integral calculus by arihant b016

Do You Know

It was in this aspect that the process of integration was treated by

Leibnitz, the symbol of ∫ being regarded as the initial letter of the word

Sum, in the same way as the symbol of differentiation d is the initial letter

in the word difference.

DefinitionIf f and g are functions of x such that g x f x′ =( ) ( ), then the function g is

called a anti-derivative (or primitive function or simply integral) of f w.r.t. x.It is written symbolically,

f x dx g x( ) ( )=∫ , whered

dxg x f x( ) ( )=

Points to Consider

1. In other words,

f x dx g x g x f x( ) ( ) ( ) ( )= ′ =∫ iff

2. ∫ = +f x dx g x c( ) ( ) , where c is constant, [ ( ( ) ) ( ) ( )]Q g x c g x f x+ ′ = ′ =

and c is called constant of integration.

Illustration 1 Ifd

dxx c n xn n[ ] ( )+ + = +1 1 , then find x dxn∫ .

Solution. As,d

dxx c n xn n[ ] ( )+ + = +1 1

⇒ ( )x cn + +1 is anti-derivative or integral of ( )n xn+ 1

∴ x dxx

nCn

n

=+

++

∫1

1

Illustration 2 Evaluate ∫ −e x x dxx (cos sin ) .

Solution. Here, e x xx (cos sin )− is the derivative of e xx cos

⇒ e x x dx e x Cx x(cos sin ) cos− = +∫

Fundamental Integration Formulae

Since,d

dxg x f x{ ( )} ( )=

⇔ f x dx g x C( ) ( )= +∫Therefore, based upon this definition and various standard

differentiation formulas, we obtain the following integration formulae

2 Integral Calculus

(i)d

dx

x

nx n

nn

+

+

= ≠ −

1

11, ⇒ x dx

x

nC nn

n

=+

+ ≠ −+

∫1

11,

(ii)d

dxx

x(log| |) = 1 ⇒ 1

xdx x C= +∫ log| | , when x ≠ 0

(iii)d

dxe ex x( ) = ⇒ e dx e Cx x= +∫

(iv)d

dx

a

aa a a

x

e

x

log, ,

= > ≠0 1 ⇒ a dx

a

aCx

x

= +∫ log

(v)d

dxx x( cos ) sin− = ⇒ sin cosx dx x C= − +∫

(vi)d

dxx x(sin ) cos= ⇒ cos sinx dx x C= +∫

(vii)d

dxx x(tan ) = sec2 ⇒ sec2 x dx x C= +∫ tan

(viii)d

dxx x( cot )− = cosec2 ⇒ cosec2 x dx x C= − +∫ cot

(ix)d

dxx x x( ) tansec sec= ⇒ sec secx x dx x Ctan = +∫

(x)d

dxx x x( ) cot− =cosec cosec ⇒ cosec cosecx x dx x Ccot = − +∫

(xi)d

dxx x(log|sin |) cot= ⇒ cot log|sin |x dx x C= +∫

(xii)d

dxx x( log|cos |) tan− = ⇒ tan log|cos |x dx x C= − +∫

(xiii)d

dxx x x(log| tan |)sec sec+ = ⇒ sec secx dx x x C= + +∫ log| tan |

(xiv)d

dxx x x(log| cot |)cosec cosec− =

⇒ cosec cosec∫ = − +x dx x x Clog| cot |

(xv)d

dx

x

a a xsin −

=

−1

2 2

1 ⇒ dx

a x

x

aC

2 2

1

−=

+−∫ sin

(xvi)d

dx

x

a a xcos −

=

−

−1

2 2

1⇒

−

−=

+−∫

1

2 2

1

a xdx

x

aCcos

(xvii)d

dx a

x

a a x

1 11

2 2tan −

=

+⇒ dx

a x a

x

aC

2 2

11

+=

+−∫ tan

(xviii)d

dx a

x

a a x

1 11

2 2cot−

=

−+

⇒−+

=

+−∫

1 12 2

1

a xdx

a

x

aCcot

(xix)d

dx a

x

a x x a

1 11

2 2sec−

=

−⇒ dx

x x a a

x

aC

2 2

11

−=

+−∫ sec

(xx)d

dx a

x

a x x a

1 11

2 2cosec−

=

−

−⇒

−

−=

+−∫

dx

x x a a

x

aC

2 2

11cosec

Chapter 1 : Indefinite Integrals 3

Points to Consider

1. k f x dx k f x dx( ) ( ) ,= ∫∫ where k is constant.

ie, The integral of the product of a constant and a function = the constant ×integral of the function.

2. ∫ ∫ ∫ ∫± ± … ± = ± ± … ±{ }f x f x f x dx f x dx f x dx f xn n1 2 1 2( ) ( ) ( ) ( ) ( ) ( ) .dx

ie, The integral of the sum or difference of a finite number of functions isequal to the sum or difference of the integrals of the various functions.

3. Geometrical interpretation

eg, y x dxx

C= = +∫2

2

∴ y f x dx F x C= = +∫ ( ) ( )

⇒ F x f x F x f x′ = ′ =( ) ( ); ( ) ( )1 1

Hence, y f x dx= ∫ ( ) denotes a family of curves such that the slope of the

tangent at x x= 1 on every member is same ie, F x f x′ =( ) ( )1 [when x1 lies in the

domain of f x( )]

Hence, anti-derivative of a function is not unique. If g x1( ) and g x2( ) are two

anti-derivatives of a function f x( ) on [ , ]a b , then they differ only by a

constant.

ie, g x g x C1 2( ) ( )− =

4. Anti-derivative of a continuous function is differentiable.

ie, If f x( ) is continuous, then

f x dx F x C∫ = +( ) ( )

⇒ F x f x′ =( ) ( )

⇒ F x′ ( )

⇒ always exists and is continuous.

5. If integrand is discontinuous at x x= 1, then its anti-derivative at x x= 1 neednot be discontinuous.

eg, x dx−∫ 1 3/ . Here, x−1 3/ is discontinuous at x = 0.

But x dx x C−∫ = +1 3 2 33

2

/ / is continuous at x = 0.

6. Anti-derivative of a periodic function need not be a periodic function.

eg, f x x( ) cos= + 1 is periodic but ( cos ) sin∫ + = + +x dx x x C1 is aperiodic.

4 Integral Calculus

O ( 0)x,

x +C21

x +C22

x +C23

Fig. 1.1

Illustration 3 Evaluate

(i)x x

xdx

2 5 1+ −∫ (ii) ( )x dx2 35+∫

Solution. (i) Ix x

xdx=

+ −∫

2 5 1

= + −

∫

x

x

x

x xdx

2

1 2

1

1 2 1 2

5 1/ / /

= + − −∫ ( )/ / /x x x dx3 2 1 2 1 25

=+

++

−− +

++ + − +x x x

C3 2 1 1 2 1 1 2 1

3 2 1

5

1 2 1 1 2 1

/ / /

/ / /

using ∫ =+

+

+x dx

x

nCn

n 1

1

⇒ I x x x C= + ⋅ − +2

5

2

35 25 2 3 2 1 2/ / /

(ii) I x dx= +∫ ( )2 35

Expanding the integrand by the binomial formula

I x x x dx= + + +∫ ( )6 4 215 75 125

Ix x x

x C= + + + +7 5 3

7

15

5

75

3125

Ix

x x x C= + + + +7

5 3

73 25 125


(i) tan2 x dx∫ (ii)dx

x xsin cos2 2∫

(iii)sin cos

sin cos

6 6

2 2

x x

x xdx

+∫ (iv)

cos cos

cos

x x

xdx

−−∫

2

1

Solution. (i) I x dx= ∫ tan2

I x dx= −∫ ( )sec2 1

I x dx dx= − ∫∫ sec2 1

⇒ I x x c= − +tan

(ii) Ix x

dx= ∫1

2 2sin cos

Ix x

x xdx=

+∫

sin cos

sin cos

2 2

2 2(as sin cos )2 2 1x x+ =

Ix

x xdx

x

x xdx= + ∫∫

sin

sin cos

cos

sin cos

2

2 2

2

2 2

I x dx x dx= + ∫∫ sec cosec2 2

I x x C= − +tan cot


(iii) Ix x

x xdx=

+∫

sin cos

sin cos

6 6

2 2

Ix x x x x x

x=

+ − +(sin cos ) sin cos (sin cos )

sin cos

2 2 3 2 2 2 2

2 2

3

xdx∫

[ ( ) ( )]using a b a b ab a b+ = + + +3 3 3 3

Ix x

x xdx=

−∫

1 3 2 2

2 2

sin cos

sin cos

= − ∫∫1

32 2sin cosx x

dx dx

=+

− +∫(sin cos )

sin cos

2 2

2 23

x x

x xdx x C

= + − +∫∫ sec cosec2 2 3x dx x dx x C

= − − +tan cotx x x C3

(iv) Ix x

xdx=

−−∫

cos cos

cos

2

1

Ix x

xdx=

− −−∫

cos ( cos )

cos

2 1

1

2

Ix x

xdx=

− + −− −∫

( cos ) (cos )

(cos )

2 1 1

1

I x dx= +∫ ( cos )2 1

I x x C= + +2 sin


(i) ∫ +x

xdx

3

2(ii) ∫ +

x

xdx

2

2 5

Solution. (i) Ix

xdx=

+∫3

2

I x xx

dx= − + −+

∫ 2 2 4

8

2

Ix

x x x C= − + − + +3

2

34 8 2log| |

(ii) Ix

xdx=

+∫2

2 5

= −+

∫ 1

5

52xdx = −

+∫xdx

x5

52 2( )

= −

+−x

xC

5

5 5

1tan

∴ I xx

C= −

+−5

5

1tan

6 Integral Calculus

Points to Consider

In rational algebraic functions if the degree of numerator is greater than orequal to degree of denominator, then always divide the numerator bydenominator and use the result of integration.

Illustration 6 Solve

(i) 5log e x dx∫ (ii) 2 4log x dx∫Solution. (i) I dxe x= ∫ 5log

= ∫ x dxelog 5 ( )log logas a be cb a=

=+

++x

Ce

e

log

(log )

5 1

5 1

∴ 55 1

5 1log

log

loge

ex

e

dxx

C=+

++

∫(ii) I dxx= ∫ 2 4log

= ∫ 2 22log xdx

= ∫ 21 2 2/ log x dx using log logb bn x

nx=

1

= ∫ 2 2log x dx

= ∫ x dx ( )logusing a ba b =

= +xC

3 2

3 2

/

/

∴ 22

34 3 2log /x dx x C= +∫

Methods of IntegrationIf the integrand is not a derivative of a simple function, then the

corresponding integrals cannot be found directly. In order to find the integralof complex problems, we follow the following rules :

Integration by Substitution(or by change of the independent variable)

If g x( ) is a continuously differentiable function, then to evaluateintegrals of the form,

I f g x g x dx= ′∫ ( ( )) . ( ) ,

we substitute g x t( ) = and g x dx dt′ =( )

The substitution reduces the integral to f t dt( ) .∫After evaluating this integral we substitute back the value of t.



(i)sin (log )x

xdx∫ (ii)

3 4

4 3

sin cos

sin cos

x x

x xdx

+−

∫

(iii)e

xdx

m xtan−

+∫1

1 2(iv) x x dxsin ( )4 72 +∫

Solution. (i) Ix

xdx= ∫

sin (log )

We know,d

dxx

x(log ) = 1

Thus, let log x t= ⇒ 1

xdx dt= …(i)

∴ I t dt t C= = − +∫ sin ( ) cos ( )

= − +cos (log )x C [using Eq. (i)]

(ii) Ix x

x xdx=

+−∫

3 4

4 3

sin cos

sin cos

We know,d

dxx x x x( sin cos ) ( cos sin )4 3 4 3− = +

Thus, let 4 3sin cosx x t− = …(i)

⇒ ( cos sin )4 3x x dx dt+ =

∴ Idt

tt C= = +∫ log| |

⇒ Ix x

x xdx=

+−∫

3 4

4 3

sin cos

sin cos

= − +log| sin cos |4 3x x C [using Eq. (i)]

(iii) Ie

xdx

m x

=+

−

∫tan 1

1 2let m x ttan− =1

I edt

m

t= ⋅∫ ⇒ m

xdx dt

1 2+=

Im

e dtt= ∫1 ⇒ 1

1

12+

=x

dxm

dt

Im

e Cm

e Ct m x= + = +−1 1 1tan

(iv) I x x dx= +∫ sin ( )4 72

Let 4 72x t+ =

⇒ 8x dx dt= , x dx dt= 1

8

∴ I tdt= ∫ sin ( )8

⇒ I t C= − +1

8cos ( )

I x C= − + +1

84 72cos ( )

8 Integral Calculus

Points to Consider

1. If f x dx g x C( ) ( ) ,= +∫ then f ax b dxa

g ax b C( ) ( )+ = + +∫1

2. If1

xdx x C= +∫ log| | , then

1 1

ax bdx

aax b C

+= + +∫ log| |

Thus, in any fundamental integral formulae given in article fundamentalintegration formulae if in place of x we have ( ),ax b+ then same formula isapplicable but we must divide by coefficient of x or derivative of (ax b+ ) ie, a.

Here is the list of some of frequently used formulae

(i) ( )( )

( ),ax b dx

ax b

a nC nn

n

+ =+

++ ≠ −

+

∫1

11

(ii)1 1

ax bdx

aax b C

+= + +∫ log| |

(iii) e dxa

e Cax b ax b+ += +∫1

(iv) a dxb

a

aCbx c

bx c+

+= ⋅ +∫

1

log

(v) sin ( ) cos ( )ax b dxa

ax b C+ = − + +∫1

(vi) cos ( ) sin ( )ax b dxa

ax b C+ = + +∫1

(vii) sec2 1( ) tan ( )ax b dx

aax b C+ = + +∫

(viii) cosec2 1( ) cot ( )ax b dx

aax b C+ = − + +∫

(ix) sec sec( ) tan ( ) ( )ax b ax b dxa

ax b C+ + = + +∫1

(x) cosec cosec( ) cot ( ) ( )ax b ax b dxa

ax b C+ + = − + +∫1

(xi) tan ( ) log|cos ( )|ax b dxa

ax b C+ = − + +∫1

(xii) cot ( ) log|sin ( )|ax b dxa

ax b C+ = + +∫1

(xiii) sec sec( ) log| ( ) tan ( )|ax b dxa

ax b ax b C+ = + + + +∫1

(xiv) cosec cosec( ) log| ( ) cot ( )|ax b dxa

ax b ax b C+ = + − + +∫1



(i) cos cos4 7x x dx∫ (ii) cos cos cosx x x dx2 5∫Solution. When calculating such integrals it is advisable to use thetrigonometric product formulae.

(i) cos cos4 7x x dx∫Here, cos cos (cos cos )4 7

1

23 11x x x x= +

∴ I x x dx= ∫ cos cos4 7

= +∫1

23 11(cos cos )x x dx

= +∫ ∫1

23

1

211cos cosx dx x dx

= + +sin sin3

6

11

22

x xC

(ii) cos cos cosx x x dx2 5∫We have, (cos cos ) cos (cos cos ) cosx x x x x x2 5

1

23 5= +

= +1

42 5 2 3 5{ }cos cos cos cosx x x x

= + + +1

44 6 2 8{ }(cos cos ) (cos cos )x x x x

∴ cos cos cos cos cos cos cosx x x x x x x2 51

42 4 6 8= + + +{ }

∴ I x x x dx= ∫ (cos cos cos )2 5

= + + +∫1

42 4 6 8(cos cos cos cos )x x x x dx

= + + + +sin sin sin sin2

8

4

16

6

24

8

32

x x x xC

Points to Consider

While solving such problems it is expedient to use the following trigonometricidentities :

1. sin cos {sin ( ) sin ( ) }mx nx m n x m n x= − + +1

2

2. cos sin {sin ( ) sin ( ) }mx nx m n x m n x= + − −1

2

3. sin sin {cos ( ) cos ( ) }mx nx m n x m n x= − − +1

2

4. cos cos {cos ( ) cos ( ) }mx nx m n x m n x= − + +1

2

10 Integral Calculus


(i) x x dx−∫ 5 (ii)8 13

4 7

x

xdx

++∫

(iii)x

xdx

3

1 2+∫ (iv)x

xdx

2

2 2

1

1

+−∫ ( )

(v)x

a bxdx

2

2( )+∫ (vi)dx

x x+ −∫ 1

Solution. (i) I x x dx= −∫ 5

Let x t− =5 2 ⇒ dx t dt= 2

∴ I t t t dt= + ⋅ ⋅∫ ( )2 5 2 = +∫2 54 2( )t t dt

= +

+25

5

3

5 3t tC

=−

+−

+2

5

5

5 5

3

5 2 3 2( ) ( )/ /x xC

(ii) Ix

xdx=

++∫

8 13

4 7

Let 4 7 2x t+ =

∴ 4 2dx t dt= , dx t dt= 1

2

∴ It

tt dt=

− +⋅∫

2 7 13 1

2

2( )

⇒ I t dt= − +∫1

22 14 132( )

= −∫1

22 12( )t dt

I tt

C= − +1

3 2

3

∴ I x x C= + − + +1

34 7

1

24 73 2 1 2( ) ( )/ /

(iii) Ix

xdx=

+∫3

1 2, let 1 2 2+ =x t

2 2dx t dt= , dx t dt=

∴ I

tt dt

t=

−

⋅

∫

2 31

2

I t t t dt= − + −∫1

83 3 16 4 2( )

= − + −

+1

8 7

3

5

75 3t

t t t C

∴ Ix

x x x=+

− + + + − +

1

8

2 1

7

3

51 2 2 1 2 1

7 25 2 3 2 1 2( )

( ) ( ) ( )/

/ / /

+ C


(iv) Ix

xdx=

+−∫

2

2 2

1

1( )

Ix

xx

dx=+

−

∫1 1

1

2

2

/(dividing Nr and Dr by x2)

∴ Idt

t= ∫ 2

let xx

t− =1 ⇒ 112

+

=

xdx dt

= − +1

tC

I

xx

Cx

xC= −

−

+ =−

+1

1 1 2

(v) Ix

a bxdx=

+∫2

2( )= +

− −

+∫1

2

2

2

2

2b

a

bx

a

b

bx adx

( )

= −+

+∫∫1

12

2 2 2bdx

a

b

bx a

bx adx

( )

Ib

xa

b

bx a a

bx adx= −

+ −+

∫1 2 22 2 2( )

= −+

− +

−∫∫1

21

2 22

bx

a

b bx adx a bx a dx

( )( )

= − + ++

+1 22 2b

xa

b bbx a

a

b bx aClog| |

( )

= − + −+

+12

3

2

bbx a bx a

a

a bxClog| |

( )

(vi) Idx

x x=

+ −∫ 1(rationalising Dr )

Ix x

x xdx=

+ ++ −∫

1

1( ) ( )

I x x dx= + +∫ ( )1

I x x C= + + +2

31

2

3

3 2 3 2( ) ( )/ /

Integral of the Form

1. f xx x

dx∫ +

−

11

12

Put xx

t+ =1 ⇒ 112

−

=x

dx dt

2. f xx x

dx−

+

∫

11

12

Put xx

t− =1 ⇒ 112

+

=x

dx dt


3.x

x kxdx

2

4 2

1

1

++ +∫

Divide numerator and denominator by x2

4.x

x kxdx

2

4 2

1

1

−+ +∫

Divide numerator and denominator by x2

Illustration 10x x x

xdx

x x

x

( ) (ln )2

4

1 1

1

+ ++∫

Solution. Ix x x

xdx

x x

x=

+ ++∫

( ) (ln )2

4

1 1

1

Put x yx = ⇒ x x dx dyx (ln )+ =1

Iy

ydy

y

yy

dyy

yy

dy=++

=+

+=

+

−

+

∫ ∫ ∫2

4

2

22

2

2

1

1

11

1

11

12

Let yy

t− =1 ⇒ 112

+

=

ydy dt

Idt

t=

+∫ 2 2=

+−1

2 2

1tant

C

=−

+ =−

− −1

2

1

2

1

2

1

2

1 1tan tan

yy

Cx

x

xx

+ C

Illustration 11 Evaluate( )

( ) tan

x dx

x x xx

2

4 2 1

1

3 11

−

+ + +

−∫ .

Solution. The given integral can be written as

Ix dx

x x xx

=−

+ + +

−∫

( / )

( / ) tan

1 1

3 11

2

2 2 1

(dividing Nr and Dr by x2)

Ix dx

x x xx

=−

+ + +

−∫

( / )

( / ) tan

1 1

1 11

2

2 1{ }

Let xx

t+ =1 ⇒ 112

−

=

xdx dt

∴ Idt

t t=

+ ⋅ −∫ ( ) tan ( )2 11…(i)

Now, make one more substitution

tan .− =1 t u Then,dt

tdu

2 1+=


∴ Eq. (i) becomes, Idu

uu C= = +∫ log| |

⇒ I t C x x C= + = + +− −log|tan | log|tan ( / )|1 1 1

Illustration 12( )

( ) ( )

/ /

/ / / /

x x dx

x x x x x x

− −+ + − + +

⋅∫7 6 5 6

1 3 2 1 2 1 2 2 1 31 1

Solution. Ix x x dx

x x x x x x=

−⋅ + + − ⋅

−7 6 7 6 5 6

7 6 1 3 2 1 2 1 2 7 61

/ / /

/ / / / /

( )

( ) ( ) /x x2 1 31+ +∫

=−

+ + − + +∫( )

( ) ( )/ / / /

1

1 1

2

3 2 2 1 2 5 3 2 1 3

x dx

x x x x x x

=

− −

+ +

− + +

∫1

1

11

11

2

1 2 1 3

xdx

xx

xx

/ /

= −+ − +∫

dt

t t( ) ( )/ /1 11 2 1 3

Putting xx

t

xdx dt

+ =

⇒ −

=

1

112

Substitute, ( )t u+ =1 6

= −−∫

6 5

3 2

u du

u u

= −−∫6

1

3u

udu, put u z− =1

= −+

∫61 3( )z

zdz

= −+ + +

∫63 3 13 2z z z

zdz

= − + + +

∫6 3 3

12z zz

dz

= − + + +

+63

3

23

3 2z zz z Clog| |

where, z xx

= + +

−1

1 1

1 6/

Illustration 13 The value of {{[ ]}} ,x dx∫ where {.} and [.] denotes frac-

tional part of x and greatest integer function) is equal to

(a) 0 (b) 1 (c) 2 (d) –1

Solution. Let I x dx= ∫ {{[ ]}}

where, [ ]x = Integer and we know { }n = 0; n ∈ Integer.

∴ I dx= =∫ 0 0

Hence, (a) is the correct answer.


Illustration 14 The value of [{ }] ;x dx∫ (where [.] and {.} denotes greatest

integer and fractional part of x is equal to

(a) 0 (b) 1 (c) 2 (d) –1

Solution. As, we know y x= { } could be shown as

∴ [{ }]x = 0

Thus, I x dx= ∫ [{ }] = =∫ 0 0dx


Illustration 15 The value ofd x

x

( ),

2

2

1

2

+

+∫ is

(a) 2 22x C+ + (b) x C2 2+ + (c) x x C2 2+ + (d) None of these

Solution. Here, Id x

x=

+

+∫( )2

2

1

2

We know, d x x dx( )2 1 2+ =

∴ Ix dx

x=

+∫2

22

Put, x t2 22+ =∴ 2 2x dx t dt=

It dt

tt C= = +∫

22

⇒ I x C= + +2 22


Illustration 16 If( )

( )log ,

x

x xdx a

x

xc

k

k

5

7 6 1+=

+

+∫ then a kand are

(a) 2 5 5 2/ , / (b) 1 5 2 5/ , / (c) 5 2 1 2/ , / (d) 2 5 1 2/ , /

Solution. Here, Ix

x xdx=

+∫( )

( )

5

7 6

=+∫dx

x x( ) ( )2 7

=+

∫dx

xx

7 25 2

11/

/

, put1 5

25 2 7 2xy

xdx dy

/ /= ⇒ − =


1

Ox

y

= −+∫

2

5 1

dy

y= − + +2

51log| |y C

=+

+2

5

1

1log

yC

=+

+2

5 1

5 2

5 2log

/

/

x

xC …(i)

where, I ax

xC

k

k=

+

+log

1(given) …(ii)

∴ From Eqs. (i) and (ii), we get

ax

xC

x

xC

k

klog log

/

/1

2

5 1

5 2

5 2+

+ =

+

+

⇒ a = 2 5/ and k = 5 2/


Illustration 17 The value ofcos cos

cos,

5 4

1 2 3

x x

xdx

+−∫ is equal to

(a) sin sinx x C+ +2 (b) sinsin

xx

C− +2

2

(c) − − +sinsin

xx

C2

2(d) None of these

Solution. Here, Ix x

xdx=

+−∫

cos cos

cos

5 4

1 2 3

=⋅

− −

∫2

9

2 2

1 2 23

212

cos cos

cos

x x

xdx

=−

∫2

2

9

2

3 43

2

2

cos cos

cos

x x

xdx

Multiplying and dividing it by cos ,3

2

xwe get

=⋅ ⋅

−∫

22

3

2

9

2

33

24

3

2

3

cos cos cos

cos cos

x x x

x xdx

=⋅ ⋅−∫

2 2 3 2 9 2

9 2

cos / cos / cos /

cos /

x x x

xdx

(using cos cos cos )3 4 33θ θ θ= −

= − ∫ 22

3

2cos cos

x xdx = − +∫ (cos cos )2x x dx

= − +

+sin

sin2

2

xx C

Hence, (c) is the correct answer.


Illustration 18 The value ofcos cos

cos

7 8

1 2 5

x x

xdx

−+∫ , is equal to

(a)sin cos2

2

3

3

x xC+ + (b) sin cosx x C− +

(c)sin cos2

2

3

3

x xC− + (d) None of these

Solution. Here, Ix x

xdx=

−+∫

cos cos

cos

7 8

1 2 5

=⋅

+∫2

15

2 21 2 5

sin sin

cos

x x

xdx

Multiplying and dividing by sin ,5

2

x

=⋅ ⋅

+ ⋅∫

215

2 2

5

25

22

5

25

sin sin sin

sin sin cos

x x x

x xx

dx

=⋅ ⋅

+ −∫

215

2 2

5

25

2

15

2

5

2

sin sin sin

sin sin sin

x x x

xx

xdx

= ⋅∫ 22

5

2sin sin

x xdx

= −∫ (cos cos )2 3x x dx

Ix x

C= − +sin sin2

2

3

3

Hence, (c) is the correct answer.

Illustration 19 If f x x dx f x c( ) cos ( ) ,= +∫1

2

2 then f x( ) can be

(a) x (b) 1 (c) cos x (d) sin x

Solution. Here, f x x dx f x C( ) cos ( )= +∫1

2

2

Differentiating both the sides, we get

f x x f x f x( ) cos ( ) ( )= ⋅ ′

ie, cos ( ( ))xd

dxf x=

⇒ f x x dx( ) cos= ∫⇒ f x x C( ) sin= +

Hence, (d) is the correct answer.


Target Exercise 1.1

Solved the following integration :

1.2 3

1

2

2 2

++∫x

x xdx

( )2.

dx

x x+ −∫ 1

3.x

x xdx

2

6 2

3

1

++∫ ( )

4.1 2

1

2

2 2

++∫x

x xdx

( )

5.( )

( )

1

1

2

2

++∫x

x xdx 6.

x

xdx

4

21 +∫

7.x

xdx

6

2

1

1

−+∫ 8.

x x

xdx

4 2

2

1

2 1

+ ++∫ ( )

9.( ) ( )x x x

x x x xdx

+ −+ +∫

1 2

10.1 2

2

1 2 1 2 3 2

2

1 2 1 2

−−

− +−

−

−

−

−

−∫x

x x x

x x

x xdx

/ / / / /

11.x

x x

x

x x

x x

x

−

− − − −−

+ +⋅

− +−

+−

∫

6

1 2

2

1 2

264

4 2 4 4

4 2 1

1 2

( )dx

12.( sin )x x x

xdx

2 2 2

21

++∫

sec13. 2x xe dx⋅∫

14.e e

e edx

x x

x x

3 5++ −∫ 15. ( )ln lne e dxa x x a∫ +

16.dx

x1 +∫ sin17. sin cos cos cosx x x x dx2 4∫

18.1

1 2

2++∫

cos

cos

x

xdx 19.

1

1

2

2

−+∫

tan

tan

x

xdx

20.cos

cos sin

22 2

x

x xdx∫ 21. 4

2

21

2cos cos sin

xx x dx⋅ ⋅∫

22.cos sin

cos sin( sin )

x x

x xx dx

−+

+∫ 2 2 2 23. ( sin cos sin )3 2 3x x x dx−∫

24. cos x dx°∫ 25.sec

sec

2 1

2 1

x

xdx

−+∫

26.cos cos

cos

x x

xdx

−−∫

2

127.

sin cos

sin(cos sin )

x x

xdx x x

++

+ >∫ 1 20

28.cos cos

cos cos

2 2x

xdx

−−∫

αα

29.sin cos

sin cos

3 3

2 2

x x

x xdx

+∫

30. sec cosec2 2x x dx∫ 31. 1 2−∫ sin x dx

32.sin cos

sin cos

6 6

2 2

x x

x xdx

+⋅∫

33. sin sin2 29

8 4

7

8 4

π π+

− +

∫

x xdx 34.

cos

cot tan

4 1x

x xdx

+−∫


35. sin sin ( ) sinα α αxx

dx− + −

∫ 2

236.

sin sin sin

cos sin

2 5 3

1 2 22

x x x

x xdx

+ −+ −∫

37.cot

cotcos cot

2 2 1

2 28 4

x

xx x dx

−−

∫ 38.

cos sin

cos(cos )

4 4

1 42 0

x x

xdx x

−+

>∫

Integration by PartsTheorem If u and v are two functions of x, then

uv dx u v dxdu

dxv dx dx= −

∫∫∫∫

ie, The integral of product of two functions = (first function) × (integral ofsecond function) – integral of (differential of first function × integral of secondfunction).

Proof For any two functions f x( ) and g x( ), we have

d

dxf x g x f x

d

dxg x g x

d

dxf x{ ( ) ( )} ( ) { ( )} ( ) { ( )}⋅ = ⋅ + ⋅

∴ f xd

dxg x g x

d

dxf x dx f x g x dx( ) { ( )} ( ) { ( )} ( ) ( )⋅ + ⋅

= ⋅∫∫

or f xd

dxg x dx g x

d

dxf x dx f x( ) { ( )} ( ) { ( )} ( )⋅

+ ⋅

= ⋅∫∫∫ g x dx( )

or f xd

dxg x dx f x g x dx g x

d

dxf x( ) { ( )} ( ) ( ) ( ) { ( )}⋅

= ⋅ − ⋅

∫∫∫ dx

Let f x ud

dxg x v( ) and { ( )}= =

So that g x v dx( ) = ∫∴ uv dx u v dx

du

dxv dx dx= ⋅ − ⋅

⋅∫ ∫∫∫

Points to Consider

While applying the above rule, care has to be taken in the selection of firstfunction (u) and selection of second function ( ).v Normally we use the following

methods :

1. If in the product of the two functions, one of the functions is not directlyintegrable (eg, log| |, sin , cos , tan , ,x x x x− − − …1 1 1 etc.) Then, we take it asthe first function and the remaining function is taken as the second function.eg, In the integration of x x dxtan ,−∫ 1 tan−1 x is taken as the first function

and x as the second function.

2. If there is no other function, then unity is taken as the second function. eg, Inthe integration of tan−∫ 1 x dx, tan−1 x is taken as first function and 1 as thesecond function.

3. If both of the function are directly integrable, then the first function is chosenin such a way that the derivative of the function thus obtained under integralsign is easily integrable.


Usually we use the following preference order for selecting the firstfunction. (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent).

In above stated order, the function on the left is always chosen as the

first function. This rule is called as ILATE.


(i) sin−∫ 1 x dx (ii) log | |e x dx∫Solution. (i) I x dx= −∫ sin 1 = ⋅−∫ sin 1 1x dx

I II

Here, we know by definition of integration by parts that order of preference

is taken according to ILATE. So, ‘ sin ’−1 x should be taken as first and ‘1’ as

the second function to apply by parts.

Applying integration by parts, we get

I x xx

x dx= ⋅ −−

⋅− ∫sin ( )1

2

1

1

= ⋅ +− ∫x xdt

tsin

/1

1 2

1

2let 1 2− =x t

− =2x dx dt

= + ⋅ +−x xt

Csin/

/1

1 21

2 1 2x dx dt= − 1

2

I x x x C= + − +−sin 1 21

∴ sin sin− −= + − +∫ 1 1 21x dx x x x C

(ii) I x dxe= ∫ log | | = ⋅∫ log | |e x dx

I II

1

Applying integration by parts, we get

= ⋅ − ⋅∫log| |x xx

x dx1

= − ∫x x dxlog| | 1

I x x x C= − +log| |


(i) x x dxcos∫ (ii) x x dx2 cos∫Solution. (i) x x dxcos∫

I x x dx= ∫I II

cos

Applying integration by parts,

I x x dxd

dxx x dx dx= −

∫ ∫∫( cos ) ( ) { (cos ) }

I x x x dx= − ⋅∫sin sin1

I x x x C= + +sin cos


(ii) I x x dx= ∫ 2

I II

cos

Applying integration by parts,

I x x dxd

dxx x dx= −

⋅∫ ∫∫2 2( cos ) ( ) { cos }

= − ⋅∫x x x x dx2 2sin (sin ) = − ∫x x x x dx2 2sin (sin )

We again have to integrate x x dxsin∫ using integration by parts,

= ⋅ − ⋅∫x x x x dx2 2sin sin

I II

= − −

∫ ∫∫x x x x dx

dx

dxx dx dx2 2sin ( sin ) ( sin )

= − − − ⋅ −∫x x x x x dx2 2 1sin { cos ( cos ) }

I x x x x x C= + − +2 2 2sin cos sin

Illustration 22 Evaluatesin cos

sin cos

− −

− −−+∫

1 1

1 1

x x

x xdx.

Solution.sin cos

sin cos

− −

− −−+∫

1 1

1 1

x x

x xdx =

− −− −

∫sin ( / sin )

/

1 12

2

x xdx

ππ

( sin cos / )Q− −+ =1 1 2θ θ π

⇒ I x dx= −−∫2

2 21

ππ( sin / )

I x dx dx= −− ∫∫4

11

πsin

I x dx x C= − +−∫4 1

πsin …(i)

Let x = sin ,2 θ then dx d d= =2 2sin cos sinθ θ θ θ θ

∴ sin sin− = ⋅∫∫ 1 2x dx dθ θ θI II

Applying integration by parts

sincos

cos− = − ⋅ + ∫∫ 1 2

2

1

22x dx dθ

θθ θ =

−⋅ +

θθ θ

22

1

42cos sin

=− ⋅

⋅ − + ⋅ ⋅ −1

21 2

1

212 2θ

θ θ θ( sin ) sin sin

=−

− + ⋅ −−1

21 2

1

211sin ( )x x x x …(ii)

From Eqs. (i) and (ii), we get

I x x x x x C=−

− + −

− +−4 1

21 2

1

211

π(sin ) ( )

= − − − − +−21 22 1

π{ }x x x x x C( ) sin


Integral of the Form e f x f x dxx { ( ) ( )}+ ′∫Theorem Prove that e f x f x dx e f x Cx x{ }( ) ( ) ( )+ ′ = +∫Proof We have, e f x f x dxx { }( ) ( )+ ′∫

= ⋅ + ⋅ ′∫∫ e f x dx e f x dxx x

II I

( ) ( )

= ⋅ − ′ ⋅ + ⋅ ′ +∫∫f x e f x e dx e f x dx Cx x x( ) ( ) ( )

= ⋅ +f x e Cx( )

Thus, to evaluate the integrals of the type e f x f x dxx { }( ) ( ) ,+ ′∫ we first

express the integral as the sum of two integrals e f x dxx ( )∫ and e f x dxx ′∫ ( )

and then integrate the integral involving e f xx ( ) as integrand by parts

taking ex as second function.

Points to Consider

The above theorem is also true, if we have ekx in place of e iex. ,

e f kx f kx dx e f kx Ckx kx{ }( ) ( ) ( )+ ′ = +∫

General Concept

e f x g x f x dxg x( ) { ( ) ( ) ( )}∫ ′ + ′

Proof { {I e f x g x dx e f x dxg x g x= ′ + ′∫ ∫( ) ( )( ) ( ) ( )

II I II123

= ⋅ − ′ ⋅ + ⋅ ′ = ⋅∫f x e f x e dx e f x dx f x eg x g x g x g x( ) ( ) ( ) ( )( ) ( ) ( ) ( )∫eg, e

x x x x x

xdxx x x( sin cos ) cos ( sin cos )+ − +

∫

2 2

2

⇒ e xx x x

xdxx x x( sin cos ) cos

sin cos+ −+

∫ 2

2

⇒ e x xx

x

x

xdx x x( sin cos ) cos

cos cos+

+

′

∫ x

⇒ ex

xCx x x( sin cos ) cos+ ⋅ +

eg, e x x dxxtan (sin )∫ − sec = −∫ ∫e x dx e x dxx xtan tansin sec

⇒ − ⋅ + −∫ ∫e x e x x dx e x dxx x xtan tan tancos cossec sec2

⇒ − ⋅e xxtan cos


(i) ex x

xdxx 1

2

+

∫

sin cos

cos(ii) e

x

xdxx2 1 2

1 2

++

∫

sin

cos


Solution. (i) I ex x

xdxx=

+

∫

12

sin cos

cos

I ex

x x

xdxx= +

∫12 2cos

sin cos

cos

I e x x dxx= +∫ { }tan sec2

I e x dx e x dxx x= ⋅ + ∫∫II I

tan (sec )2

I x e x e dx e x dx Cx x x= ⋅ − ⋅ + ⋅ +∫∫tan sec sec2 2

I e x Cx= +tan

(ii) I ex

xdxx=

++

∫ 2 1 2

1 2

sin

cos=

+

∫ ex x

xdxx2

2

1 2

2

sin cos

cos

= +

∫ ex

x x

xdxx2

2 2

1

2

2

2cos

sin cos

cos= +

∫ e x x dxx2 21

2sec tan

= ⋅ + ⋅∫∫ e x dx e x dxx x2 2 21

2II Itan sec

= ⋅ − ⋅ + ⋅∫∫tan xe

xe

dx e x dxx x

x2

22

2 2

2 2

1

2sec sec

I e x Cx= ⋅ +1

2

2 tan

Illustration 24 Evaluate ex

xdxx 1

1 2

2−

+

∫ .

Solution. I ex

xdxx=

−+

∫

1

1 2

2

=− +

+∫ ex x

xdxx ( )

( )

1 2

1

2

2 2

I ex

x

x

xdxx=

++

−+

∫1

1

2

1

2

2 2 2 2( ) ( )

=+

−+

∫ ex

x

xdxx 1

1

2

12 2 2( )as

d

dx x

x

x

1

1

2

12 2 2+

= −

+

( )

=+

+e

xC

x

1 2

∴ Ie

xC

x

=+

+1 2

Integrals of the Form e bx dx e bx dxax axsin , cos∫∫Let I e bx dxax= ∫ (sin )

Then, I bx e dxax= ⋅∫ sin

I II

I bxe

ab bx

e

adx

ax ax

= ⋅

− ⋅∫sin cos


Ia

bx eb

abx

e

ab bx

e

adxax

ax ax

= ⋅ − ⋅ − − ⋅

∫1

sin cos ( sin )

= ⋅ − ⋅ − ⋅∫1

2

2

2abx e

b

abx e

b

abx e dxax ax axsin cos sin

Ia

bx eb

abx e

b

aIax ax= ⋅ − ⋅ −1

2

2

2sin cos

∴ Ib

aI

e

aa bx b bx

ax

+ = ⋅ ⋅ −2

2 2

1( sin cos )

⇒ Ia b

a

e

aa bx b bx

ax2 2

2 2

+

= −( sin cos )

or Ie

a ba bx b bx C

ax

=+

− +2 2

( sin cos )

Thus, e bx dxe

a ba bx b bx Cax

ax

sin ( sin cos )=+

− +∫ 2 2

Similarly, e bx dxe

a ba bx b bx Cax

ax

cos ( cos sin )=+

+ +∫ 2 2

Aliter Use Euler’s equation

Let P e bx dxax= ∫ cos and Q e bx dxax= ∫ sin

Hence, P iQ e e dx e dxax ibx a ib x+ = ⋅ =∫ ∫ +( )

P iQa ib

ea ib

a be bx i bxa ib x ax+ =

+=

−+

++12 2

( ) (cos sin )

=+ − −

+( cos sin ) ( sin cos )ae bx be bx i ae bx be bx

a b

ax ax ax ax

2 2

∴ Pe a bx b bx

a b

ax

=+

+( cos sin )

2 2

Qe a bx b bx

a b

ax

=−

+( sin cos )

2 2


(i) e x dxx cos2∫ (ii) sin (log )x dx∫Solution. (i) I e x dxx= ⋅∫ cos2 = ⋅

+

∫ e

xdxx 1 2

2

cos

I e dx x e dxx x= + ⋅∫∫1

2

1

22cos

I e Ix= +1

2

1

21 …(i)

where I x e dxx1 2= ⋅∫ cos


I x e dx x e x e dxx x x1 2 2 2 2= ⋅ = ⋅ − − ⋅∫∫ cos cos sin

I II

= ⋅ + ⋅∫e x x e dxx xcos sin2 2 2

I II

= ⋅ + ⋅ − ⋅∫e x x e x e dxx x xcos {sin cos }2 2 2 2 2

= ⋅ + ⋅ −e x x e Ix xcos sin2 2 2 4 1

∴ I e x x ex x1

1

52 2 2= +{ }cos sin …(ii)

From Eqs. (i) and (ii), we get

I e e x x ex x x= + ⋅ + ⋅1

2

1

2

1

52 2 2{ }cos sin

I e e x x Cx x= + + +1

2

1

102 2 2{ }cos sin

(ii) I x dx= ∫ sin (log )

Let log x t= ⇒ x e dx e dtt t= =or

∴ I t e dt t e t e dtt t t= ⋅ = ⋅ − ⋅∫∫ (sin ) sin cosI II I II

I t e t e t e dtt t t= ⋅ − ⋅ − − ⋅∫sin {cos sin }

I e t e t It t= ⋅ − ⋅ −sin cos

∴ I e t t Ct= − +1

2(sin cos )

Ix

x x C= − +2

{ }sin (log ) cos (log )

Illustration 26 Evaluatex dx

x x x

2

2( sin cos )+⋅∫

Solution. Let Ix

x x xdx=

+∫2

2( sin cos )

Multiplying and dividing it by ( cos )x x , we get

I x xx x

x x xdx= ⋅

+∫ ( )( cos )

( sin cos )sec

I II

2

I x xx x

x x xdx= ⋅

+∫seccos

( sin cos )2

−

+

∫∫d

dxx x

x x

x x xdx dx( )

cos

( sin cos )sec

2

= ⋅−

+x x

x x xsec

1

( sin cos )

− ⋅ + ⋅−

+∫ ( tan )( sin cos )

x x x xx x x

dxsec sec1

=−

++

+⋅ +

x x

x x x

x x x

x x x x

sec

( sin cos )

( sin cos )

cos ( sin cos )2dx∫


=−

++ ∫

x x

x x xx dx

secsec

( sin cos )

2

Ix x

x x xx C=

−+

+ +sec

( sin cos )tan

Integration of the Type (sin cos )m nx x dx⋅∫(i) Where m n, belongs to natural number.

(ii) If one of them is odd, then substitute for term of even power.

(iii) If both are odd, substitute either of them.

(iv) If both are even, use trigonometric identities only.

(v) If m and n are rational numbers andm n+ −

2

2is a negative

integer, then substitute cot x p= or tan x p= which so ever is

found suitable.

Illustration 27 Evaluate sin cos3 5x x dx⋅∫ .

Solution. I x x dx= ⋅∫ sin cos3 5

Let cos sinx t x dx dt= ⇒ − =I t t dt= − − ⋅∫ ( )1 2 5

I t dt t dtt t

C= − = − +∫∫ 7 58 6

8 6

Ix x

C= − +cos cos8 6

8 6

Aliter I R R dR x R= − =∫ 3 2 21( ) , if sin , cos x dx dR=

I R dR R dR R dR= − + ∫∫∫ 3 5 72

Ix x x

C= − + +sin sin sin4 6 8

4

2

6 8

Points to Consider

This problem can also be handled by successive reduction or by trigonometricalidentities. Answers will be in different form but identical with modified constantof integration.

Illustration 28 Evaluate sin cos ./ /− −⋅∫ 11 3 1 3x x dx

Solution. Here, sin cos/ /− −⋅∫ 11 3 1 3x x dx ie,− − −

= −

11

3

1

32

23

∴ Ix

x xdx x x dx=

⋅=

−

−−∫

cos

sin sin(cot ) )

/

//

1 3

1 3 41 3 2 2(cosec∫


I x x x dx= +−∫ (cot ) ( cot )/1 3 2 21 cosec (let cot ,x t x dx dt= − =cosec2 )

= − +−∫ t t dt1 3 21/ ( ) = − +−∫ ( )/ /t t dt1 3 5 3

= − +

+3

2

3

8

2 3 8 3t t C/ / = − +

+3

2

3

8

2 3 8 3(cot ) (cot )/ /x x C


(i)1

sin ( ) cos ( )x a x bdx

− −∫ (ii)1

cos ( ) cos ( )x a x bdx

− −∫

Solution. (i) Ix a x b

dx=− −∫

1

sin ( ) cos ( )

Ia b

a b

dx

x a x b=

−−

⋅− −∫

cos ( )

cos ( ) sin ( ) cos ( )

=−

⋅− − −

− −∫1

cos ( )

cos ( ) ( )

sin ( ) cos ( )a b

x b x a

x a x bdx

{ }

=−

⋅− ⋅ −− −

+1

cos ( )

cos ( ) cos ( )

sin ( ) cos ( )

sin (

a b

x b x a

x a x b

x − ⋅ −− −

∫b x a

x a x bdx

) sin ( )

sin ( ) cos ( )

=−

− + −∫1

cos ( )cot ( ) tan ( )

a bx a x b dx{ }

=−

− − − +1

cos ( )log|sin ( )| log|cos ( )|

a bx a x b C{ }

=−

−−

+1

cos ( )log

sin ( )

cos ( )a b

x a

x bCe

(ii) Ix a x b

dx=− −∫

1

cos ( ) cos ( )

=−

−− −∫

1

sin ( )

sin ( )

cos ( ) cos ( )a b

a b

x a x bdx

=−

− − −− −∫

1

sin ( )

sin ( ) ( )

cos ( ) cos ( )a b

x b x a

x a x bdx

{ }

=−

− −− −

−−1

sin ( )

sin ( ) cos ( )

cos ( ) cos ( )

cos (

a b

x b x a

x a x b

x b) sin ( )

cos ( ) cos ( )

x a

x a x bdx

−− −

∫

=−

− − −∫1

sin ( )tan ( ) tan ( )

a bx b x a dx{ }

=−

− − + − +1

sin ( )[ log|cos ( )| log|cos ( )|]

a bx b x a C

=−

−−

+1

sin ( )log

cos ( )

cos ( )a b

x a

x bC

Illustration 30 Evaluatesin ( )

sin ( )

x a

x bdx

++∫ .

Solution. Let Ix a

x bdx=

++∫

sin ( )

sin ( )

Put x b t+ = ⇒ dx dt=


∴ It b a

tdt=

− +∫

sin ( )

sin=

−+

−

∫sin cos ( )

sin

cos sin ( )

sin

t a b

t

t a b

tdt

= − + − ∫∫cos ( ) sin ( ) cot ( )a b dt a b t dt1

= − + − +t a b a b t Ccos ( ) sin ( ) log|sin |

= + − + − + +( ) cos ( ) sin ( ) log|sin ( )|x b a b a b x b C

Some Special Integrals

(i)dx

x a a

x

aC

2 2

11

+=

+−∫ tan

(ii)dx

x a a

x a

x aC

2 2

1

2−=

−+

+∫ log

(iii)dx

a x a

a x

a xC

2 2

1

2−=

+−

+∫ log

(iv)dx

a x

x

aC

2 2

1

−=

+−∫ sin

(v)dx

a xx x a C

2 2

2 2

+= + + +∫ log| |

(vi)dx

x ax x a C

2 2

2 2

−= + − +∫ log| |

(vii) a x dx x a x ax

aC2 2 2 2 2 11

2

1

2− = − +

+−∫ sin

(viii) a x dx x a x a x a x C2 2 2 2 2 2 21

2

1

2+ = + + + + +∫ log| |

(ix) x a dx x x a a x x a C2 2 2 2 2 2 21

2

1

2− = − − + − +∫ log| |

Some Important Substitutions

Expression Substitution

a x2 2+ x a a= tan cotθ θor

a x2 2− x a a= sin cosθ θor

x a2 2− x a a= sec or cosecθ θa x

a x

a x

a x

−+

+−

or x a= cos 2θ

x

xx x

−−

− −α

βα βor ( ) ( ) x = +α θ β θcos sin2 2


integral calculus by arihant b016

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