nd
INTEGRAL CALCULUS FINALS REVIEWER (2 Sem ‘11-‘12)
4. ∫sec xdx 3
= ∫sec xsecxdx
INTEGRATION TECHNIQUES
2
I. Integration by Parts
dv = sec2xdx v = tanx
u = secx du = secxtanxdx
∫udv = uv – ∫vdu
= secxtanx ─ ∫secxtan xdx 2
1. ∫lnxdx
= secxtanx ─ ∫secx(sec x – 1)dx 2
*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 dx du = x v=x *substitute these values sa formula na ∫udv = uv – ∫vdu
∫lnxdx = xlnx ─ ∫x ▪
dx x
= secxtanx ─ ∫(sec x – secx)dx 3
= secxtanx ─ ∫sec xdx + ∫secx 3
*transpose ∫sec3xdx kasi same siya nung sa other side
2 ∫sec xdx = secxtanx + ∫secx 1 ∫sec3xdx = 2 (secxtanx + ln(secx + tanx)) + c 3
5. ∫e cos2xdx x
u = ex
= xlnx – x + c
du = e dx
2. ∫x lnxdx 2
2
u = lnx dx du = x
dv = x dx x3 v= 3
1 1 dx ∫ x2lnx dx = 3 x3lnx ─ 3 ∫ x3 ▪ x 1 3 1 2 = 3 x lnx ─ 3 ∫ x dx u = x2 du = 2xdx
*transpose ∫exsin2xdx kasi same siya nung sa other side
1
1
5 x 1 x 1 4 ∫e cos2xdx = 2 e sin2x + 2 cos2x 2 1 ∫excos2xdx = 5 ex sin2x + 2 cos2x + c 6. ∫(x + sinx) dx 2
= ∫(x + 2xsinx + sin x)dx 3 x 2 = 3 + 2∫xsinxdx + ∫sin xdx 3 x 1 = 3 + 2∫xsinxdx + 2 ∫(1 – cos2x)dx 3 x 1 1 = 3 + 2 x ─ 2 sin2x + 2∫xsinxdx 3 x 1 1 = 3 + 2 x ─ 4 sin2x+ 2∫xsinxdx 2
3. ∫x e dx 3 x
u = x3 du = 3x2dx
dv = exdx v = ex
∫ x3exdx = x3ex ─ 3 ∫ x2exdx u = x2 du = 2xdx
dv = exdx v = ex
= x e ─ 3 * x e – 2∫ xe dx ] x
dv = exdx v = ex
u=x du = dx
= x e ─ 3x e + 6[ xe – ∫e dx ] 3 x 2 x x x = x e ─ 3x e + 6xe – 6e + c x
1
∫excos2xdx + 4 ∫excos2xdx = 2 exsin2x + 4 excos2x
3
2 x
dv = sin2xdx 1 v = ─ 2 cos2x
1 x 1 1 x 1 x = 2 e sin2x – 2 ─2 e cos2x ─ ─ 2 ∫e cos2xdx 1 x 1 x 1 x = 2 e sin2x + 4 e cos2x ─ 4 ∫e cos2xdx
1 3 1 3 2 x = 3 x lnx ─ 3 x + 3 ▪ 3 3 1 3 1 3 2x = 3 x lnx ─ 3 x + 9 + c 1 3 2 = 3 x lnx ─ 1 + 3 + c 1 3 1 = 3 x lnx ─ 3 + c
3 x
u = ex du = e dx
dv = dx v=x
2 x
1 x 1 x = 2 e sin2x – 2 ∫e sin2xdx x
1 3 1 3 2 = 3 x lnx ─ 3 [x – 2∫ x dx]
3 x
dv = cos2xdx 1 v = 2 sin2x
x
x
3
2
u=x du = dx
dv = sinxdx v = ─ cosx
x 1 1 = 3 + 2 x ─ 4 sin2x+ 2 *─xcosx +∫cosxdx] 3 x 1 1 = 3 + 2 x ─ 4 sin2x+ 2 *─xcosx + sinx + 3 x 1 1 = 3 + 2 x ─ 4 sin2x ─ 2xcosx + 2sinx + c
7.
∫
3
t dt 2 1 + 3t 2
=
∫ t1▪t+ dt3t
2
u = t2
dv = (1 + 3t2)-1/2tdt 1 1 v = 6 ▪ 2(1 + 3t2)1/2= 3 (1 + 3t2)1/2
du = 2tdt
1 2 2 2 2 = 3 t 1 + 3t ─ 3 ∫ 1 + 3t tdt
n
use u du
3 3 2 2 4/3 2 3 = ─8 (1 – x ) x + 7 ─ 7 x 3 4 3 2 4/3 2 = ─8 (1 – x ) 7 x + 7 3 2 4/3 2 = ─ 56 (1 – x ) (4x + 3) + c -11 11. ∫Sec x dx
1 u = Sec-1x
u = 1 + 3t2 du = 6tdt
1 2 2 1 2 2 2 3/2 = 3 t 1 + 3t ─ 3 ▪ 6 ▪ 3 (1 + 3t ) 1 2 2 2 2 3/2 = 3 t 1 + 3t ─ 27 (1 + 3t ) + c
du = ─
=─
8. ∫xCsc xdx -1
u = Csc-1x dx du = ─ x x2 ─ 1
dv = xdx x2 v= 2
∫
xdx 2 x ─1
1 2 -1 1 2 -1/2 = 2 x Csc x + 2 ∫ (x – 1) xdx use undu 1 2 -1 1 2 1/2 = 2 x Csc x + 2 (x – 1) + c 9. ∫sin x dx y=
x y2 = x 2ydy = dx
-11 = xSec x +
1 x
1 ─ x2 x2 dx x2
12.
3
= xln x ─ 3
= xln x ─ 3 ∫ln xdx 3
2
u = ln2x 2lnxdx du = x
= ─2 x cos x + 2sin x + c 10. ∫x (1 – x ) dx ∫x x(1 – x ) dx 2
u=x
du = 2xdx
v=x
∫x ▪ ln xxdx
x back sa mga y 2
dv = dx
2
dv = sinydy v = ─cosy
= 2 * ─ycosy + ∫cosydy ] = ─2ycosy + 2siny + c
2 1/3
use undu
∫ln3xdx u = ln3x 3ln2xdx du = x
u=y du = dy
v=x
dx x2
xdx 2 1─x 1 2 -11 2 1/2 = xSec x + 2 ▪ 1 (1 – x ) + c -11 2 = xSec x + 1 – x + c
= 2∫ysinydy
3
1 x2 ─ 1
∫
=∫ siny ▪ 2ydy
*substitute
1 x
=─1 1 2 x▪x 1─x dx =─ 1 ─ x2
1 2 -1 1 2 dx = 2 x Csc x + 2 ∫ x ▪ 2 x x ─1 1 2 -1 1 = 2 x Csc x + 2
dv = dx dx x2
2 1/3
= xln x ─ 3 [xln x ─ 2 ∫ x ▪ 3
2 1/3
dv = (1 – x ) xdx 1 3 v = ─2 ▪ 4 (1 – x2)4/3
3 2 2 4/3 3 2 4/3 = ─8 x (1 – x ) + 8 ∫ (1 – x ) xdx 3 2 2 4/3 3 1 3 2 7/3 = ─8 x (1 – x ) + 4 ▪ 2 ▪ 7 (1 – x ) 3 2 9 2 4/3 2 7/3 = ─8 x (1 – x ) + 56 (1 – x ) 3 2 4/3 2 3 2 = ─8 (1 – x ) x + 7 (1 ─ x )
2
dv = dx v=x
lnxdx x
= xln x ─ 3xln x + 6 ∫lnxdx 3
2
u = lnx dx du = x
use undu
dv = dx v=x
dx 3 2 = xln x ─ 3xln x + 6 *xlnx ─ ∫ x ▪ x ] = xln x ─ 3xln x + 6 *xlnx ─ ∫ dx] 3 2 = xln x ─ 3xln x + 6xlnx ─ 6x + c 3
2
13.
π 2
0
5
c. sinxcosx recall the identity sin2x = 2sinxcosx. Just transpose 2 to the other 1 side. So you’ll get 2 sin2x = sinxcosx
4
sin xdx sin xsinxdx u = sin4x du = 4sin3xcosxdx
dv = sinxdx v = ─cosxdx
= ─sin xcosx + 4∫ sin xcos xdx 4
3
=
2
sin 2x dx 5 sin 2x 2 6 8 sin 2xcos 2x
= ─sin xcosx + 4∫ sin x(1 – sin x)dx 4
3
2
= ─sin xcosx + 4∫ (sin x – sin x) dx 4
3
5
use undu
= ─sin xcosx + 4∫sin xdx – 4∫sin xdx 4
3
5
7
∫1
7
=8
∫sinsin2x2xdx 3
6
*transpose ∫sin xdx kasi same siya nung sa other side
cos 2x
5
4∫sin xdx +∫sin xdx = ─sin xcosx + 4∫sin xdx 5
5
4
3
5∫sin xdx = ─sin xcosx + 4∫sin xdx 5
4
= 8 ∫sin 2xcos 2xdx 4
dy *represent 2x as y. so y = 2x. And dy = 2dx. So dx = 2
3
u = sin2x du = 2sinxcosxdx
dv = sinxdx v = ─cosxdx
6
1 4 6 = 8 ▪ 2 ∫sin ycos ydy *change the limits. To do that, substitute x sa y = 2x. x
1
∫sin5xdx = 5 (─sin4xcosx + 4∫sin3xdx)
0 0
y π
4 6 2 cos x cos x = ─sin xcosx + 4─ 4 + 6 ] 0 4
ADDITIONAL FORMULA: WALLIS’ FORMULA π *only works when the upper and lower limits are 2 and 0. π 2
0
m
n
sin xcos xdx =
[(m-1)(m-3)…2 or 1+▪*(n-1)(n-3)…2 or 1+ •α (m+n)(m+n-2)(m+n-4)…2 or 1
π where: α = 2 , if both m and n are EVEN
*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.
0
=
2
7/2
(1 ─ cos 2x) dx 4 2 tan 2xcsc 4xsinxcosx
∫ sin 2x
2
π 2
0
4
6
sin ycos ydy
*use Wallis’ formula [(4-1)(4-3)][(6-1)(6-3)(6-5)] π = 4 ▪ (6+4)(6+4-2)(10-4)(10-6)(10-8) ▪ 2 (3▪1)(5▪3▪1) π = 4 ▪ 10▪8▪6▪4▪2 ▪ 2 3π = 27 II. Substitution Methods
= 1, if other wise
π 4
=4
π 4 π 2
7/2
(sin 2x) dx 1 1 4 2 2 cos 2x ▪4sin 2xcos 2x ▪ 2 sin2x 4
*okay so isa-isahin natin yung mga chuchu sa denominator: a. tan42x sinx sin42x recall sa identities na tanx = cosx . Kaya naging tan42x= cos42x yay 1 b. csc24x sin24x 1 recall the trigonometric transformation formula sinxcosx = 2 sin2x. So ang main agenda mo is to get sin24x 1 sinxcosx = 2 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
1 sin2xcos2x = 2 sin4x *square both sides
1 sin22xcos22x = 4 sin24x
1 *transpose 4
4sin22xcos22x = sin24x
tadaaaaa yay you
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)u du 1/2
= ∫(u – u )du 2 5/2 2 3/2 =5u –3u +c 5/2 3/2 6u – 10u = +c 15 2 3/2 = 15 u (3u – 5) + c 2 3/2 = 15 (1 + x) [3(1 + x) – 5] + c 2 3/2 = 15 (1 + x) (3x – 2) + c 3/2
1/2
∫
3
2
x dx x xdx 2. (x2 + a2)3 (x2 + a2)3 u = x2 + a2 x2 = u – a2 2xdx = du *substitute all x’s with u’s
1 =2
B. Algebraic Substitution 1. ∫x 1 + x dx
2
u= 1+x u2 = 1 + x x = u2 – 1 dx = 2udu
3
= 2∫(u – 1) ▪ u ▪ udu
∫(u - ua )du
2
= 2∫u (u – 1)du 2
1 -2 2 -3 = 2 ∫(u – a u )du -1
= 2∫ (u – u )du 5 3 u u = 2 5 ─ 3 + c 5 3 3u ─ 5u +c = 2 15 2 3 2 = 15 u (3u – 5) + c 2 3/2 = 15 (1 + x) [3(1 + x) – 5] + c 2 2/3 = 15 (1 + x) (3x – 2) + c 4
2 -2
1 u au = 2 -1 ─ -2 + c 2 1 1 a = 2 ─ u + 2u2 + c 2 1 -2u + a = 2 2u2 +c *substitute the value of u back to x2 + a2 2
2
2
1 -2(x + a ) + a = 2 2(x2 + a2)2 + c 1 2 2 2 = 4(x2 + a2)2 [a – 2(x + a )] + c 1 2 2 = ─ 4(x2 + a2)2 (2x + a ) + c
2.
y +3 x xdx 3. (3 - 2y)2/3 dy (x2 + a2)3 u = 3 – 2y 2y = 3 – u 2dy = ─du
3-u+6 2 2/3 du u
∫
1 =─2 1 -2/3 = ─ 4 ∫(9 – u)u du 1 -2/3 1/3 = ─ 4 ∫(9u – u )du 1/3 4/3 1 3 ▪ 9u 3u =─4 1 ─ 4 +c 3 1/3 = ─16 u (36 – u) + c 3 1/3 = ─16 (3 – 2y) (2y + 33) + c
2
dy 1/3 y - y dy
3 3 2 2
u = y1/3 u3 = y 3u2du = dy
2
∫
2
2
∫ u du = 3∫u(u - 1) udu = 3∫ u - 1 u du = 3 u3 - u 2
2
2
2
= 3ln(u – 1) 3 2/3 = 2 ln(y – 1)
]
3 3 2 2
3 3 = 2 (ln2 – ln1) = 2 ln2 7 dx 3. 0 1+3 x+1 u=3 x+1 u3 = x + 1 3u2du = dx
∫
2
u du =3 1+u
*divide u2 by 1 + u
1 = 3∫u - 1 + u + 1 du
**change the limits. To do that, substitute x sa u = 3 x + 1 x 0 7 1 2 y 2
u = 3 2 + u + ln(u + 1) 1 3 = 32 + ln2
]
1 2
4.
2x
x
x
e dx e ▪ e dx x x 1+e 1+e
ln2 0
=─
u = 1 + ex u2 = 1 + ex ex = u2 – 1 2udu = exdx
∫
ydy 2 -1/2 ydy 2 ─ ∫(1 – y ) 1-y
1 **change the limits. To do that, substitute x sa x = y x 5/3 5/4 3/5 4/5 u
2
∫(u - u1)udu
=2
1 2 1/2 = 2 ▪ 1 (1 – y)
]
= 2∫(u – 1)du *change the limits. To do that, substitute x sa u = 1 + ex 2
x
0
ln2
2
u 3
3
u = 2 3 - u
]
3.
2
1 2 2 = 33 (3 3 - 2 2 ) - ( 3 - 2 ) = 3
2
2
u = 5y du = 5dy
*use this when you see equations like this:
1 ─5
dx 1 dy and substitute x = y & dx = ─ y2 x ax + bx + c
=─
∫1
dx 2 2x - x dy 2 y
∫1
=─
y
2
2
u = asecθ
2
1 2 y -1
2
2
x = 2asin θ
2
x = 2atan θ
x - 2ax
x = 2asec θ
2
u = asecθ du = atanθsecθdθ
∫ a tanθsecθdθ = a ∫ sec θ - 1 1 tanθsecθdθ = a ∫ tan θ 1 secθdθ = a ∫ tanθ tanθsecθdθ = a a2sec2θ - a2 2
2
dx 2 x x -1 dy 2 y 2
u = atanθ
∫u du- a
2
2
∫1
2
2
2 x -1+c
5/4
2
2ax + x
= ─ ∫(2y – 1) dy 1 2 1/2 = ─ 2 ▪ 1 (2y – 1) + c
5/3
u =asinθ
2ax - x
1.
Substitute these:
2
u –a
-1/2
2.
+c
2
a +u
dy 2y - 1
=─
2
25 - x x
a –u
∫ ∫
du u2 - a2
If you see this combination:
2y - 1 2 y y dy 2 y =─ 1 2 y 2y -1 =─
a=1
D. Trigonometric Substitution
2 1 2 y -y dy 2 y
y =─
1 dy *substitute x = y & dx = ─ y2
∫
1 5+ = ─ 5 ln
2
∫x
16 1 1 - 25 = 5
∫x 2xdx- x dy = ─∫ 25y - 1
C. Reciprocal Substitution
1.
4/5
9 1 - 25 ─
=
3
3/5
1 = a ∫cscθdθ 1 = a ln |cscθ – cotθ| + c
2
2
*going back to u = asecθ…i-draw mo sa right triangle hyp *so diba cscθ, which is opp so u
magiging
1 = a ln
. and cotθ
u2 - a2
u a 2 2 – 2 2 u -a u -a u-a 2 2 +c u -a
*use this when you see trigo functions
a
=
1 = a ln
u2 - a2
E. Half-Angle Substitution
+c
1 z = tan2 (nx)
1 2dz dx = n ▪ 1 + z2
2z tan(nx) = 1 - z2
2z sin(nx) = 1 + z2
2
1-z cos(nx) = 1 + z2
*i-square yung fraction para mawala yung square root at may ma-cancel hihi
1.
1 (u - a)(u - a) = a ln (u - a)(u + a) + c
∫1 + sinxdx+ cosx =
1 u-a = a ln u + a + c 2.
2 0
x
2
=∫ x
2
2
u = asinθ x = 2sinθ dx = 2cosθdθ
= ∫ (2sinθ)
2
2
4 - (2sinθ) ▪ 2cosθdθ
= 2▪4 ∫ sin θ 4 - 4sin θ cosθdθ 2
2
= 8 ∫ sin θ ▪ 2 2
2.
2
1 - sin θ cosθdθ
= 16 ∫ sin θ cos θ cosθdθ 2
2
2
0 0
u π/2
= 16
0
2
x
2
2
=
2
2
2
2
2
2
=
4
4a sin θ - 4a sin θ sinθcosθdθ
∫ sin4θ 4a2sin2θ(1 - sin2θ) sinθcosθdθ 3 4 2 = 16a ∫ sin θ ▪ 2asinθ cos θ sinθcosθdθ 4 6 2 = 32a ∫ sin θcos θdθ = 16a
3
2asin2θ
π/2
1
6
Tan Tan
z 2
]
2
= 32a sin θcos θdθ *use wallis’ 0 4 4 32a (5 x 3 x 1)(1) π 5πa = 8x6x4x2 ▪2 = 8
Tan 2 2
3.
0
∫
=
π tan2 -1 2
du a2 + u 2
2
π/2 0
1 tan2 (nx) -1
1
π/2
*change the limits. To do that, substitute x sa x = 4
2 2
-1
]
= ∫ (2asin θ) 2a(2asin θ) - (2asin θ) ▪ 4asinθcosθdθ
∫ sin4θ
2
2 2 2 1 tanx -1 = Tan 2 2 2
2
2ax - x dx
2
2
1
=
x = 2asin θ dx = 4asinθcosθdθ
= 4a▪4a
1 2dz 2 ▪ 1 + z2 1 - z2 3 + 1 + z2
2
sin θcos θdθ
2
n=2
∫3 + 3z2dz+ 1 - z 1 2dz 1 dz = 2 ∫4 + 2z = 2 ∫ 2 + z
2
2
= ln (1 + z) + c 1 = ln (1 + tan2 x) + c π/2 dx 3 + cos2x 0
1 =2
2 π 2
*use Wallis’ Formula (1)(1) π = 16 ▪ (4)(2) ▪ 2 = π 2a
dz ∫22dz + 2z = ∫1 + z
∫
2
x
0
=
=
= 16 ∫ sin θcos θdθ *change the limits. To do that, substitute x sa x = 2sinθ
3.
∫
2
(2) - (x) dx
2dz 1 + z2 2z 1 - z2 1 +1 + z2 + 1 + z2 2dz 1 + z2 1 + z2 + 2z + 1 - z2 1 + z2
=
4 - x dx
2
n=1
]
π/2 0
π/2 0
- Tan
-1tan0
2
= π 4 2
dx 12 + 13cosx
2dz 1 + z2 1 - z2 12 + 13▪ 1 + z2
∫12 + 12z 2z+ 13 - 13z dz dx = 2∫25 - z = 2∫(5) - (z) =
2
2
2
2
2
z = 5sinθ, dz = 5cosθdθ
2
∫ 5cosθdθ = 2∫25(1 - sin θ) 2 cosθdθ 2 dθ 2 = 5 ∫ cos θ = 5 ∫cosθ = 5 ∫ secθdθ 5cosθdθ = 2 25 - 25sin2θ
2.
2
2 = 5 ln (secθ + tanθ) 2 1 sinθ 2 1 + sinθ = 5 ln cosθ + cosθ = 5 ln cosθ *going back to z = 5sinθ…i-draw mo sa right triangle hyp *so diba sinθ, which is opp so z magiging5 . and cosθ 25 - z2 = 5
2 = 5 ln
25 - z 5
2
2 = 5 ln
25 - z
5+z 2 25 - z
1 *yung 2 i-move mo sa harap
1 *change the limits. To do that, substitute x sa z = tan2 x
]
1 0
1 3 = 5 ln 2
= -lnx + ln(x + 4) + 3ln(x + 3) + lnc = ln
c(x + 4)(x + 3) x
C D A B ∫x(x(x- 2)dx + 3) = ∫ x + x + x + 3 + (x + 3) dx 2
2
2
2
B D = Alnx + x + Cln(x + 3) + x + 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 7 2 7 5 A = 27 , B = ─ 9 , C = ─ 27 , D = ─9
7 2 7 5 = 27 lnx ─ 9 x ─ 27 ln(x + 3) ─ 9(x + 3) + c C. Quadratic, Distinct Factors 1.
+ 4) + B ∫x +x 4x+ 2+ 5 dx = ∫A(2x x + 4x + 5 dx 2
2
*(2x + 4) came from the derivative of x2 + 4x + 5
III. Partial Fractions
∫
∫
2x + 4 dx = A x2 + 4x + 5 dx + B x2 + 4x + 5 du *dun sa A, u .
A. Linear, Distinct Factors 1.
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
1.
2 1 (5 + z)(5 + z) = 5 ▪ 2 ln (5 + z)(5 - z) 1 5+z = 5 ln 5 - z
2
B. Linear, Repeated Factors
*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 2 with 5 .
2 5+z ^ 2 ▪ 12 = 5 ln 2
3
= Alnx + Bln(x + 4) + Cln(x + 3)
2
z 1+5
2
12 3x + 8x - 12 ∫x3x+ 7x+ 8x+-12x = ∫x(x + 4)(x + 3) dx A B C = ∫ x + x + 4 + x + 3 dx
(2x + 11)dx ∫(2xx ++ 11)dx x - 6 = ∫(x + 3)(x - 2) A B = ∫x + 3 + x - 2 dx 2
*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
∫ dx = Aln(x + 4x + 5) + B∫(x + 2) + (1)
dx 2 = Aln(x + 4x + 5) + B x2 + 4x + 4 + (5 - 4) 2 2
2
-1
2
= Aln(x + 4x + 5) + B Tan (x + 2) + c
= 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 3
c(x - 2) = ─ln(x + 3) + 3ln(x – 2)+ lnc = ln (x + 3)
*remember yung exponent pwede itanspose transpose. We’ll do it sa 3ln(x – 2) to magiging ln(x – 2)3
Equating Coefficients: x + 2 = A(2x + 4) + B 1 x: 1 = 2A A = 2 1 c: 2 =4(2 ) + B B = 0
1 2 = 2 ln(x + 4x + 5) + c
du
u2 + a2
3
C. Quadratic, Distinct Factors 1.
∫x (x(x+- 3)dx 4x + 5) A B C(2x + 4) + D E(2x + 4) + F = ∫ x + x + x + 4x + 5 + (x + 4x + 5) dx 2
2
∫
2
2
∫
∫
∫
(2x + 4)dx dx E (x2 + 4x + 5)2 + F (x2 + 4x + 5)2
∫ y(xR – xL)dy
Ax‾ =
∫ (xR2 – xL2)dy
ANALYSIS OF POLAR CURVES
B 2 -1 = Alnx + x + Cln(x + 4x + 5) + DTan (x + 2) +
I. Symmetry ox: F(r,θ) =
∫
E dx 2 2 2 x + 4x + 5 + F [(x + 4x + 4) + (5 - 4)] dx *Now let’s focus on F [(x + 2)2 + (1)]2 .
∫
u=x+2 u = asinθ x + 2 = tanθ
Ay‾ =
2
dx (2x + 4)dx dx = A x + B x-2dx + C x2 + 4x + 5 + D x2 + 4x + 4 + (5 - 4) +
∫
∫ (xR – xL)dy
A=
2
2
∫
B. Horizontal Element
oxy: F(r,θ) =
F(r , -θ) F(-r, π - θ) F(r , π - θ) F(-r , - θ)
a=1 ox: F(r,θ) =
2
dx = sec θdθ x+2 tanθ = 1 *draw this
F(-r , θ) F(r, π + θ)
II. Intersection w/ the pole
Bianca
set r = 0 and solve for θi III. Intersection with axes θ
2
2
2
∫ ∫ ∫ dθ F = F∫sec θ = F∫ cos θdθ = 2 ∫ (1 + cos2θ)dθ sec θdθ sec θdθ sec θdθ 2 2 2 2 4 =F (tan θ + 1) = F (sec θ) = F sec θ 2
2
F 1 = 2 *θ + 2 sin2θ] 1 *recall that 2 sin2θ = sinθcosθ. According to the drawing of the triangle, sinθ =
x+2 2
and cos θ =
1 2
. So
x + 4x + 5 x + 4x + 5 x+2 1 multiply them to get x2 + 4x + 5 . So yun yung value ng 2 sin2θ. *as for θ, recall that x + 2 = tanθ. So θ = Tan-1(x + 2). *so the final equation is:
B 2 -1 = Alnx + x + Cln(x + 4x + 5) + DTan (x + 2) + E F x+2 -1 2 2 x + 4x + 5 + 2 Tan (x + 2) + x + 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‾ =
∫ (ya
2
–
2 yb )dx
0°
90°
180°
270°
360°
r IV. Critical Points dr set dθ = 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 a 0<|b|<1
with a loop
a 0<|b|=1
cardioid
a 1<|b|<2
with a dent
a |b |≥2
convex
B. Rose Curves r = asin(nθ)
r = acos(nθ)
1.
5. r = 2 at 360° or back to 0° ulit Then TA-DAAA!
r = 2 – 2sinθ (cardioid) Intersection with the pole: r = 2 – 2sinθ 0 = 2 – 2sinθ 2sinθ = 2 -1 θ = sin (1) π θ=2
Intersection with the axes: Compute for the Area: θ
0°
90°
180°
270°
360°
r
2
0
2
4
2
Critical Points: dr dθ = -2cosθ 0 = -2cosθ -1
θ = cos (0) π 3π θ=2, 2 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)
1 A=2
*isang half lang yung area na kukunin mo. Eh since sobrang carbon copy yung other half of the heart, multiply the Area by 2.
1 A=2▪2
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.
π/2 -π/2
2
(2 – 2sinθ) dθ
A = ∫ (4 + 8sinθ + 4sin θ)dθ 2
A = 4∫ (1+ 2sinθ + sin θ)dθ 2
1 1 A = 4 θ ─ 2cosθ + 2 θ - 2 sin2θ
3 1 A = 42 θ - 2cosθ - 4 sin2θ
π/2
] -π/2 = 6π
2.
2. r = 0 at 90°. As you plot it, gawa ka na ng curve
∫ r2dθ
r = 3sin2θ Intersection with the pole: r = 3sin2θ 0 = 3sin2θ 0 = sin2θ -1 2θ = sin (0) 2θ = 0, π, 2π, 3π, etc. π 3π θ = 0, 2 , π, 2 , etc. Critical Points: dr dθ = 6sin2θ = 0 0 = sin2θ -1 2θ = sin (0) π 3π 5π 7π 2θ = 2 , 2 , 2 , 2 π 3π 5π 7π θ=4, 4 , 4 , 4 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
VOLUMES & CENTROIDS OF PLANE AREAS
2.
V=π
a
(x2)2 = x
2
r dh
x4 = x x4 – x = 0
Vx‾ = ∫ XCdv
x(x3 – 1) = 0
Vy‾ = ∫ YCdv
x = 0, y = 0 and x = 1, y = 1
CONDITIONS:
2
2
XRIGHT: y = x x = y
1. element must be parallel to the axis
XLEFT: y = x x = y
1
V=π
3. the axis should be a boundary
2
2
2
V = π∫ (y + 2y 2
2
V = π∫ (ya – yb) dx 2
3
0
2
7
x x V = π (x - 2 + 7 )
]
1 0
1
3
b
1.
a
2
V = 2π
6
4
5
3
y 2y - 5 - 3 - y
]
1 0
29π = 30
xydx
a
b
V = 2π
xydy
a
(when using a horizontal element)
2
(R – r )dh 1.
2
3
*substitute y = 2x sa y = x to get the points of intersection 2
3
2
3
(2x) = x 4x – x = 0 2
x (4 – x) = 0 2
x =0
4–x=0
x = 0, then y = 0
x = 4, then y = 8
3
Find V if the area bounded by y = 1 – x , ox and oy is revolved about ox. 3
3
y=1–x x =1–yx= 1
V = 2π
0
2
1-y
V = 2π∫ (1 - y) ydy 1/3
*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
2
3
xydy V = 2π∫ (xL – xR)ydy
0 4 0
2
+ 1 – y – 2y – 1)dx
(when using a vertical element)
(1 ─ 2x + x )d
9π = 14
Find V if the axes bounded 2 3 by y = x and y = 2x is revolved about ox.
V=π
) dx
2
b
2
B. Method of Circular Ring V=π
3/2
2
C. Method of Cylindrical Shell
V = π∫ (1 – x – 0) dx V=π
1/2
y 4y V = π 2 + 3
V = π∫ r dx
r=y +1
(R – r )dh
0
V = π∫ (( y + 1) – (y + 1) Find V, ‾x , ‾y of the solid generated by rendering the 3 region bounded by y = 1 – x , ox and oy about ox.
2
2
R = y + 1 (kasi about x = -1)
2. r must be parallel to the axis
1.
2
Points of Intersection:
A. Method of Circular Disk b
2
Find the volume of y = x and y = x about x + 1 = 0.
(R – r )dh
3
2
u(1 – u ) ▪ 3u du
1
*you may interchange the limits by turning the equation to negative
V = π∫ [(2x) – (x ) ]dx 2
3/2 2
V = π∫ (4x – x )dx 2
3
3
4
4x x V=π 3 - 4
]
1
V = ─6π
3
3
u (1 – u ) du
0
V = ─6π∫ (u – u ) du 3
4 0
64π = 3
4
6
7
u u V = ─6π 4 – 7
]
1 0
9π = 14
LENGTH OF ARC
S = 2a∫ 2 + 2cosθ dθ
∫ 1 + ( ) y in terms of x dx S = ∫ 1 + (dy ) x in terms of y dx dy S = ∫ (dt ) + ( dt ) parametric dr S = ∫ r + (dθ ) r in terms of θ dy 2 dx
S=
S = 2 2 a∫ 1 + cosθ dθ *let’s focus on 1 + cosθ. Recall that:
2
2
2
1.
1 2 cos θ = 2 (1 + cos2θ)
2
2
2cos θ = 1 + cos2θ
2
2θ 2cos 2 = 1 + cosθ
π
Find the length of the curve y = lncosx from x = 0 to x = 4 .
dy sinx dx = ─ cosx S=
∫
θ S = 2 2 ▪ 2 a∫cos2 dθ sinx
(
)
1 + ─cosx
2
dx =
∫
θ S = 4a(2)sin2
2
1 +tan x dx
S = ∫ sec x dx = = ∫ secxdx 2
]
S = ln(secx + tanx) 2.
π/4 0
]
= ln(1 +
2)
Find the length of the curve x = 2(2t + 3) , y = 3(t + 1) from t = 0 to t = 1.
2
1.
SA = 2π∫xdS
about oy
Find the area of the surface of revolution generated by 3 revolving y = x between x = 0 and x = a about ox.
dy dt = 6(t + 1)
SA = 2π∫x
3
SA = 2π∫x
3
2
2
(6 2t + 3 ) + (6(t + 1)) dt
u = 1 + 9x
S = ∫ 36(2t + 3) + 36(t + 1) dt 2
2
S = 6∫ (t + 1)dt 1
= 15
Find the total arc length of the cardioid r = a(1 + cosθ). dr dθ = a(─sinθ) 2
2
*minultply sa 2 kasi 2 parts yung cardioid
S =2
∫
2
2
2
2
a (1 + cosθ) + a sin θ
S = 2a ∫ 1 + 2cosθ + cos θ + sin θ dθ 2
2
2
*recall cos θ + sin θ = 1
dx
4
n
1 + 9x dx u du 4
π 3/2 SA = 27 [ (1 + 9x) + 1]
S = 6∫ (t + 1) dt
S = 2 ∫ (a(1 + cosθ)) + (─asinθ)
2
3
du = 36x du
]
2
0
dy
( )
1 + dx
1 2 4 3/2 SA = 2π ▪ 36 ▪ 3 (1 + 9x )
S = 6∫ 2t + 3 + t + 2t + 1 dt
]
= 8aError! Bookmark not defined.
about ox
SA = 2π∫ydS
2
0
SA = 2π∫ydS
dx 1/2 3 dt = 22 (2t + 3) (2) = 6 2t + 3
S = 3(t + 1)
π
AREA OF A SURFACE OF REVOLUTION
3/2
S=∫
3.
2θ 2cos 2 dθ
S = 2 2 a∫
y = lncosx
2
a 0
SECOND THEOREM OF PAPPUS V = 2πAd where: A = area d = perpendicular distance of the centroid from the axis of revolution
1.
2
2
Find the volume if area bounded by y = x and y = x is revolved about x = -1. A = ∫(ya – yb)dx A = ∫(x
1/2
2
– x )dx 3
2 3/2 x A=3x ─ 3
]
1 0
Ax‾ =
∫ x(ya – yb)dx
Ax‾ =
∫ x(x1/2 – x2)dx
Ax‾ =
∫ (x3/2 – x3)dx 4
2 5/2 x Ax‾ = 5 x – 4
]
1 =3
1 0
1 3 3 ‾x = 20 9 ‾x = 20 1 9 29π V = 2πAd = 2π3 20 + 1 = 30 * may + 1 kasi yung yung axis nasa x = -1.
