INTEGRAL CALCULUS FINALS REVIEWER (2 nd 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 = dx x v = x *substitute these values sa formula na ∫udv = uv – ∫vdu ∫lnxdx = xlnx ─ ∫x ▪ dx x = xlnx – x + c 2. ∫x 2 lnxdx u = lnx dv = x 2 dx du = dx x v = x 3 3 ∫ x 2 lnx dx = 1 3 x 3 lnx ─ 1 3 ∫ x 3 ▪ dx x = 1 3 x 3 lnx ─ 1 3 ∫ x 2 dx u = x 2 dv = dx du = 2xdx v = x = 1 3 x 3 lnx ─ 1 3 [x 3 – 2∫ x 2 dx] = 1 3 x 3 lnx ─ 1 3 x 3 + 2 3 ▪ x 3 3 = 1 3 x 3 lnx ─ 1 3 x 3 + 2x 3 9 + c = 1 3 x 3 lnx ─ 1 + 2 3 + c = 1 3 x 3 lnx ─ 1 3 + c 3. ∫x 3 e x dx u = x 3 dv = e x dx du = 3x 2 dx v = e x ∫ x 3 e x dx = x 3 e x ─ 3 ∫ x 2 e x dx u = x 2 dv = e x dx du = 2xdx v = e x = x 3 e x ─ 3 * x 2 e x – 2∫ xe x dx ] u = x dv = e x dx du = dx v = e x = x 3 e x ─ 3x 2 e x + 6[ xe x – ∫e x dx ] = x 3 e x ─ 3x 2 e x + 6xe x – 6e x + c 4. ∫sec 3 xdx = ∫sec 2 xsecxdx u = secx dv = sec 2 xdx du = secxtanxdx v = tanx = secxtanx ─ ∫secxtan 2 xdx = secxtanx ─ ∫secx(sec 2 x – 1)dx = secxtanx ─ ∫(sec 3 x – secx)dx = secxtanx ─ ∫sec 3 xdx + ∫secx *transpose ∫sec 3 xdx kasi same siya nung sa other side 2 ∫sec 3 xdx = secxtanx + ∫secx ∫sec 3 xdx = 1 2 ( ) secxtanx + ln(secx + tanx) + c 5. ∫e x cos2xdx u = e x dv = cos2xdx du = e x dx v = 1 2 sin2x = 1 2 e x sin2x – 1 2 ∫e x sin2xdx u = e x dv = sin2xdx du = e x dx v = ─ 1 2 cos2x = 1 2 e x sin2x – 1 2 ─ 1 2 e x cos2x ─ ─ 1 2 ∫e x cos2xdx = 1 2 e x sin2x + 1 4 e x cos2x ─ 1 4 ∫e x cos2xdx *transpose ∫e x sin2xdx kasi same siya nung sa other side ∫e x cos2xdx + 1 4 ∫e x cos2xdx = 1 2 e x sin2x + 1 4 e x cos2x 5 4 ∫e x cos2xdx = 1 2 e x sin2x + 1 2 cos2x ∫e x cos2xdx = 2 5 e x sin2x + 1 2 cos2x + c 6. ∫(x + sinx) 2 dx = ∫(x 2 + 2xsinx + sin 2 x)dx = x 3 3 + 2∫xsinxdx + ∫sin 2 xdx = x 3 3 + 2∫xsinxdx + 1 2 ∫(1 – cos2x)dx = x 3 3 + 1 2 x ─ 1 2 sin2x + 2∫xsinxdx = x 3 3 + 1 2 x ─ 1 4 sin2x+ 2∫xsinxdx u = x dv = sinxdx du = dx v = ─ cosx = x 3 3 + 1 2 x ─ 1 4 sin2x+ 2 *─xcosx +∫cosxdx] = x 3 3 + 1 2 x ─ 1 4 sin2x+ 2 *─xcosx + sinx + = x 3 3 + 1 2 x ─ 1 4 sin2x ─ 2xcosx + 2sinx + c
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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