integral calculus finals reviewer
TRANSCRIPT
INTEGRAL CALCULUS FINALS REVIEWER (2nd
Sem ‘11-‘12)
INTEGRATION TECHNIQUES
I. Integration by Parts
∫udv = uv – ∫vdu
1. ∫lnxdx *First, determine u and dv. Yung dv, dapat laging kasama yung “dx” sa formula. For u, an easier way to find that is by using the code “LIPET”: Logarithm, Inverse, Polynomial, Exponential, Trigonometric. Kumbaga parang ‘yan yung hierarchy ng pagpipilian mo kung ano yung gagawin mong u. Logarithm being the highest and Trigonometric the lowest
u = lnx dv = dx
du = dxx v = x
*substitute these values sa formula na ∫udv = uv – ∫vdu
∫lnxdx = xlnx ─ ∫x ▪ dxx
= xlnx – x + c
2. ∫x2lnxdx
u = lnx dv = x2dx
du = dxx v =
x3
3
∫ x2lnx dx =
13 x
3lnx ─
13 ∫ x
3 ▪
dxx
= 13 x
3lnx ─
13 ∫ x
2dx
u = x2 dv = dx du = 2xdx v = x
= 13 x
3lnx ─
13 [x
3 – 2∫ x
2dx]
= 13 x
3lnx ─
13 x
3 +
23 ▪
x3
3
= 13 x
3lnx ─
13 x
3 +
2x3
9 + c
= 13 x
3
lnx ─ 1 +
23 + c
= 13 x
3
lnx ─
13 + c
3. ∫x3e
xdx
u = x3 dv = exdx du = 3x2dx v = ex
∫ x3e
xdx = x
3e
x ─ 3 ∫ x
2e
xdx
u = x2 dv = exdx du = 2xdx v = ex
= x3e
x ─ 3 * x
2e
x – 2∫ xe
xdx ]
u = x dv = exdx du = dx v = ex
= x3e
x ─ 3x
2e
x + 6[ xe
x – ∫e
xdx ]
= x3e
x ─ 3x
2e
x + 6xe
x – 6e
x + c
4. ∫sec3xdx
= ∫sec2xsecxdx
u = secx dv = sec2xdx du = secxtanxdx v = tanx
= secxtanx ─ ∫secxtan2xdx
= secxtanx ─ ∫secx(sec2x – 1)dx
= secxtanx ─ ∫(sec3x – secx)dx
= secxtanx ─ ∫sec3xdx + ∫secx
*transpose ∫sec3xdx kasi same siya nung sa other side
2 ∫sec3xdx = secxtanx + ∫secx
∫sec3xdx =
12 ( )secxtanx + ln(secx + tanx) + c
5. ∫excos2xdx
u = ex dv = cos2xdx
du = exdx v = 12 sin2x
= 12 e
xsin2x –
12 ∫e
xsin2xdx
u = ex dv = sin2xdx
du = exdx v = ─ 12 cos2x
= 12 e
xsin2x –
12
─
12 e
xcos2x ─
─
12 ∫e
xcos2xdx
= 12 e
xsin2x +
14 e
xcos2x ─
14 ∫e
xcos2xdx
*transpose ∫exsin2xdx kasi same siya nung sa other side
∫excos2xdx +
14 ∫e
xcos2xdx =
12 e
xsin2x +
14 e
xcos2x
54 ∫e
xcos2xdx =
12 e
x
sin2x +
12 cos2x
∫excos2xdx =
25 e
x
sin2x +
12 cos2x + c
6. ∫(x + sinx)2dx
= ∫(x2 + 2xsinx + sin
2x)dx
= x
3
3 + 2∫xsinxdx + ∫sin2xdx
= x
3
3 + 2∫xsinxdx + 12 ∫(1 – cos2x)dx
= x
3
3 + 12
x ─
12 sin2x + 2∫xsinxdx
= x
3
3 + 12 x ─
14 sin2x+ 2∫xsinxdx
u = x dv = sinxdx du = dx v = ─ cosx
= x
3
3 + 12 x ─
14 sin2x+ 2 *─xcosx +∫cosxdx]
= x
3
3 + 12 x ─
14 sin2x+ 2 *─xcosx + sinx +
= x
3
3 + 12 x ─
14 sin2x ─ 2xcosx + 2sinx + c
7. ∫ t3dt
1 + 3t2
= ∫ t2▪t
dt
1 + 3t2
u = t2 dv = (1 + 3t2)-1/2tdt
du = 2tdt v = 16 ▪ 2(1 + 3t2)1/2=
13 (1 + 3t2)1/2
= 13 t
21 + 3t
2 ─
23 ∫ 1 + 3t
2 tdt use undu
u = 1 + 3t2 du = 6tdt
= 13 t
21 + 3t
2 ─
23 ▪
16 ▪
23 (1 + 3t
2)
3/2
= 13 t
21 + 3t
2 ─
227 (1 + 3t
2)
3/2 + c
8. ∫xCsc-1
xdx
u = Csc-1x dv = xdx
du = ─ dx
x x2 ─ 1 v =
x2
2
= 12 x
2Csc
-1x +
12 ∫ x
2 ▪
dx
x x2 ─ 1
= 12 x
2Csc
-1x +
12 ∫ xdx
x2 ─ 1
= 12 x
2Csc
-1x +
12 ∫ (x
2 – 1)
-1/2xdx use undu
= 12 x
2Csc
-1x +
12 (x
2 – 1)
1/2 + c
9. ∫sin x dx
y = x y2 = x 2ydy = dx
=∫ siny ▪ 2ydy
= 2∫ysinydy
u = y dv = sinydy du = dy v = ─cosy
= 2 * ─ycosy + ∫cosydy ]
= ─2ycosy + 2siny + c
*substitute x back sa mga y
= ─2 x cos x + 2sin x + c
10. ∫x3(1 – x
2)
1/3dx ∫x
2x(1 – x
2)
1/3dx
u = x2 dv = (1 – x2)1/3xdx
du = 2xdx v = ─12 ▪
34 (1 – x2)4/3
= ─38 x
2(1 – x
2)
4/3 +
38 ∫ (1 – x
2)
4/3xdx use undu
= ─38 x
2(1 – x
2)
4/3 +
34 ▪
12 ▪
37 (1 – x
2)
7/3
= ─38 x
2(1 – x
2)
4/3 +
956 (1 – x
2)
7/3
= ─38 (1 – x
2)
4/3
x
2 +
37 (1 ─ x
2)
= ─38 (1 – x
2)
4/3
x
2 +
37 ─
37 x
2
= ─38 (1 – x
2)
4/3
4
7 x2 +
37
= ─ 3
56 (1 – x2)
4/3(4x
2 + 3) + c
11. ∫Sec-11
x dx
u = Sec-11x dv = dx
du = ─
dxx2
1x
1x2 ─ 1
v = x
= ─
dxx2
1x
1 ─ x2
x2
= ─
dxx2
1x ▪
1x 1 ─ x2
= ─ dx
1 ─ x2
= xSec-11
x + ∫ xdx
1 ─ x2 use undu
= xSec-11
x + 12 ▪
21 (1 – x
2)
1/2 + c
= xSec-11
x + 1 – x2 + c
12. ∫ln3xdx
u = ln3x dv = dx
du = 3ln2xdx
x v = x
= xln3x ─ 3 ∫x ▪
ln2xdxx
= xln3x ─ 3 ∫ln
2xdx
u = ln2x dv = dx
du = 2lnxdx
x v = x
= xln3x ─ 3 [xln
2x ─ 2 ∫ x ▪
lnxdxx
= xln3x ─ 3xln
2x + 6 ∫lnxdx
u = lnx dv = dx
du = dxx v = x
= xln3x ─ 3xln
2x + 6 *xlnx ─ ∫ x ▪
dxx ]
= xln3x ─ 3xln
2x + 6 *xlnx ─ ∫ dx]
= xln3x ─ 3xln
2x + 6xlnx ─ 6x + c
13. π2
0sin
5xdx sin
4xsinxdx
u = sin4x dv = sinxdx du = 4sin3xcosxdx v = ─cosxdx
= ─sin4xcosx + 4∫ sin
3xcos
2xdx
= ─sin4xcosx + 4∫ sin
3x(1 – sin
2x)dx
= ─sin4xcosx + 4∫ (sin
3x – sin
5x) dx use undu
= ─sin4xcosx + 4∫sin
3xdx – 4∫sin
5xdx
*transpose ∫sin5xdx kasi same siya nung sa other side
4∫sin5xdx +∫sin
5xdx = ─sin
4xcosx + 4∫sin
3xdx
5∫sin5xdx = ─sin
4xcosx + 4∫sin
3xdx
u = sin2x dv = sinxdx du = 2sinxcosxdx v = ─cosxdx
∫sin5xdx =
15 ( )─sin
4xcosx + 4∫sin
3xdx
= ─sin4xcosx + 4
─
cos4x
4 + cos
6x
6 ]
π2
0
ADDITIONAL FORMULA: WALLIS’ FORMULA
*only works when the upper and lower limits are π2 and 0.
π2
0sin
mxcos
nxdx =
[(m-1)(m-3)…2 or 1+▪*(n-1)(n-3)…2 or 1+(m+n)(m+n-2)(m+n-4)…2 or 1 • α
where: α = π2 , if both m and n are EVEN
= 1, if other wise
*yung “2 or 1”, ibig sabihin yung subtraction blah, yung value nun diba paliit nang paliit. Basta until maging 2 OR 1 ka magsstop.
1. π4
0
(1 ─ cos22x)
7/2dx
tan42xcsc
24xsinxcosx
= ∫ (sin22x)
7/2 dx
sin42x
cos42x ▪
14sin
22xcos
22x ▪
12 sin2x
*okay so isa-isahin natin yung mga chuchu sa denominator: a. tan42x
recall sa identities na tanx = sinxcosx . Kaya naging tan42x=
sin42xcos42x yay
b. csc24x 1
sin24x
recall the trigonometric transformation formula sinxcosx = 12
sin2x. So ang main agenda mo is to get sin24x
sinxcosx = 12 sin2x
*i-double mo yung angle ng right side. so pag dinouble mo yung angle sa right side, double the angle sa left as well
sin2xcos2x = 12 sin4x
*square both sides
sin22xcos22x = 14 sin24x *transpose
14
4sin22xcos22x = sin24x tadaaaaa yay you
c. sinxcosx recall the identity sin2x = 2sinxcosx. Just transpose 2 to the other
side. So you’ll get 12 sin2x = sinxcosx
= ∫ sin72x dx
18
sin52x
sin22xcos
62x
= 8 ∫sin72x dx
sin
32x
cos62x
= 8 ∫sin42xcos
62xdx
*represent 2x as y. so y = 2x. And dy = 2dx. So dx = dy2
= 8 ▪ 12 ∫sin
4ycos
6ydy
*change the limits. To do that, substitute x sa y = 2x.
x 0 π4
y 0 π
2
= 4 π2
0sin
4ycos
6ydy
*use Wallis’ formula
= 4 ▪ [(4-1)(4-3)][(6-1)(6-3)(6-5)]
(6+4)(6+4-2)(10-4)(10-6)(10-8) ▪ π2
= 4 ▪ (3▪1)(5▪3▪1)10▪8▪6▪4▪2 ▪
π2
= 3π2
7
II. Substitution Methods
A. Substitution of Functions
1. ∫x 1 + x dx u = 1 + x x = u – 1 dx = du
*substitute all x’s with u’s
= ∫(u – 1)u1/2
du
= ∫(u3/2
– u1/2
)du
= 25 u
5/2 –
23 u
3/2 + c
= 6u
5/2 – 10u
3/2
15 + c
= 2
15 u3/2
(3u – 5) + c
= 2
15 (1 + x)3/2
[3(1 + x) – 5] + c
= 2
15 (1 + x)3/2
(3x – 2) + c
2. ∫ x3dx
(x2 + a
2)
3 x
2xdx
(x2 + a
2)
3
u = x2 + a2
x2 = u – a2
2xdx = du *substitute all x’s with u’s
= 12 ∫(u - a
2)du
u3
= 12 ∫(u
-2 – a
2u
-3)du
= 12
u
-1
-1 ─ a
2u
-2
-2 + c
= 12
─
1u +
a2
2u2 + c
= 12
-2u + a
2
2u2 + c
*substitute the value of u back to x2 + a2
= 12
-2(x
2 + a
2) + a
2
2(x2 + a
2)
2 + c
= 1
4(x2 + a
2)
2 [a2 – 2(x
2 + a
2)] + c
= ─ 1
4(x2 + a
2)
2 (2x2 + a
2) + c
3. ∫ y +3(3 - 2y)
2/3 dy x
2xdx
(x2 + a
2)
3
u = 3 – 2y
2y = 3 – u 2dy = ─du
= ─ 12 ∫
3 - u + 62
u2/3 du
= ─ 14 ∫(9 – u)u
-2/3du
= ─ 14 ∫(9u
-2/3 – u
1/3)du
= ─ 14
3 ▪ 9u
1/3
1 ─ 3u
4/3
4 + c
= ─3
16 u1/3
(36 – u) + c
= ─3
16 (3 – 2y)1/3
(2y + 33) + c
B. Algebraic Substitution
1. ∫x 1 + x dx
u = 1 + x u2 = 1 + x x = u2 – 1 dx = 2udu
= 2∫(u2 – 1) ▪ u ▪ udu
= 2∫u2(u
2 – 1)du
= 2∫ (u4 – u
2)du
= 2
u
5
5 ─ u
3
3 + c
= 2
3u
5 ─ 5u
3
15 + c
= 2
15 u3 (3u
2 – 5) + c
= 2
15 (1 + x)3/2
[3(1 + x) – 5] + c
= 2
15 (1 + x)2/3
(3x – 2) + c
2. 3 3
2 2
dyy - y
1/3 dy
u = y1/3
u3 = y 3u2du = dy
= 3∫u2du
u3 - u
= 3∫ u2du
u(u2 - 1)
= 3∫ uduu
2 - 1
= 3ln(u2 – 1)
= 32 ln(y
2/3 – 1) ]
3 3
2 2
= 32 (ln2 – ln1) =
32 ln2
3. 7
0
dx
1 + 3
x + 1
u = 3 x + 1 u3 = x + 1 3u2du = dx
= 3∫u2du
1 + u *divide u2 by 1 + u
= 3∫
u - 1 +
1u + 1 du
**change the limits. To do that, substitute x sa u = 3 x + 1
x 0 7
y 1 2
= 3
u2
2 + u + ln(u + 1) ] 1
2
= 3
1
2 + ln32
4. ln2
0
e2x
dx
1 + ex
ex ▪ e
xdx
1 + ex
u = 1 + ex u2 = 1 + ex
ex = u2 – 1 2udu = exdx
= 2∫(u2 - 1)udu
u
= 2∫(u2 – 1)du
*change the limits. To do that, substitute x sa u = 1 + ex
x 0 ln2
u 2 3
= 2
u
3
3 - u ] 3
2
= 3
1
3 (3 3 - 2 2 ) - ( 3 - 2 ) = 2 2
3
C. Reciprocal Substitution
*use this when you see equations like this:
dx
x ax2 + bx + c
and substitute x = 1y & dx = ─
dyy
2
1. ∫ dx
x 2x - x2 *substitute x =
1y & dx = ─
dyy2
= ─∫dyy
2
1y
2y -
1y
2
= ─∫dyy
2
1y
2y - 1y
2
= ─∫dyy
2
1y
2 2y -1
= ─∫ dy
2y - 1
= ─ ∫(2y – 1)-1/2
dy
= ─ 12 ▪
21 (2y – 1)
1/2 + c
= ─ 2x - 1 + c
2. 5/3
5/4
dx
x2
x2 - 1
= ─∫dyy
2
1
y 2 1
y2 - 1
= ─∫ ydy
1 - y2 ─ ∫(1 – y
2)
-1/2ydy
**change the limits. To do that, substitute x sa x = 1y
x 5/3 5/4
u 3/5 4/5
= 12 ▪
21 (1 – y)
1/2] 3/5
4/5
= 1 - 9
25 ─ 1 - 1625 =
15
3. ∫ dx
x 2x - x2
= ─∫ dy
25y2 - 1
u = 5y a = 1 du = 5dy
─15 ∫ du
u2 - a2
= ─ 15 ln
5 + 25 - x2
x + c
D. Trigonometric Substitution
If you see this combination: Substitute these:
a2 – u
2 u =asinθ
a2 + u
2 u = atanθ
u2 – a
2 u = asecθ
2ax - x2 x = 2asin
2θ
2ax + x2 x = 2atan
2θ
x2 - 2ax x = 2asec
2θ
1. ∫ duu
2 - a
2
u = asecθ du = atanθsecθdθ
= a∫tanθsecθdθa
2sec
2θ - a
2
= aa
2 ∫tanθsecθdθsec
2θ - 1
= 1a ∫tanθsecθdθ
tan2θ
= 1a ∫secθdθ
tanθ
= 1a ∫cscθdθ
= 1a ln |cscθ – cotθ| + c
*going back to u = asecθ…i-draw mo sa right triangle
*so diba cscθ, which is hypopp so
magiging u
u2 - a2 . and cotθ
= a
u2 - a2
= 1a ln
u
u2 - a
2 –
a
u2 - a
2 + c
= 1a ln
u - a
u2 - a
2 + c
*i-square yung fraction para mawala yung square root at may ma-cancel hihi
= 1a ln
(u - a)(u - a)(u - a)(u + a) + c
= 1a ln
u - au + a + c
2. 2
0x
24 - x
2 dx
= ∫ x2(2)
2 - (x)
2 dx
u = asinθ x = 2sinθ dx = 2cosθdθ
= ∫ (2sinθ)2 4 - (2sinθ)
2 ▪ 2cosθdθ
= 2▪4 ∫ sin2θ 4 - 4sin
2θ cosθdθ
= 8 ∫ sin2θ ▪ 2 1 - sin
2θ cosθdθ
= 16 ∫ sin2θ cos
2θ cosθdθ
= 16 ∫ sin2θcos
2θdθ
*change the limits. To do that, substitute x sa x = 2sinθ
x 0 2
u 0 π
2
= 16 π/2
0 sin
2θcos
2θdθ
*use Wallis’ Formula
= 16 ▪ (1)(1)(4)(2) ▪
π2 = π
3. 2a
0x
22ax - x
2 dx
x = 2asin2θ dx = 4asinθcosθdθ
= ∫ (2asin2θ)
2 2a(2asin
2θ) - (2asin
2θ)
2 ▪
4asinθcosθdθ
= 4a▪4a2 ∫ sin
4θ 4a
2sin
2θ - 4a
2sin
4θ sinθcosθdθ
= 16a3 ∫ sin
4θ 4a
2sin
2θ(1 - sin
2θ) sinθcosθdθ
= 16a3 ∫ sin
4θ ▪ 2asinθ cos
2θ sinθcosθdθ
= 32a4 ∫ sin
6θcos
2θdθ
*change the limits. To do that, substitute x sa x =
2asin2θ
= 32a4
π/2
0 sin
6θcos
2θdθ *use wallis’
= 32a
4 (5 x 3 x 1)(1)
8 x 6 x 4 x 2 ▪ π2 =
5πa4
8
E. Half-Angle Substitution
*use this when you see trigo functions
z = tan12 (nx) dx =
1n ▪
2dz1 + z
2
tan(nx) = 2z
1 - z2 sin(nx) =
2z1 + z
2
cos(nx) = 1 - z
2
1 + z2
1. ∫ dx1 + sinx + cosx n = 1
=
2dz1 + z2
1 +2z
1 + z2 + 1 - z2
1 + z2
= ∫
2dz1 + z2
1 + z2 + 2z + 1 - z2
1 + z2
= ∫ 2dz2 + 2z = ∫ dz
1 + z
= ln (1 + z) + c
= ln (1 + tan12 x) + c
2. π/2
0
dx3 + cos2x n = 2
=∫
12 ▪
2dz1 + z2
3 + 1 - z2
1 + z2
= 12 ∫ 2dz
3 + 3z2 + 1 - z
2
= 12 ∫ 2dz
4 + 2z2 =
12 ∫
dz2 + z
2 du
a2 + u
2
= 1
2 2 Tan
-1 z
2 ]
π/2
0
= 1
2 2 Tan
-1tan
12 (nx)
2 ]
π/2
0
= 1
2 2 Tan
-1tanx
2 ]
π/2
0
= 1
2 2
Tan-1
tanπ2
2 - Tan
-1tan0
2 =
π
4 2
3. π/2
0
dx12 + 13cosx
=∫
2dz1 + z2
12 + 13▪ 1 - z2
1 + z2
= ∫ 2z12 + 12z
2 + 13 - 13z
2
= 2∫ dz25 - z
2 = 2∫ dx(5)
2 - (z)
2 z = 5sinθ, dz = 5cosθdθ
= 2∫ 5cosθdθ25 - 25sin
2θ
= 2∫ 5cosθdθ25(1 - sin
2θ)
= 25 ∫cosθdθ
cos2θ =
25 ∫ dθ
cosθ = 25 ∫ secθdθ
= 25 ln (secθ + tanθ)
= 25 ln
1
cosθ + sinθcosθ =
25 ln
1 + sinθ
cosθ
*going back to z = 5sinθ…i-draw mo sa right triangle
*so diba sinθ, which is hypopp so
magigingz5 . and cosθ
= 25 - z2
5
= 25 ln
1 +
z5
25 - z2
5
= 25 ln
5 + z
25 - z2
*square and get the square root of the fraction. Squinare and kinuha yung sqrt para parang walang damage na nangyari. It was as if you raised the fraction to the first power. Pero diba pag may exponent yung base ng ln, pwede mo siyang i-lagay and imultiply
with 25 .
= 25 ln
5 + z
25 - z2 ^
2 ▪ 12 *yung
12 i-move mo sa harap
= 25 ▪
12 ln
(5 + z)(5 + z)(5 + z)(5 - z)
*change the limits. To do that, substitute x sa z = tan12 x
= 15 ln
5 + z5 - z ]
1
0
= 15 ln
32
III. Partial Fractions
A. Linear, Distinct Factors
1. ∫(2x + 11)dxx
2 + x - 6 = ∫ (2x + 11)dx
(x + 3)(x - 2)
= ∫
A
x + 3 + B
x - 2 dx *multiply the whole equation
with the denominator of the original fraction
∫ (2x + 11)dx = ∫ ( )A(x - 2) + B(x + 3) dx
= ∫ A(x - 2)dx + ∫B(x + 3)dx
= Aln(x – 2) + Bln(x + 3) + lnc *”lnc” yung ginamit para lang maging mas pretty/simplified yung kalabasang equation later hihi
when x = 2: when x = -3: 2(2) + 11 = A(0) + B(2 + 3) 2(-3) + 11 = A(-3 – 2) + B(0) 15 = 5B 5 = -5A B = 3 A = -1
= ─ln(x + 3) + 3ln(x – 2)+ lnc = lnc(x - 2)
3
(x + 3)
*remember yung exponent pwede itanspose transpose. We’ll do it sa 3ln(x – 2) to magiging ln(x – 2)3
2. ∫ 3x2 + 8x - 12
x3 + 7x
2 + 12x = ∫ 3x
2 + 8x - 12
x(x + 4)(x + 3) dx
= ∫
A
x + B
x + 4 + C
x + 3 dx
= Alnx + Bln(x + 4) + Cln(x + 3) Equating Coefficients: 3x2 + 8x – 12 = A(x + 4)(x + 3) + Bx(x + 3) + Cx(x + 4) 3x2 + 8x – 12 = A(x2 + 7x + 12) + B(x2 + 3x) + C(x2 + 4x)
x2: 3 = A + B + C x: 8 = 7A + 3B + 4C c: -12 = 12A A = -1 *using system of equations: (3 = -1 + B + C) – 3 -12 = -3B – 3C 8 = -7 + 3B + 4C 15 = 3B + 4C 3 = C 3 = A + B + C B = 3 – (-1) – 3 B = 1
= -lnx + ln(x + 4) + 3ln(x + 3) + lnc = lnc(x + 4)(x + 3)
3
x
B. Linear, Repeated Factors
1. ∫ (x - 2)dxx
2(x + 3)
2 = ∫
A
x + Bx
2 + C
x + 3 + D
(x + 3)2 dx
= Alnx + Bx + Cln(x + 3) +
Dx + 3 + E
Equating Coefficients: x – 2 = Ax(x + 3)2 + B(x + 3)2 + Cx2(x + 3) + Dx2 x – 2 = A(x3 + 6x2 + 9x) + B(x2 + 6x + 9) + C(x3 + 3x2) + Dx2
x3: 0 = A + C x2: 0 = 6A + B + 3C + D x: 1 = 9A + 6B c: -2 = 9B
A = 7
27 , B = ─ 29 , C = ─
727 , D = ─
59
= 7
27 lnx ─ 29 x ─
727 ln(x + 3) ─
59(x + 3) + c
C. Quadratic, Distinct Factors
1. ∫ x + 2x
2 + 4x + 5 dx = ∫A(2x + 4) + B
x2 + 4x + 5 dx
*(2x + 4) came from the derivative of x2 + 4x + 5
= A∫ 2x + 4x
2 + 4x + 5 dx + B∫ dx
x2 + 4x + 5
*dun sa A, duu .
= Aln(x2 + 4x + 5) + B∫ dx
x2 + 4x + 4 + (5 - 4)
= Aln(x2 + 4x + 5) + B∫ dx
(x + 2)2 + (1)
2 du
u2 + a2
= Aln(x2 + 4x + 5) + B Tan
-1(x + 2) + c
Equating Coefficients: x + 2 = A(2x + 4) + B
x: 1 = 2A A = 12
c: 2 =4(12 ) + B B = 0
= 12 ln(x
2 + 4x + 5) + c
C. Quadratic, Distinct Factors
1. ∫ (x - 3)dxx
2(x
2 + 4x + 5)
2
= ∫
A
x + Bx
2 + C(2x + 4) + D
x2 + 4x + 5 +
E(2x + 4) + F(x
2 + 4x + 5)
2 dx
= A∫dxx + B∫x
-2dx + C∫ (2x + 4)dx
x2 + 4x + 5 + D∫ dx
x2 + 4x + 4 + (5 - 4) +
E∫ (2x + 4)dx(x
2 + 4x + 5)
2 + F∫ dx(x
2 + 4x + 5)
2
= Alnx + Bx + Cln(x
2 + 4x + 5) + DTan
-1(x + 2) +
Ex
2 + 4x + 5 + F∫ dx
[(x2 + 4x + 4) + (5 - 4)]
2
*Now let’s focus on F∫ dx[(x + 2)2 + (1)]2 .
u = x + 2 a = 1 u = asinθ x + 2 = tanθ dx = sec
2θdθ
tanθ = x + 2
1 *draw this
Bianca
= F∫ sec2θdθ
(tan2θ + 1)
2 = F∫sec2θdθ
(sec2θ)
2 = F∫sec2θdθ
sec4θ
= F∫ dθsec
2θ = F∫ cos
2θdθ =
F2 ∫ (1 + cos2θ)dθ
= F2 *θ +
12 sin2θ]
*recall that 12 sin2θ = sinθcosθ. According to the drawing of the
triangle, sinθ = x + 2
x2 + 4x + 5 and cos θ =
1
x2 + 4x + 5 . So
multiply them to get x + 2
x2 + 4x + 5 . So yun yung value ng 12 sin2θ.
*as for θ, recall that x + 2 = tanθ. So θ = Tan-1(x + 2). *so the final equation is:
= Alnx + Bx + Cln(x
2 + 4x + 5) + DTan
-1(x + 2) +
Ex
2 + 4x + 5 +
F2
Tan
-1(x + 2) +
x + 2x
2 + 4x + 5 + G
*just solve for the values of A, B, C, D, E and F and you’ll get the final answer
AREAS & CENTROIDS OF PLANE AREAS
A. Vertical Element
A = ∫ (ya – yb)dx
Ax‾ = ∫ x(ya – yb)dx
Ay‾ = ∫ (ya2 – yb
2)dx
B. Horizontal Element
A = ∫ (xR – xL)dy
Ay‾ = ∫ y(xR – xL)dy
Ax‾ = ∫ (xR2 – xL
2)dy
ANALYSIS OF POLAR CURVES
I. Symmetry
ox: F(r,θ) = F(r , -θ)
F(-r, π - θ)
oxy: F(r,θ) = F(r , π - θ)
F(-r , - θ)
ox: F(r,θ) = F(-r , θ)
F(r, π + θ)
II. Intersection w/ the pole
set r = 0 and solve for θi
III. Intersection with axes
IV. Critical Points
set drdθ = 0 and solve for θC
V. Divisions
use θi & θC
VI. Additional Points
SOME COMMON POLAR POLES
A. Limacons : r = a ± bsinθ or r = a ± bcosθ
0 < | ab | < 1 with a loop
0 < | ab | = 1 cardioid
1 < | ab | < 2 with a dent
| ab | ≥ 2 convex
B. Rose Curves
r = asin(nθ) r = acos(nθ)
θ 0° 90° 180° 270° 360°
r
1. r = 2 – 2sinθ (cardioid) Intersection with the pole:
r = 2 – 2sinθ 0 = 2 – 2sinθ 2sinθ = 2 θ = sin
-1(1)
θ = π2
Intersection with the axes:
Critical Points:
drdθ = -2cosθ
0 = -2cosθ
θ = cos-1
(0)
θ = π2 ,
3π2
Plot the points on the graph and draw the heart:
1. r = 2 when the angle’s nasa 0°. So plot the first point sa (2,0)
2. r = 0 at 90°. As you plot it, gawa ka na ng curve
3. r = 2 at 180°. Ilagay mo yung point 2 units TOWARDS 180°. Kumbaga sa side ng 180° yung “2”.
4. r = 4 at 270°. Like 180°, plot “4” sa side ng 270.
5. r = 2 at 360° or back to 0° ulit Then TA-DAAA!
Compute for the Area:
A = 12 ∫ r
2dθ
*isang half lang yung area na kukunin mo. Eh since sobrang carbon copy yung other half of the heart, multiply the Area by 2.
A = 12 ▪ 2
π/2
-π/2 (2 – 2sinθ)
2dθ
A = ∫ (4 + 8sinθ + 4sin2θ)dθ
A = 4∫ (1+ 2sinθ + sin2θ)dθ
A = 4
θ ─ 2cosθ +
12
θ -
12 sin2θ
A = 4
3
2 θ - 2cosθ - 14 sin2θ ]
π/2
-π/2
= 6π
2. r = 3sin2θ Intersection with the pole:
r = 3sin2θ 0 = 3sin2θ 0 = sin2θ 2θ = sin
-1(0)
2θ = 0, π, 2π, 3π, etc.
θ = 0, π2 , π,
3π2 , etc.
Critical Points: drdθ = 6sin2θ = 0
0 = sin2θ 2θ = sin
-1(0)
2θ = π2 ,
3π2 ,
5π2 ,
7π2
θ = π4 ,
3π4 ,
5π4 ,
7π4
Intersection with axes & Critical Points: θ 0° 45° 90° 135° 180° 225° 270° 315° 360°
r 0 3 0 -3 0 3 0 -3 0
θ 0° 90° 180° 270° 360°
r 2 0 2 4 2
VOLUMES & CENTROIDS OF PLANE AREAS
A. Method of Circular Disk
V = π b
ar
2dh
Vx‾ = ∫ XCdv
Vy‾ = ∫ YCdv
CONDITIONS:
1. element must be parallel to the axis
2. r must be parallel to the axis
3. the axis should be a boundary
1. Find V, x‾ , y‾ of the solid generated by rendering the region bounded by y = 1 – x
3, ox and oy about ox.
V = π∫ r2dx
V = π∫ (ya – yb)2 dx
V = π∫ (1 – x3
– 0)2 dx
V = π 0
1(1 ─ 2x
3 + x
6)d
V = π (x - x
2
2 + x
7
7 ) ] 1
0
= 9π14
B. Method of Circular Ring
V = π b
a(R
2 – r
2)dh
1. Find V if the axes bounded by y
2 = x
3 and y = 2x is
revolved about ox.
*substitute y = 2x sa y2 = x3 to get the points of intersection
(2x)2 = x
3 4x
2 – x
3 = 0
x2(4 – x) = 0
x2 = 0 4 – x = 0
x = 0, then y = 0 x = 4, then y = 8
V = π 4
0(R
2 – r
2)dh
V = π∫ [(2x)2 – (x
3/2)
2]dx
V = π∫ (4x2 – x
3)dx
V = π
4x
3
3 - x
4
4 ] 4
0
= 64π
3
2. Find the volume of y = x2 and y
2 = x about x + 1 = 0.
Points of Intersection:
(x2)2 = x
x4 = x
x4 – x = 0
x(x3 – 1) = 0
x = 0, y = 0 and x = 1, y = 1
XRIGHT: y = x2 x = y XLEFT: y
2 = x x = y
2
R = y + 1 (kasi about x = -1) r = y2 + 1
V = π 1
0(R
2 – r
2)dh
V = π∫ ( )( y + 1)2 – (y
2 + 1)
2 dx
V = π∫ (y + 2y1/2
+ 1 – y4 – 2y
2 – 1)dx
V = π
y
2
2 + 4y
3/2
3 - y
5
5 - 2y
3
3 - y ] 1
0
= 29π30
C. Method of Cylindrical Shell
V = 2π b
axydx
(when using a vertical element)
V = 2π b
axydy
(when using a horizontal element)
1. Find V if the area bounded by y = 1 – x3, ox and oy is
revolved about ox.
y = 1 – x3 x
3 = 1 – y x =
31 - y
V = 2π 1
0xydy V = 2π∫ (xL – xR)ydy
V = 2π∫ (1 - y)1/3
ydy
*use algebraic substitution & change limits
u = (1 – y)1/3 y = 1 – u3
u3 = 1 – y dy = 3u2du
*change the limits. To do that, substitute y sa u = (1 – y)1/3
V = 2π 0
1 u(1 – u
3)
▪ 3u
2du
*you may interchange the limits by turning the equation to negative
V = ─6π 1
0 u
3(1 – u
3)
du
V = ─6π∫ (u3 – u
6)
du
V = ─6π
u
4
4 – u
7
7 ]
1
0
= 9π14
LENGTH OF ARC
S = ∫ 1 + ( )dydx
2 y in terms of x
S = ∫ 1 + ( )dxdy
2 x in terms of y
S = ∫ ( )dxdt
2 + ( )dy
dt 2 parametric
S = ∫ r2 + ( )dr
dθ 2 r in terms of θ
1. Find the length of the curve y = lncosx from x = 0 to x = π4 .
y = lncosx
dydx = ─
sinxcosx
S = ∫ 1 + ( )─sinxcosx
2 dx = ∫ 1 +tan
2x dx
S = ∫ sec2x dx = = ∫ secxdx
S = ln(secx + tanx) ] π/4
0
= ln(1 + 2 )
2. Find the length of the curve x = 2(2t + 3)3/2
, y = 3(t + 1)2
from t = 0 to t = 1.
dxdt = 2
3
2 (2t + 3)1/2
(2) = 6 2t + 3
dydt = 6(t + 1)
S = ∫ (6 2t + 3 )2 + (6(t + 1))
2 dt
S = ∫ 36(2t + 3) + 36(t + 1)2 dt
S = 6∫ 2t + 3 + t2 + 2t + 1 dt
S = 6∫ (t + 1)2 dt
S = 6∫ (t + 1)dt
S = 3(t + 1)2 ]
1
0
= 15
3. Find the total arc length of the cardioid r = a(1 + cosθ).
drdθ = a(─sinθ)
S = 2 ∫ (a(1 + cosθ))2 + (─asinθ)
2
*minultply sa 2 kasi 2 parts yung cardioid
S =2 ∫ a2(1 + cosθ)
2 + a
2sin
2θ
S = 2a ∫ 1 + 2cosθ + cos2θ + sin
2θ dθ
*recall cos2θ + sin
2θ = 1
S = 2a∫ 2 + 2cosθ dθ
S = 2 2 a∫ 1 + cosθ dθ
*let’s focus on 1 + cosθ. Recall that:
cos2θ =
12 (1 + cos2θ)
2cos2θ = 1 + cos2θ
2cos2θ2 = 1 + cosθ
S = 2 2 a∫ 2cos2θ2 dθ
S = 2 2 ▪ 2 a∫cosθ2 dθ
S = 4a(2)sinθ2 ]
π
0
= 8aError! Bookmark not defined.
AREA OF A SURFACE OF REVOLUTION
SA = 2π∫ydS about ox
SA = 2π∫xdS about oy
1. Find the area of the surface of revolution generated by revolving y = x
3 between x = 0 and x = a about ox.
SA = 2π∫ydS
SA = 2π∫x3
1 + ( )dydx
2 dx
SA = 2π∫x3
1 + 9x4 dx u
ndu
u = 1 + 9x4 du = 36x
3du
SA = 2π ▪ 1
36 ▪ 23 (1 + 9x
4)
3/2] a
0
SA = π27 [ (1 + 9x)
3/2 + 1]
SECOND THEOREM OF PAPPUS V = 2πAd
where: A = area d = perpendicular distance of the centroid from the axis of revolution
1. Find the volume if area bounded by y2 = x and y = x
2 is
revolved about x = -1.
A = ∫(ya – yb)dx
A = ∫(x1/2
– x2)dx
A = 23 x
3/2 ─
x3
3 ] 1
0
= 13
Ax‾ = ∫ x(ya – yb)dx
Ax‾ = ∫ x(x1/2
– x2)dx
Ax‾ = ∫ (x3/2
– x3)dx
Ax‾ = 25 x
5/2 –
x4
4 ] 1
0
13 x‾ =
320
x‾ = 9
20
V = 2πAd = 2π
1
3
9
20 + 1 = 29π30
* may + 1 kasi yung yung axis nasa x = -1.