 7 20
9 7 102
( )x
x xc
−
− −+
18. –x
xx2 2−−+
4
2ln
−−+−x
xx222x4 2
– sin–1
+3
1x2+ c
19. x tan −1 x. �n (1 + x2) + (tan −1 x)2 − 2x tan −1 x
+ �n (1 + x2) − �n x1 22
+
+ c
20. – cxsinab
xcos+
+
21.2
3 �n 3
3
x
1x + −
1
3 3x − ( )
1
3 13x + + c
22. ln (x esinx) -1
2ln (1 - x2 e2 sinx) + c
23. x; x2 + 2x cos α + 1
24. tan x ln (1 + sin2x) – 2x + 2 tan–1 ( 2 .tan x) + c.
25.3
2 tan-1 x -
1
2 �n (1 + x) +
1
4 �n (1 + x2) +
x
x1 2+ + c
26.z
m
m
m
+
+
1
6 1( ) + c, where z = 2
x3m + 3
x2m + 6
xm
EXERCISE–7Part : (A) Only one correct option
1. If f(x) is a function satisfying f
x
1+ x2 f(x) = 0 for all non-zero x, then ∫
θ
θ
eccos
sin
dx)x(f equals
(A) sinθ + cosecθ (B) sin2 θ (C) cosec2 θ (D) none of these
2. The value of the integral ∫ +α+
1
0
2 1cosx2x
dx, where 0 < α <
2
π, is equal to
(A) sin α (B) α sin α (C) α
αsin2
(D) 2
αsin α
3. If ∫100
0
)x(f dx = a, then ( )∑ ∫=
+−
100
1r
1
0
dxx1rf =
(A) 100 a (B) a (C) 0 (D) 10 a
4. If f(x) is an odd function defined on
−2
T,
2
T and has period T, then φ(x) = ∫
x
a
dt)t(f is
(A) a periodic function with period 2
T(B) a periodic function with period T
(C) not a periodic function (D) a periodic function with period 4
T
5. If f(π) = 2 and ∫π
′′+0
))x(f)x(f( sin x dx = 5 then f(0) is equal to, (it is given that f(x) is continuous in [0, π])
(A) 7 (B) 3 (C) 5 (D) 1
6. If f(0) = 1, f(2) = 3, f′(2) = 5 and f′(0) is finite, then ∫ ′′1
0
f.x (2x) dx is equal to
(A) zero (B) 1 (C) 2 (D) none of these
7. ∞→nlim
n/1
n
)1n(sin.......
n2
3sin.
n2
2sin.
n2sin
π−πππ is equal to
(A) 2
π(B) e4/π (C) e2/π (D) none of these
8. f(x) = Minimum {tanx, cot x} ∀ x ∈
π2
,0 . Then ∫π 3/
0
)x(f dx is equal to
(A) �n
2
3(B) �n
2
3(C) �n ( 2 ) (D) �n ( 3 )
9. If A = ∫π
+0
2)2x(
xcos dx, then ∫
π
+
2
01x
x2sin dx is equal to
(A) 2
1 +
2
1
+π – AA (B)
2
1
+π – AA (C) 1 +
2
1
+π – AA (D) A –
2
1 –
2
1
+π
10. ∫π
π−+
2/
2/
2 1x2cos8
dx|x| has the value
(A) 6
2π(B)
12
2π(C)
24
2π(D) none of these
11. ∞→nLt ∑
+= −
n3
1n2r22 nr
n is equal to
(A) log 3
2(B) log
2
3(C) log
3
2(D) log
2
3
12. If ∫y
a
2dttcos = ∫2x
at
tsindt , then the value of
dx
dy is
(A) ycosx
xsin22
2
(B) 2
2
ycosx
xsin2(C)
−
2
ysin21x
xsin22
2
(D) none of these
13. If f(x) =
=
+=
whereelse,1
.....3,2,1n,1n
nxwhere,0
, then the value of ∫2
0
dx)x(f
(A) 1 (B) 0 (C) 2 (D) ∞
14. ∫ −
1
0
4/3)x1(
dxx =
(A) 16
15(B) –
5
16(C) –
16
3(D) none
15. Let Ι1 = ∫
+
2
12x1
dx and Ι
2 = ∫
2
1x
dx, then
(A) Ι1 > Ι
2(B) Ι
2 > Ι
1(C) Ι
1 = Ι
2(D) Ι
1 > 2Ι
2
16. The value of ∫ −]x[
0
])x[x( dx is
(A) 2
1[x] (B) 2[x] (C)
]x[2
1(D) none of these
17. The value of the integeral ∫−
−−
++
+
3
1
21
2
1
x
1xtan
1x
xtan dx is equal to
(A) π (B) 2π (C) 4π (D) none of these
18. The value of ∫π
+2/
0
|xcotxtan|log dx is
(A) π log 2 (B) –π log 2 (C) 2
π log 2 (D) –
2
π log 2
19. If ∫1
0
x2
e (x – α) dx = 0, then
(A) 1 < α < 2 (B) α < 0 (C) 0 < α < 1 (D) α = 0
20. Suppose for every integer n, ∫+
=1n
n
2ndx)x(f . The value of ∫−
4
2
dx)x(f is :
(A) 16 (B) 14 (C) 19 (D) 21
21. Let A = 0
1
∫e d t
t
t
1 + dt then
e
t a
t
a
a −
− − −∫1
1
dt has the value :
(A) Ae–a (B) – Ae–a (C) – ae–a (D) Aea
22.( )
∫−
5
2/5
4
32
x
x25 dx equals to :
(A)3
π(B)
3
2π(C)
6
π(D) none
23. The function f(x) = ∫x
0t
dt satisfies [IIT - 1996]
(A) f(x + y) = f(x) + f(y) (B)
y
xf = f(x) + f(y) (C) f(xy) = f(x) + f(y) (D) none of these
24. The value ofcos2
1
x
adx
x+−∫π
π
, a > 0 is
(A) π (B) aπ (C) π/2 (D) 2π
25. The integral
−∫1 2
1 2
/
/
[ ]x nx
x+
+−
�
1
1 d
x equals:
(A) − 1/2 (B) 0 (C) 1 (D) 2 ln (1/2)
26. If Ι (m, n) = ∫ +1
0
nm )t1(t dt, then the expression of Ι(m, n) in terms of Ι(m + 1, n – 1) is
[IIT - 2003]
(A) 1m
2n
+ –
1m
n
+ Ι (m + 1, n – 1) (B)
1m
n
+ Ι (m + 1, n – 1)
(C) 1m
2n
− –
1m
n
+ Ι (m + 1, n – 1) (D)
1m
n
+ Ι (m +1, n – 1)
27. If ∫1
xsin
2t (f(t)) dt = (1 – sinx), then f
3
1 is [IIT - 2005]
(A) 1/3 (B) 1/ 3 (C) 3 (D) 3
28. ∫−
++++++0
2
23 )}1xcos()1x(3x3x3x{ dx is equal to [IIT - 2005]
(A) – 4 (B) 0 (C) 4 (D) 6
Part : (B) May have more than one options correct
29. The value of integral ∫π
0
)x(sinxf dx is
(A) π ∫π
0
)x(sinf dx (B) π ∫π 2/
0
dx)x(sinf (C) 0 (D) none of these
30. If f(x) is integrable over [1, 2], then ∫2
1
)x(f dx is equal to
(A) ∞→nlim
n
1 ∑
=
n
1rn
rf (B) ∞→n
lim n
1 ∑
+=
n2
1nrn
rf
(C) ∞→nlim
n
1 ∑
=
+n
1rn
nrf (D) ∞→n
lim n
1 ∑
=
n2
1rn
rf
31. If f(x) = ∫ +x
0
44 )tsint(cos dt, f (x + π) will be equal to
(A) f(x) + f(π) (B) f(x) + 2 f(π) (C) f(x) + f
π2
(D) f(x) + 2f
π2
32. The value of ( )2 3 3
1 2 2
2
20
1x x
x x x
+ +
+ + +∫
( ) dx is:
(A)π4
+ 2 ln2 − tan−1 2 (B)π4
+ 2 ln2 − tan−11
3(C) 2 ln2 − cot−1 3 (D) −
π4
+ ln4 + cot−1 2
33. Given f is an odd function defined everywhere, periodic with period 2 and integrable on every interval. Let
g(x) = ∫x
0
f(t) dt. Then:
(A) g(2n) = 0 for every integer n (B) g(x) is an even function(C) g(x) and f(x) have the same period (D) none
34. If Ι = ∫π
+
2/
03 xsin1
dx, then
(A) 0 < Ι < 1 (B) Ι > 2
π(C) Ι < π2 (D) Ι > 2π
35. If In = ( )
dx
xn
1 20
1
+∫ ; n ∈ N, then which of the following statements hold good?
(A) 2n In + 1
= 2 −n + (2n − 1) In
(B) I2 =
π8
1
4+
(C) I2 =
π8
1
4− (D) I
3 =
π16
5
48−
EXERCISE–8
1.
0
π
∫ e xcos2 cos3 (2n + 1) x d
x, n ∈ I
2. If f, g, h be continuous functions on [0, a] such that f (a − x) = f (x), g (a − x) = − g (x)
and 3 h (x) − 4 h (a − x) = 5, then prove that,
0
a
∫ f (x) g (x) h (x) = 0.
3. Assuming
0
π
∫ log sin x d x = − π log 2, show that,
0
π
∫ θ3 log sin θ d θ =3
2
π
0
π
∫ θ2 log ( )2 sin θ d θ.
4. Show that fa
x
x
a
x
xdx a f
a
x
x
a
dx
x( ).
lnln . ( ).
0 0
∞ ∞
∫ ∫+ = +
5. Prove that0
x
∫ f t dt
u
( )
0
∫
du =0
x
∫ f (u).(x − u) du. 6. Prove that
( )dx
x
dx
xn
nn
1 11
0
1
0 + −= ∫∫
∞
/ (n > 1)
7. Prove that Limitn → ∞
1
ncos cos cos ...... cos2 2 2 2
2
2
2
3
2 2
p p p p
n n n
π π π π+ + + +
=
p r
rr
p +
=∏
41
8.
0
π
∫( )
x dx
a x b x2 2 2 2
2cos sin+
9. Evaluate 0
1
∫ x − t. cos π t dt where ‘x’ is any real number
10.
−∫
1 3
1 3 cos tan− −
+
+
−
+
12
12
2
1
2
1
1
x
x
x
x
ex
dx
11. Evaluate, I =
0
1
∫ 2 sin (p t) sin (q
t) d
t, if:
(i) p & q are different roots of the equation, tan x = x.
(ii) p & q are equal and either is root of the equation tan x = x.
12. Prove that ∫ +
x
01x
xsin dx ≥ 0 for x ≥ 0.
13. Let f(x) be a continuous functions ∀ x ∈ R, except at x = 0 such that ∫a
0
dx)x(f , a ∈ R+ exists. If
g(x) = ∫a
xt
)t(f dt, prove that ∫
a
0
dx)x(g = ∫a
0
dx)x(f
14. If f(x) = x
xsin ∀ x ∈ (0, π], prove that,
2
π ∫
π
−π
2/
0
x2
f)x(f dx = ∫π
0
dx)x(f
15. Letd
dxF x
e
x
x
( )sin
= , x > 0. If2
2
1
4e
x
xsin
∫ dx = F (k) − F (1) then one of the possible values of k is ______.
16.−∫
1 3
1 3
x
x
4
41− cos−1 2
1 2
x
x+
dx. [IIT - 1995, 5 + 2 + 2 ]
17. Evaluate ∫π
0
|xcos|e
+
xcos
2
1cos3xcos
2
1sin2 sinx dx. [IIT - 2005, 2]
18. The value of 5050
∫
∫
−
−
1
0
10150
1
0
10050
dx)x1(
dx)x1(
is [IIT - 2006, (6, 0)]
ANSWER
EXERCISE–7
1. D 2. C 3. B 4. B 5. B 6. C
7. B 8. D 9. A 10. B 11. B 12. B
13. C 14. D 15. B 16. A 17. B 18. A
19. C 20. C 21. B 22. A 23. C 24. C
25. A 26. A 27. C 28. C 29. AB 30. BC
31. AD 32. AD 33. ABC 34. BC 35. AB
EXERCISE–8
1. 0 8. 33
222
ba4
ba )( +π
9. 2
2
π− cos πx for 0 < x < 1 ;
2
2
π for x ≥ 1 & 2
2
π− for x ≤ 0
10.π
2 311. (i) 0 (ii)
p
p
2
21 +
15. 16 16.π4
�n ( )2 312 3
2
+ + −π π
17.5
24
−
+
1
2
1sine
2
1
2
1cose 18. 5051
ASSERTION AND REASON Some questions (Assertion–Reason type) are given below. Each question contains Statement
– 1 (Assertion) and Statement – 2
(Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. So select the correct choice :
Choices are :
(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1.
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False.
(D) Statement – 1 is False, Statement – 2 is True.
IINNDDEEFFIINNIITTEE && DDEEFFIINNIITTEE IINNGGEEGGRRAATTIIOONN 129. Let F(x) be an indefinite integral of cos
2x.
Statement-1: The function F(x) satisfies F(x + π) = F(x) ∀ real x
Statement-2: cos2(x + π) = cos
2x.
130. Statement-1: ∫|x| dx can not be found while
1
1
| x |dx−
∫ can be found.
Statement-2: |x| is not differentiable at x = 0.
131. Statement-1: 4
1
1 x
+ ∫ dx = tan
–1 (x
2) + C Statement-2:
2
1dx
1 x+∫ = tan
–1x + C
132. Statement-1: If y is a function of x such that y(x – y)2 = x then
2dx 1log(x y) 1
x 3y 2 = − − −∫
Statement-2: dx
x 3y−∫ = log (x – 3y) + c
133. Statement–1 : f(x) = logsecx –
2x
2 Statement–2 : f(x) is periodic
134. Statement–1 :
9/ 211/ 2 11
11
x 2dx ln x 1 x c
111 x= + + +
+∫
Statement–2 : 2
2
dxln | x 1 x | c
1 x= + + +
+∫
135. Statement–1 :
101
0
tan x dx 10 tan1− = − ∫ ; where [x] = G.I.F.
Statement–2 : [tan–1
x] = 0 for 0 < x < tan 1 and [tan–1
x] = 1 for tan 1 ≤ x < 10.
136. Statement–1 :
/ 2
30
dx
41 tan x
ππ
=+
∫
Statement–2 : ( )a a
0 0
f (x) dx f a x dx= +∫ ∫
/ 2 / 2
3 30 0
dx dx
41 tan x 1 cot x
π ππ
= =+ +
∫ ∫
a a
0 0
f (x) dx f (a x) dx= −∫ ∫ .
137. Statement–1 : 2
0
1 sin xdx 0
π
− =∫ Statement–2 :
0
cos x dx 0
π
=∫ .
138. Statement–1 : ( )x 2 xe tan x sec x dx e tan x c+ = +∫
Statement–2 : ( )x xe f (x) f (x) dx e f (x) c′+ = +∫ .
139. Statement–1 : If f(x) satisfies the conditions of Rolle's theorem in [α, β], then f (x)dx
β
α
′ =β − α∫
Statement–2 : If f(x) satisfies the conditions of Rolle's theorem in [α, β], then f (x)dx 0
β
α
′ =∫
140. Statement–1 :
4
0
[| sin x | | cos x |]dx
π
+∫ , where [⋅] denotes G.I.F. equals 8π.
Statement–2 : If f(x) = |sinx| + |cosx|, then 1 ≤ f(x) ≤ 2 .
141. Let f(x) be a continuous function such that
n 1
3
n
f (x) dx n ,
+
=∫ n∈I
Statement–1 :
3
3
f (x)dx 27−
=∫ Statement–2 :
2
2
f (x)dx 27−
=∫
142. Let In = ( )e
n
1
nx∫ � dx, n ∈ N
Statement–I : I1. I2, I3 . . . is an increasing sequence. Statement–II : n� x is an increasing function.
143. Let f be a periodic function of period 2. Let g(x) =
x
0
f (t)∫ dt and h(x) = g(x + 2) – g(x).
Statement–1 : h is a periodic function. Statement–2 : g(x + 2) – g(x) = g(2).
144. Statement–1 : ( )x
xe1 x log x dx e log x c
x+ = +∫
Statement–2 : ( )x xe f (x) f (x) dx e f (x) c′+ = +∫ .
145. Statement–1 : If I1 = 1
2
x
dt
1 t+∫ and
1/ x
2 2
1
dtI , x 0
1 t>
+∫ then I1 = I2.
Statement–2 : { }2
2
min . x [x], x [ x] dx 0−
− − − − =∫
146. Statement–1 : 8 < 6
4
2x dx 12<∫ .
Statement–2 : If m is the smallest and M is the greatest vlaue of a function f(x) in an interval (a, b),
then the vlaue of the integral
b
a
f (x)dx∫ is such that for a < b, we have M(b – a) ≤
b
a
f (x)dx M(b a)≤ −∫ .
147. Statement–1 :
axax e
e sin bxdxA
=∫ (asinbx – bcosbx)+c Then A is 2 2
a b+
Statement–2 : x
2
1 sin x cos xe dx
cos x
+
∫ = ex tanx + c
148. Statement–1 :
2
2
d(x 1)
2
+
λ +∫ is equal to
22 x 2 c+ +
Statement–2 :
a / 2
11
xdx
1 x+∫ is 2/11 ln |x +
111 x+ | + c
149. Statement–1 :
/ 3
3
/ 6
1
1 tan x
π
π+∫ is π/12 Statement–2 :
b b
a a
f (x)dx f (a b x)= + −∫ ∫ dx
150. Statement–1 : If f satisfies f(x + y) = f(x) + f(y) ∀ x , y ∈R then
5
5
f (x) dx−
∫ = 0
Statement–2 : If f is an odd function then
a
a
f (x) dx−
∫ = 0
151. Statement–1 : If f(x) is an odd function of x then
x
a
f (t)dt∫ is an even function of (n)
Statement–2 : If graph of y = f(x) is symmetric about y–axis then f(x) is always an even function.
152. Statement–1 : Area bounded by y = {x}, {x} is fractional part of x = 0, x = 2 and x–axis is 1.
Statement–2 : Area bounded by y = |sinx|, x = 0, x = 2π is 2 sq. unit.
153. Statement-1: 2 2 2n
1 1 1lim ....
33n4n 1 4n 2→∞
π+ + + =
− −
Statement-2:
1n
nr 1 0
1 rlim f f (x)dx
n n→∞=
=
∑ ∫ , symbols have their usual meaning.
154. Statement-1: If In = ∫tann x dx, then 5 (I4 + I6) = tan
5x .
Statement-2: If In = ∫ tan4x dx, then
n 1tan x
n
−
- In-2 = In, n∈N.
155. Statement-1: If a > 0 and b2 – 4ac < 0, then the value of the integral
2
dx
ax bx c+ +∫ will be of the type µ tan-
1
x Ac
B
+ +
, where A, B, C, µ are constants.
Statement-2: If a > 0, b2 – 4ac < 0 then ax2 + bx + c can be written as sum of two squares.
156. Statements-1:
2 xx
2 3/ 2 2
x x 1 ee dx c
(x 1) x 1
− += +
+ +∫ Statements-2:
xe (f (x) f (x)dx′+∫ = ex f(x) + c
157. Statements-1:
2
24 2 1
x 2dx
x 2(x 5x 4) tan
x
−
−
++ +
∫ = log |tan-1
(x + 2/x)| + c
Statements-2: 1
2 2
dx 1 xtan c
a x a a
−= ++∫
158. Statements-1: 2
xln
xe c(ln x) ln x
= +∫ Statements-2: ∫ex (f(x) + f′(x)) dx = ex f(x) + c.
159. Statements-1: 43 4
1 1 1dx 1 c
2 xx 1 x= − + +
+∫ Statements-2: For integration by parts we have to follow ILATE rule.
160. Statements-1: A function F(x) is an antiderivative of a function f(x) if F ′(x) = f(x)
Statements-2: The functions x2 + 1, x
2 − π, x
2 + 2 are all antiderivatives of the function 2x.
161. Statements-1: ∫b
ax
x dx = b – a , a < b
Statements-2: If f(x) is a function continuous every where in the interval (a, b) except x = c then b c b
a a c
f (x)dx f (x)dx f (x)dx= +∫ ∫ ∫
162. Statements-1:
3
3
1
4 3 x dx 2 30≤ + ≤∫
Statements-2: m and M be the least and the maximum value of a continuous function
y = f(x) in [a, b] then
b
a
m(b a) f (x)dx M(b a)− ≤ ≤ −∫
163. Statements-1: 2
1
x
0
1 e dx e< <∫
Statements-2: if f(x) ≤ g(x) ≤ h(x) in (a, b) then
b b b
a a a
f (x)dx g(x)dx h(x)dx≤ ≤∫ ∫ ∫
164. Statements-1:
1
4
0
1 x dx 1.2+ <∫
Statements-2: For any functions f(x) and g(x), integrable on the interval (a,b), then
b b b
2 2
a a a
f (x)g(x)dx f (x)dx g (x)dx≤∫ ∫ ∫
165. Statements-1:
1
2
1
1dx 2
x−
= −∫
Statements-2: If F(x) is antiderivative of a continuous function (a, b) then
b
a
f (x)dx F(b) F(a)= −∫
166. Statements-1: 2
cos x
(1 sin x)+ can be integrated by substitution it sinx = t.
Statements-2: All integrands are integrated by the method of substitution only.
167. Statement-1 : x
2
1 sin x cose dx
cos x
+
∫ = xe tan x c+∫
Statement-2 : ∫ex (f(x) + f ′ (x)dx = e
x f′(x) + c
168. Statements-1: x 2 x x x1 1
e (x 1)cos (x.e )dx x.e sin 2(x.e ) C2 4
+ = + +∫
Statements-2: ( ) { }f (x) '(x)dx, (x) tφ φ φ =∫ equals f (t)dt∫ .
169. Statements-1: log xdx x log x x c= − +∫
Statements-2: du
uvdx u vdx vdx dxdx
= +
∫ ∫ ∫ ∫
170. Statements-1:
2 xx
2 2
x 4x 2 ee dx C
x 4x 4 (x 2)
+ += +
+ + + ∫ Statements-2: ( )x xe f (x) f '(x) dx e f (x) C+ = +∫
171. Statements-1:
1 12 2
1 0
sin x x x2
3 | x | 3 | x |−
−= −
+ +∫ ∫ Statements-2:
a a a
a 0 0
f (x) dx f (x)dx f ( x)dx−
= = + −∫ ∫ ∫
172. Statements-1: The value of
1
3
0
(1 x)(1 x )dx+ +∫ can not exceed 15
8
Statements-2: If m ≤ f(x) ≤ M ∀ x ∈ [a, b] then
b
a
m(b a) f (x)dx (b a)M− ≤ ≤ −∫
173. Statements-1:
/ 2 5/ 2
5/ 2 5/ 2
0
(sin x)dx
(sin x) (cos x) 4
ππ
=+∫ Statements-2: Area bounded by y = 3x and y = x
2 is
9
2= sq. units
174. Statements-1:
9 x
e
x 10
10x 10 log 10
10 x
+
+∫ dx = log|10 x + x10
| + c Statements-2: f (x)
dx log| f (x) | cf (x)
′= +∫
175. Statements-1:
x
2 x
e (1 x)dx
cos (xe )
+∫ = tan (xe
x) + c Statements-2:
2sec xdx tan x c= +∫
176. Statement-1 : f(x) =
x
2
1
ln t dt(x 0),
1 t t>
+ +∫ then f(x) = - 1
fx
Statements-2: f(x) =
x
1
ln t dt
t 1+∫ , then f(x) + 1 1
fx 2
=
(ln x)2
177. Statement-1 :
1 12 2
1 0
sin x x 2xdx dx
3 | x | 3 | x |−
− −=
− −∫ ∫ .
Statements-2: Since sin x
3 | x |− is an odd function. So, that
1
1
sin x0
3 | x |−
=−∫
.
178. Statements-1 :
n t
0
| sin x |dx
π+
∫ = (2n + 1) – COSt (0 ≤ t ≤ π)
Statements-2:
b c b
a a c
f (x)dx f (x) dx f (x) dx= +∫ ∫ ∫ and
na a
0 0
f (x)dx n f (x) dx=∫ ∫ if f(a + x) = f(x)
179. Statements-1: The value of the integral 2
1
x
0
e dx∫ belongs to [0, 1]
Statements-2: If m & M are the lower bound and the upper bounds of f(x) over [a, b] and f is integrable, then m (b − a) ≤ b
a
f (x) dx∫ ≤ M(b – a).
180. Statements-1: 1
0
[cot x]dx
∞
−
∫ = cot1, where [⋅] denotes greatest integer function.
Statements-2:
b
a
f (x) dx∫ is defined only if f(x) is continuous in (a, b) [⋅] function is discontinuous at all integers
181. Statements-1: ( )4
2 2
4
1 x x 1 x x dx−
+ + − − +∫ = 0 Statements-2:
a
a
f (x)dx 0−
=∫ if f(x) is an odd function.
182. Statements-1: All continuous functions are integrable
Statements-2: If a function y = f(x) is continuous on an interval [a,b] then its definite integral over [a, b] exists.
183. Statements-1: If f(x) is continuous on [a, b], a ≠ b and if
b
a
f (x)dx 0=∫ , then f(x) = 0 at least once in [a, b]
Statements-2: If f is continuous on [a, b], then at some point c in [a, b] f(c) =
b
a
1f (x)dx
b a− ∫
184. Statements-1:
4
4
| x 2 |dx 50−
+ =∫ Statements-2:
b c b
0 a c
f (x)dx f (x)dx f (x)dx= +∫ ∫ ∫ where C∈ (A, B)
185. Statements-1:
2
2
1 xlog dx 0
1 x−
+ =
− ∫ Statements-2: If f is an odd function
a
a
f (x) dx 0−
=∫
186. Statement-1 If ax
0
1e dx
a
∞− =∫ then
m ax
m 1
0
m!x e dx
a
∞−
+=∫ Statement-2 :
nkx
n
d(e )
dx = k
n e
kx and
n n
n n 1
d 1 ( 1) n!
dx x x +
− =
187. Statement-1 :
10
0
{x [x]dx 5− =∫ Statements-2:
na a
a 0
f (x)dx n f (x)dx=∫ ∫
188. Statements-1:
0
| cos x | dx 2
π
=∫ Statements-2:
b c b
a a c
f (x)dx f (x)dx f (x)dx= +∫ ∫ ∫ where a < c < b.
189. Statements-1:
cos x
cos x cosx
0
edx
e e
π
−= π
+∫ Statements-2:
b b
a a
f (x)dx f (a b x)dx= + −∫ ∫
190. Statements-1:
1000
x [x]
0
e dx 1000(e 1)− = −∫ Statements-2:
n 1
x [x] x [x]
0 0
e dx n e dx− −=∫ ∫
191. Statements-1: tan x
0
dx
1 2 2
ππ
=+∫
Statements-2:
b b
a a
f (x) dx f (a b x) dx= + −∫ ∫
ANSWER 129. D 130. B 131. D 132. C 133. A 134. A 135. A 136. C
137. D 138. A 139. A 140. D 141. D 142. D 143. A 144. A
145. C 146. A 147. D 148. C 149. A 150. A 151. C 152. C
153. D 154. C 155. A 156. C 157. A 158. A 159. B 160. B
161. A 162. A 163. A 164. A 165. D 166. C 167. C 168. A
169. C 170. A 171. A 172. A 173. B 174. A 175. A 176. D
177. A 178. A 179. D 180. A 181. A 182. B 183. A 184. A
185. A 186. A 187. C 188. A 189. D 190. A 191. A
Que. from Compt. Exams
(Indefinite Integral)
1. ∫ =−− )cos()cos( bxax
dx
(a) cbx
axba +
−
−−
)sin(
)sin(log)(cosec (b) c
bx
axba +
−
−−
)cos(
)cos(log)(cosec
(c) cax
bxba +
−
−−
)sin(
)sin(log)(cosec (d) c
ax
bxba +
−
−−
)cos(
)cos(log)(cosec
2. =+++
∫bxax
dx [AISSE 1989]
(a) cbxaxab
++−+−
])()[()(3
2 2/32/3 (b) cbxaxba
++−+−
])()[()(3
2 2/32/3
(c) cbxaxba
++++−
])()[()(3
2 2/32/3 (d) None of these
3. ∫ =+
+dx
xx
xx
cos5sin4
sin3cos3 [EAMCET 1991]
(a) )cos5sin4log(41
3
41
27xxx +− (b) )cos5sin4log(
41
3
41
27xxx ++
(c) )cos5sin4log(41
3
41
27xxx −− (d) None of these
4. If ∫ +−=+ acxdxxx )2sin(2
1)2cos2(sin , then the value of a and c is [Roorkee 1978]
(a) 4/π=c and ka = (an arbitrary constant) (b) 4/π−=c and 2/π=a
(c) 2/π=c and a is an arbitrary constant (d) None of these
5. ∫ =−
−−dx
x
xx
)1(
22
3
[AI CBSE 1985]
(a) cx
x
x+−
−
+
21
1log
2
(b) cx
x
x++
+
−
21
1log
2
(c) cx
x
x++
−
+
21
1log
2
(d) cx
x
x+−
+
−
21
1log
2
6. ∫ =−
−dx
xx
xx22
88
cossin21
cossin [IIT 1986]
(a) cx +2sin (b) cx +− 2sin2
1 (c) cx +2sin
2
1 (d) cx +− 2sin
7. =+∫ 2
2
)( bxa
dxx [IIT 1979]
(a)
+−++
bxab
abxa
b
ax
b
1)log(
212
2 (b)
+++−
bxab
abxa
b
ax
b
1)log(
212
2
(c)
++++
bxab
abxa
b
ax
b
1)log(
21 2
2 (d)
+−+−+
bxab
abxa
b
a
b
ax
b
1)log(
21 2
2
8. =++
∫ − 21222 )(tan)1( xqpx
dx
(a) cxqpxqq
+++ −− ])(tantanlog[1 21221 (b) cxqpxq +++ −− ])(tantanlog[ 21221
(c) cxqpq
++ − 2/3122 )tan(3
2 (d) None of these
9. ∫ =+
dx
x
x
3
5
1
[IIT 1985]
(a) cx ++ 2/33 )1(9
2 (b) cxx ++++ 1 2/32/33 )1(
3
2)1(
9
2
(c) cxx ++−+ 2/132/33 )1(3
2)1(
9
2 (d) None of these
10. ∫+− 2cossin xx
dx equals [MP PET 2002]
(a) cx
+
+−
82tan
2
1 π (b) c
x+
+
82tan
2
1 π (c) c
x+
+
82cot
2
1 π (d) c
x+
+−
82cot
2
1 π
11. =+∫ x
ceb
dxa [MP PET 1988; BIT Ranchi 1979]
(a) cceb
e
b
ax
x
+
+log (b) c
e
ceb
b
ax
x
+
+log (c) c
ceb
e
a
bx
x
+
+log (d) c
e
ceb
a
bx
x
+
+log
12. =∫ dxxsin [Roorkee 1977]
(a) cxx +− ]cos[sin2 (b) cxxx +− ]cos[sin2
(c) cxx ++ ]cos[sin2 (d) cxxx ++ ]cos[sin2
13. ∫ =−
dxx
x2/32
2
)9(
(a) cx
x
x+−
−
−
3sin
9
1
2 (b) c
x
x
x++
−
−
3sin
9
1
2 (c) c
x
xx+
−−−
2
1
93sin (d) None of these
14. =+
−∫ dx
x
xx
2
2
1
1
(a) cxx +−+− ]1[sin2
1 421 (b) cxx +−+− ]1[sin2
1 221
(c) cxx +−+− 421 1sin (d) cxx +−+− 221 1sin
15. If cxfab
dxxxxf +−
=∫ ))(log()(2
1cossin)(
22, then =)(xf
(a) xbxa
2222 cossin
1
+ (b)
xbxa2222 cossin
1
− (c)
xbxa2222 sincos
1
+ (d)
xbxa2222 sincos
1
−
16. ∫ =+ xx
dx22 cos5sin4
[AISSE 1986]
(a) cx
+
−
5
tan2tan
5
1 1 (b) cx
+
−
5
tantan
5
1 1 (c) cx
+
−
5
tan2tan
52
1 1 (d) None of these
17. ∫ =+−
+dx
xx
x
1
124
2
[MP PET 1991]
(a) cx
x+
+−2
1 1tan (b) c
x
x+
+−2
1 1cot (c) c
x
x+
−− 1tan
21 (d) c
x
x+
−− 1cot
21
18. ∫ =dxx 2)(log [IIT 1971, 77]
(a) cxxxxx +−− 2log2)(log 2 (b) cxxxxx +−− log2)(log 2
(c) cxxxxx ++− 2log2)(log 2 (d) cxxxxx ++− log2)(log 2
19. The value of ∫−
dxx
ax )( 22
will be [UPSEAT 1999]
(a)
−−− −
a
axaax
)(tan)(
22122 (b)
−+− −
a
axaax
)(tan)(
22122
(c) ][tan)( 221222 axaax −+− − (d) cax +− /tan 1
20. =∫ dxxx 2sec2tan 3 [IIT 1977]
(a) cxx +− 2sec2
12sec
6
1 3 (b) cxx ++ 2sec2
12sec
6
1 3
(c) cxx +− 2sec3
12sec
9
1 2 (d) None of these
21. =∫− dxxx 1sin [MP PET 1991]
(a) cxx
xx
+−+
− − 21
2
14
sin4
1
2 (b) cx
xx
x+−+
+ − 21
2
14
sin4
1
2
(c) cxx
xx
+−−
− − 21
2
14
sin4
1
2 (d) cx
xx
x+−−
+ − 21
2
14
sin4
1
2
22. ∫ =−
dxx
xa
(a) ca
xa
a
x
a
xa +
−+−1sin (b) cxa
a
x
a
x+−+− 221sin
(c) cxaa
x
a
xa +
−−− 221sin (d) cxa
a
x
a
x+−−− 221sin
23. If
∈
4
3,
4
ππx , then ∫ =
−
−dxxe
x
xx x cos2sin1
cossin sin
(a) ce x +sin (b) ce xx +−cossin
(c) ce xx ++cossin (d) ce xx +−sincos
24. If CeBAxdxee
ee x
xx
xx
+−+=−
+∫ −
−
)49log(49
64 2 , then A, B and C are [IIT 1990]
(a) +=== 3log2
3,
35
36,
2
3CBA constant
(b) +=== 3log2
3,
36
35,
2
3CBA constant
(c) +−=−=−= 3log2
3,
36
35,
2
3CBA constant
(d) None of these
25. The value of ∫ dxx3sec will be [UPSEAT 1999]
(a) [ ])tanlog(sectansec2
1xxxx ++
(b) [ ])tanlog(sectansec3
1xxxx ++
(c) [ ])tanlog(sectansec4
1xxxx ++
(d) [ ])tanlog(sectansec8
1xxxx ++
26. ∫ =+
−dxe
x
x x
3)1(
1 [IIT 1983; MP PET 1990]
(a) cx
ex
++
−2)1(
(b) cx
ex
++ 2)1(
(c) cx
ex
++ 3)1(
(d) cx
ex
++
−3)1(
27. If ∫= dxxeIx 2sin , then for what value of K, +−= )2cos22(sin xxeKI x constant [MP PET 1992]
(a) 1 (b) 3 (c) 5 (d) 7
28. The value of ∫ −− 223 xx
dx will be [UPSEAT 1999]
(a)
−
+
x
x
1
3log
4
1 (b)
−
+
x
x
1
3log
3
1
(c)
−
+
x
x
1
3log
2
1 (d)
+
−
x
x
3
1log
29. =+∫ dxxx 32 [AISSE 1985]
(a) cxxx
++−+ 2/52/3 )32(15
1)32(
3
(b) cxxx
++++ 2/52/3 )32(15
1)32(
3
(c) cxxx
++++ 2/52/3 )32(6
1)32(
2
(d) None of these
30. ∫ =
−
+θ
θθ
θθθ d
sincos
sincoslog2cos [IIT 1994]
(a)
−
+−
θθ
θθθθ
sincos
sincoslog)sin(cos 2
(b)
−
++
θθ
θθθθ
sincos
sincoslog)sin(cos 2
(c)
+
−−
θθ
θθθθ
sincos
sincoslog
2
)sin(cos 2
(d) θθπ
θ 2seclog2
1
4tanlog2sin
2
1−
+
31. ∫ =+
dxxxx
x2
2
)cossin( [MNR 1989; RPET 2000]
(a) xxx
xx
cossin
cossin
+
+ (b)
xxx
xxx
cossin
cossin
+
−
(c) xxx
xxx
cossin
cossin
+
− (d) None of these
32. If ∫= dxbxeuax cos and ∫= dxbxev
ax sin , then =++ ))(( 2222vuba
(a) axe2 (b) axeba
222)( +
(c) axe
2 (d) axeba
222 )( −
33. If ,)(log∫= dxxIn
n then =+ −1nn nII
[Karnataka CET 2003]
(a) nxx )(log (b) n
xx )log(
(c) 1)(log −nx (d) n
xn )(log
34. ∫ =
+ dx
xe
x
42sin2/ π
[Roorkee 1982]
(a) cx
e x +2
cos2/ (b) cx
e x +2
cos2 2/
(c) cx
ex +
2sin2/ (d) c
xe
x +2
sin2 2/
35. If Axxdxxx
x+−−−=
+−
+∫ )2ln(7)3ln(9
65
322
, then =A [MP PET 1992]
(a) +− )2ln(5 x constant (b) +−− )3ln(4 x constant
(c) Constant (d) None of these
36. ∫ =+ x
dx
cos2
(a) cx
+
−
2tan
3
1tan2 1 (b) c
x+
−
2tan
3
1tan
3
2 1
(c) cx
+
−
2tan
3
1tan
3
1 1 (d) None of these
37. ∫ ++dx
xx
x
124 equal to [MP PET 2004]
(a)
+−
3
12tan
3
1 21 x
(b)
+−
3
12tan
3
1 21 x
(c) )12(tan3
1 21 +−x (d) None of these
38. ∫ =+ )2sin(sin xx
dx [IIT 1984]
(a) )cos21log(3
2)cos1log(
2
1)cos1log(
6
1xxx +−++−
(b) )cos21log(3
2)cos1log(2)cos1log(6 xxx +−++−
(c) )cos21log(3
2)cos1log(
2
1)cos1log(6 xxx ++++−
(d) None of these
39. If Axxxdxxx
x ae +−
+−=+−
+ −
∫122
5
2tan
2
1)1()1(log
)1)(1(
32,
where A is any arbitrary constant, then the value of ‘a’ is [MP PET 1998]
(a) 5/4 (b) – 5/3
(c) – 5/6 (d) – 5/4
40. If ,2
2
1
1log
)1()4(
)12(22
2
Cx
x
x
x
xx
dxxba
+
+
−
−
+=
−−
+∫ then the values of a and b are respectively [Roorkee 2000]
(a) 1/2, 3/4 (b) –1, 3/2
(c) 1, 3/2 (d) –1/2, ¾
(Definite Integral)
1. If I is the greatest of the definite integrals
,cos1
0
21 ∫
−= dxxeI x dxxeI x 21
02 cos
2
∫−=
,1
03
2
∫−= dxeI
x ,1
0
2/4
2
∫−= dxeI
x then
(a) 1II = (b) 2II =
(c) 3II = (d) 4II =
2. Let )(xf be a function satisfying )()( xfxf =′ with 1)0( =f and )(xg be the function satisfying .)()( 2xxgxf =+ The value of
integral ∫1
0)()( dxxgxf is equal to
[AIEEE 2003; DCE 2005]
(a) )7(4
1−e (b) )2(
4
1−e
(c) )3(2
1−e (d) None of these
3. If ∫=x
mm dxxI
1)(log satisfies the relation ,1−−= mm lIkI then
(a) ek = (b) ml =
(c) e
k1
= (d) None of these
4. Let f be a positive function. Let
{ }dxxxfxIk
k∫ −−=
11 )1( , { }dxxxfI
k
k∫ −−=
12 )1(
when .012 >−k Then 21 / II is [IIT 1997 Cancelled]
(a) 2 (b) k
(c) 2/1 (d) 1
5. If ∫∫ +=1
0,)()(
x
x
dttftxdttf then the value of )1(f is
[IIT 1998; AMU 2005]
(a) 1/2 (b) 0
(c) 1 (d) –1/2
6. ∫−
1
0 4
7
1
dx
x
x is equal to [AMU 2000]
(a) 1 (b) 3
1
(c) 3
2 (d)
3
π
7. If n is any integer, then ∫ =+π
0
3cos)12(cos
2
dxxnex
[IIT 1985; RPET 1995; UPSEAT 2001]
(a) x (b) 1 (c) 0 (d) None of these
8. The value of the definite integral ∫ +
1
03 16x
dxx lies in the interval ].,[ ba The smallest such interval is
(a)
17
1,0 (b) [0, 1]
(c)
27
1,0 (d) None of these
9. Let a,b,c be non-zero real numbers such that ∫∫ +++=+++2
0
281
0
28 ))(cos1())(cos1( dxcbxaxxdxcbxaxx
Then the quadratic equation 02 =++ cbxax has
[IIT 1981; CEE 1993]
(a) No root in (0, 2)
(b) At least one root in (0, 2)
(c) A double root in (0, 2)
(d) None of these
10. If ∫−=x
dttxf1
,||)( ,1−≥x then [MNR 1994]
(a) f and f ′ are continous for 01 >+x
(b) f is continous but f ′ is not continous for 01 >+x
(c) f and f ′ are not continous at 0=x
(d) f is continous at 0=x but f ′ is not so
11. Let ∫=x
dttfxg0
)()( where ]1,0[,1)(2
1∈≤≤ ttf and
2
1)(0 ≤≤ tf for ]2,1(∈t , then [IIT Screening 2000]
(a) 2
1)2(
2
3<≤− g (b) 2)2(0 <≤ g
(c) 2
5)2(
2
3≤< g (d) 4)2(2 << g
12. The value of ∫−>
+
π
π,0,
1
cos 2
adxa
xx
is
[IIT Screening 2001; AIEEE 2005]
(a) π (b) πa
(c) 2
π (d) π2
13. If ∫ −−=
+=
)(
)(1 )}1({,
1)(
af
afx
x
dxxxxgIe
exf , and ∫ −
−=)(
)(2 ))}1({
af
afdxxxgI , then the value of
1
2
I
I is
[AIEEE 2004]
(a) 1 (b) –3
(c) –1 (d) 2
14. Let RRf →: and RRg →: be continuous functions, then the value of the integral
∫− =−−−+2/
2/)]()([)]()([
π
πdxxgxgxfxf
[IIT 1990; DCE 2000; MP PET 2001]
(a) π (b) 1
(c) 1− (d) 0
15. The numbers P, Q and R for which the function RxQePexfxx ++= 2)( satisfies the conditions ,1)0( −=f 31)2(log =′f and
∫ =−4log
0 2
39])([ dxRxxf are given by
(a) ,2=P ,3−=Q 4=R (b) ,5−=P ,2=Q 3=R
(c) ,5=P ,2−=Q 3=R (d) ,5=P ,6−=Q 3=R
16.
+
∑∫∑∫
=
+
=−−
10
1
12
2
2710
1
2
12
27sinsin
n
n
nn
n
ndxxdxx equals
[MP PET 2002]
(a) 227 (b) 54−
(c) 36 (d) 0
17. Let ∫ =1
0,1)( dxxf ∫ =
1
0)( adxxfx and ∫ =
1
0
22 ,)( adxxfx then the value of ∫ =−1
0
2 )()( dxxfax [IIT 1990]
(a) 0 (b) 2a
(c) 12 −a (d) 222 +− aa
18. Given that ,))()((2))()((0 222222
2
∫∞
+++=
+++ accbbacxbxax
dxx π then the value of ∫∞
++0 22
2
)9)(4( xx
dxx is
[Karnataka CET 1993]
(a) 60
π (b)
20
π
(c) 40
π (d)
80
π
19. If ∫ +=1
0,)1(),( dtttnml nm then the expression for ),( nml in terms of )1,1( −+ nml is [IIT Screening 2003]
(a) )1,1(11
2−+
+−
+nml
m
n
m
n
(b) )1,1(1
−++
nmlm
n
(c) )1,1(11
2−+
++
+nml
m
n
m
n
(d) )1,1(1
−++
nmln
m
20. 5
444....321
limn
n
n
++++
∞→=
++++−
∞→ 5
333....321
limn
n
n
[AIEEE 2003]
(a) 30
1 (b) Zero
(c) 4
1 (d)
5
1
21. If ,0,5
2)(
5
0
2
>=∫ ttdxxxft
then =
25
4f
[IIT Screening 2004]
(a) 5
2 (b)
2
5
(c) 5
2− (d) None of these
22. For which of the following values of m, the area of the region bounded by the curve 2xxy −= and the line mxy = equals
2
9
[IIT 1999]
(a) 4− (b) 2−
(c) 2 (d) 4
23. Area enclosed between the curve 32 )2( xxay =− and line ax 2= above x-axis is [MP PET 2001]
(a) 2aπ (b)
2
32
aπ
(c) 22 aπ (d) 2
3 aπ
24. What is the area bounded by the curves 922 =+ yx and xy 8
2 = is [DCE 1999]
(a) 0 (b)
−+ −
3
1sin9
2
9
3
22 1π
(c) π16 (d) None of these
25. The area bounded by the curves 1|| −= xy and 1|| +−= xy is [IIT Screening 2002]
(a) 1 (b) 2
(c) 22 (d) 4
26. The volume of spherical cap of height h cut off from a sphere of radius a is equal to [UPSEAT 2004]
(a) )3(3
2 hah −π
(b) )2)(( 22ahhaha −−−π
(c) 3
3
4h
π (d) None of these
27. If for a real number ][, yy is the greatest integer less than or equal to ,y then the value of the integral ∫2/3
2/
]sin2[
π
π
dxx is
[IIT 1999]
(a) π− (b) 0
(c) 2
π− (d)
2
π
28. If ,2
sin)( Bx
Axf +
=
π 2
2
1=
′f and ∫ =
1
0,
2)(
π
Adxxf then the constants A and B are respectively [IIT 1995]
(a) 2
π and
2
π (b)
π
2 and
π
3
(c) π
4 and 0 (d) 0 and
π
4−
29. If ∫∞
−−=0
1 ,dxxeInx
n then ∫∞
−− =0
1dxxe
nxλ
(a) nIλ (b) nIλ
1
(c) n
nI
λ (d) n
nIλ
30. ∫=4/
0tan
π
dxxI nn , then ][lim 2−
∞−+ nn
nIIn equals
[AIEEE 2002]
(a) 1/2 (b) 1
(c) ∞ (d) 0
31. The area bounded by the curves xy ln= , ||ln xy = , |ln| xy = and |||ln| xy = is [AIEEE 2002]
(a) 4 sq. unit (b) 6 sq. unit
(c) 10 sq. unit (d) None of these
32. dxx
xn
∫
+π
0sin
2
1sin
, )( Nn ∈ equals [Kurukshetra CEE 1998]
(a) πn (b) 2
)12(π
+n
(c) π (d) 0
33. If ∫ =−1
0,0)(
2
dxxex α then
[MNR 1994; Pb. CET 2001; UPSEAT 2000]
(a) 21 << α (b) 0<α
(c) 10 << α (d) None of these
34. ∫π
π
10
|sin| dxx is [AIEEE 2002]
(a) 20 (b) 8
(c) 10 (d) 18
35. ∫− +
+π
πdx
x
xx2cos1
)sin1(2 is [AIEEE 2002]
(a) 4/2π (b) 2π
(c) 0 (d) 2/π
36. On the interval ,4
7,
3
5
ππ the greatest value of the function ∫ =−=
x
dtttxf3/5
)sin2cos6()(π
(a) 12233 ++ (b) 12233 −−
(c) Does not exist (d) None of these
37. If ∫∫∫ ===2
13
1
02
1
01
232
2,2,2 dxIdxIdxI xxx , ∫=2
14
3
2 dxI x , then [AIEEE 2005]
(a) 43 II = (b) 43 II >
(c) 12 II > (d) 21 II >
38. If xx
fxf =
−
13)(2 , then dxxf∫
2
1)( is equal to
[J & K 2005]
(a) 2ln5
3 (b) )2ln1(
5
3+
−
(c) 2ln5
3− (d) None of these
89 of 89
39. If 03 =∫b
adxx and
3
22 =∫ dxxb
a, then the value of a and b will be respectively [AMU 2005]
(a) 1, 1 (b) 1,1 −−
(c) 1,1 − (d) 1,1−
40. The sine and cosine curves intersects infinitely many times giving bounded regions of equal areas. The area of one of such
region is [DCE 2005]
(a) 2 (b) 22
(c) 23 (d) 24
(Indefinite Integral)
1 b 2 b 3 a 4 a 5 d
6 b 7 d 8 a 9 c 10 d
11 a 12 b 13 a 14 a 15 a
16 c 17 c 18 c 19 a 20 a
21 a 22 a 23 a 24 d 25 a
26 b 27 c 28 a 29 a 30 d
31 c 32 c 33 a 34 d 35 c
36 b 37 b 38 a 39 d 40 a
(Definite Integral)
1 d 2 d 3 b 4 c 5 a
6 b 7 c 8 a 9 b 10 a
11 b 12 c 13 d 14 d 15 d
16 d 17 a 18 a 19 a 20 d
21 a 22 b 23 b 24 b 25 b
26 a 27 c 28 c 29 c 30 b
31 a 32 c 33 c 34 d 35 b
36 b 37 d 38 b 39 d 40 b
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