INTEGRAL CALCULUS REVIEWER (2 nd Sem 2011–2012) Integration – process of a function whose derivative or differential is given Integrand – the given function Integral – the required function THEOREM: Two functions having the same derivatives differ at most by a constant Proof: Let f(x) + g(x) be the function such that f’(x) = g’(x) INDEFINITE INTEGRAL – if f(x) is a function whose derivative is f(x), the relation between the two is given by: ∫ f(x) dx = F(x) + c Where: ∫ = integral sign f(x) = integrand F(x) = particular integrand c = constant of integration F(x) + c = indefinite integral of…. PROPERTIES: 1. ∫ du = u + c 2. ∫ (du + dv – dw) = ∫ du + ∫ dv – ∫ dw 3. ∫ Rdu = ∫ du c POWER FORMULAS : ∫ x n dx = x n+1 n + 1 + c if n ≠ –1 ∫ x –1 dx = lnx + c if n = –1 EXAMPLES: 1. ∫ (3x 2 – 6 x – 9 x 4 ) dx ∫ 3x 2 dx – ∫ 6x 1/2 dx – ∫ 9x –4 dx 3∫ x 2 dx – 6∫ x 1/2 dx – 9∫ x –4 dx n = 2 n = 1 2 n = -4 3 ▪ x 3 3 – 6 ▪ x 1/2 3 2 – 9 ▪ x -3 -3 + c X 3 – 4x 3/2 + 3x -3 + c 2. ∫ (x 2 - 3)(x 2 + 2) 3 x dx ∫ (x 4 – x 2 – 6)x -1/3 dx ∫ (x 4 ▪ x -1/3 – x 2 ▪ x -1/3 – 6 ▪ x -1/3 ) dx ∫ x 11/3 dx – ∫ x 5/3 dx – 6 ∫ x- 1/3 dx n = 11 3 n = 5 3 n =─ 1 3 3x 14/3 14 – 3x 8/3 8 – 9x 2/3 + c 3. ∫ 3 3x 2 - 6 x x 1/3 - 5 3 2x 3 dx *put all x’s outside radicals ∫ 3 3 x 2/3 - 6x 5/6 - 5 3 2 x dx *bring out constant denominator and place variable denominator in the numerator 1 3 2 ∫ ( 3 3 x 2/3 – 6x 5/6 – 5 ) ▪ x –1 dx 1 3 2 ∫ ( 3 3 x 2/3 ▪ x –1 – 6x 5/6 ▪ x –1 – 5 ▪ x –1 ) dx 1 3 2 ∫ ( 3 3 x –1/3 – 6x –1/6 – 5 x –1 ) dx n = – 1 3 n = – 1 6 n = –1 1 3 2 3 3 ▪ 3x 2/3 2 ─ 6 ▪ 6x 5/6 5 ─ 5 lnx + c CONSTANT INTEGRATION 1. If dy = (2X – 5)dx and y = 2 when x = –1, find y when x = 4. x -1 4 y 2 ? dy = (2x-5)dx ∫ dy = ∫ (2x-5)dx y + c = x 2 – 5x + c *Hindi pwedeng both sides may constant (c) so you have to choose which side to put 1 c 1 st option: y + c = x 2 – 5x *substitute x & y to get c 2 + c = (-1) 2 – 5(-1) c = 4 *then substitute c & x to y + c = x 2 – 5x y + 4 = (4) 2 – 5(4) y = -8
A reviewer for Integral Calculus. A list of formulas is provided. Sample problems are also included from easy to hard.
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INTEGRAL CALCULUS REVIEWER (2nd
Sem 2011–2012)
Integration – process of a function whose derivative or differential is given
Integrand – the given function
Integral – the required function
THEOREM: Two functions having the same derivatives differ at most by a constant
Proof:
Let f(x) + g(x) be the function such that f’(x) = g’(x)
INDEFINITE INTEGRAL – if f(x) is a function whose derivative is f(x), the relation between the two is given by:
∫ f(x) dx = F(x) + c
Where:
∫ = integral sign
f(x) = integrand
F(x) = particular integrand
c = constant of integration
F(x) + c = indefinite integral of….
PROPERTIES:
1. ∫ du = u + c
2. ∫ (du + dv – dw) = ∫ du + ∫ dv – ∫ dw
3. ∫ Rdu = ∫ du c
POWER FORMULAS:
∫ xndx =
xn+1
n + 1 + c if n ≠ –1
∫ x–1
dx = lnx + c if n = –1
EXAMPLES:
1. ∫ (3x2 – 6 x –
9x
4 ) dx
∫ 3x2dx – ∫ 6x
1/2dx – ∫ 9x
–4dx
3∫ x2dx – 6∫ x
1/2dx – 9∫ x
–4dx
n = 2 n = 12 n = -4
3 ▪ x
3
3 – 6 ▪ x
1/2
32
– 9 ▪ x
-3
-3 + c
X3 – 4x
3/2 + 3x
-3 + c
2. ∫ (x
2 - 3)(x
2 + 2)
3
x dx
∫ (x4 – x
2 – 6)x
-1/3 dx
∫ (x4
▪ x-1/3
– x2
▪ x-1/3
– 6 ▪ x-1/3
) dx
∫ x11/3
dx – ∫ x5/3
dx – 6 ∫ x-1/3
dx
n = 113 n =
53 n =─
13
3x14/3
14 – 3x
8/3
8 – 9x2/3
+ c
3. ∫ 3
3x2 - 6 x x
1/3 - 5
3
2x3
dx
*put all x’s outside radicals
∫ 3
3 x2/3
- 6x5/6
- 5
3
2 x dx
*bring out constant denominator and place variable denominator in the numerator
13
2 ∫ (
33 x
2/3 – 6x
5/6 – 5 ) ▪ x
–1 dx
13
2 ∫ (
33 x
2/3▪ x
–1 – 6x
5/6 ▪ x
–1 – 5 ▪ x
–1 )
dx
13
2 ∫ (
33 x
–1/3 – 6x
–1/6 – 5 x
–1 )
dx
n = –13 n = –
16 n = –1
13
2
3
3 ▪ 3x2/3
2 ─ 6 ▪ 6x
5/6
5 ─ 5 lnx + c
CONSTANT INTEGRATION
1. If dy = (2X – 5)dx and y = 2 when x = –1, find y when x = 4.
x -1 4
y 2 ?
dy = (2x-5)dx
∫ dy = ∫ (2x-5)dx
y + c = x2 – 5x + c
*Hindi pwedeng both sides may constant (c) so you have to choose which side to put 1 c
1st
option: y + c = x
2 – 5x
*substitute x & y to get c
2 + c = (-1)2 – 5(-1)
c = 4 *then substitute c & x to y + c = x2 – 5x
y + 4 = (4)2 – 5(4)
y = -8
1st
option: y = x
2 – 5x + c
*substitute x & y to get c
2 = (-1)2 – 5(-1) + c
c = -4 *then substitute c & x to y = x2 – 5x + c
y = (4)2 – 5(4) + (-4)
y = -8
2. Find the equation of the curve if the slope at pt (2,3)
is given by 2x + 12y - 3 .
slope = dydx =
2x + 12y - 3 (2y – 3)dy = (2x + 1)dx
∫ (2y – 3)dy = ∫ (2x + 1)dx
y2 – 3y = x
2 + x + c
*substitute pt (2,3)
32 – 3(3) = 2
2 + 2 + c
c = -6 Equation: y
2 – 3y = x
2 + x – 6 (hyperbola)
3. If at any point (x,y) on a curve d
3y
dx3 = 2 and (1,3) is the
pt. of inflection at which the slope of the inflectional tangent line is -2, find the equation of the curve.
d3y
dx3 = 2
ddx
d
2y
dx2 = 2
d
d
2y
dx2 = 2dx
∫ d
d
2y
dx2 = ∫2dx
d2y
dx2 = 2x + c1
ddx
dy
dx = 2x + c1
d
dy
dx = (2x + c1)dx
∫ d
dy
dx = ∫(2x + c1)dx
dydx = x
2 + c1x + c2
dy = (x
2 + c1x + c2)dx
∫dy = ∫ (x2 + c1x + c2)dx
y = x
3
3 + c1x
2
2 + c2x + c3
SYSTEM OF EQUATIONS: a. (1,3) is a point on a curve. So we substitute it
to the last equation.
3 = 13 +
c1
2 + c2 + c3
16 = 3c1 + 6c2 + 6c3
b. slope = dydx = -2 at x = 1. Substitute these
values to the second equation. -2 = 1
2 + c1(1) + c2
c1 + c2 = -3 d
2y
dx2 = 2x + c1
0 = 2 + c1` c1 = -2
*substitute c1to the other equations to get the other 2 constants
c1= -2, c2 = -1 and c3 = 143
*substitute these values to the last equation
y = x
3
3 – x2 – x +
143
4. Find the area under the parabola y = 8 – x
2 – 2x,
above the x-axis. *complete the square
x2
+ 2x + __ = 8 – y + __ x
2 + 2x + 1 = 9 – y
(x + 1)2 = –(y – 9)
*it is a parabola the opens downward
V (-1,9) dA = (yA – yB)dx
*dx = xLEFT – xRIGHT
dA = (8 – x2 – 2x)dx
∫dA = ∫ (8 – x2 – 2x)dx
A = 8x – x
3
3 – x2
+ c
On the x-axis, y = 0.
*substitute y in y = 8 – x2 – 2x 0 = 8 – x
2 – 2x
x2 + 2x – 8 = 0
(x + 4) (x – 2) = 0 When x = -4, A = 0
*0 yung area pag x = -4 kasi hindi wala pang area na nabubuo sa point na yun.
*Substitute these values to A = 8x – x3
3 – x2 + c
0 = 8(-4) – (-4)
3
3 – (-4)2 + c
c = 803
When x = 2, and c = 803
*Substitute these values to A = 8x – x3
3 – x2 + c. We
did this again tapos with x = 2 kasi may area nang
macocover sa point na yun .
A = 8(2) – 2
3
3 – 22 +
803 = 36 sq. units
5. An art collector purchased for $1000 a painting by an
artist whose works are currently increasing with respect to the time according to the formula
dudt = 5t
2/3 + 10t + 50
where u dollars is the anticipated value of the painting in t years after its purchase. If this formula is used for the next 6 years, what is its anticipated value 4 years from now?
dudt = 5t
2/3 + 10t + 50
∫ du = ∫ (5t2/3
+ 10t + 50)dt
u = 5t
5/3
53
+ 5t2 + 50t + c
u = 3t5/3
+ 5t2 + 50t + c
u 1000 ?
t 0 4
When u = 1000 and t = 0. *Substitute to u = 3t5/3 + 5t2 + 50t + c
c = 1000 When t = 4 and c = 1000
*Substitute to u = 3t5/3 + 5t2 + 50t + c
u = $1,286.89
6. A woman in a hot air balloon dropped her binoculars 150ft above the ground and is rising at the rate of 10ft/s. (a) How long will it take the binoculars to reach the ground? (b) With what speed will it strike the ground?
a = g = dvdt = -32ft/s
2
*negative yung acceleration/gravity kasi opposite siya ng direction ng velocity ng hot air balloon
dvdt = -32
∫ dv = ∫ -32dt
v = -32t + c1
When t = 0 and v = 10ft/s
*substitute these values to v = -32t + c1
10 = -32(0) + c1 c1= 10
v = dsdt = -32t + c1
∫ ds = ∫ (-32t + c1)dt
s = -16t2 + c1t + c2
When t = 0 and s = 0
*substitute these values to s = -16t2 + c1t + c2
0 = -16(0)2 + c1(0) + c2
c2 = 0
*Substitute c1 and c2 to s = -16t2 + c1t + c2
s = -16t2 + 10t + 0
When s = -150
*Substitute s to s = -16t2 + 10t + 0. Negative yung s kasi opposite siya ng initial direction
-150 = -16t2 + 10t
16t2 – 10t – 150 = 0
*get t by using the quadratic formula
t = 3.4 seconds *you will get 2 answers here. ‘yung isa negative. Siyempre, ineneglect natin ‘yung negative dahil bawal maging nega ‘yung time
Differentiate s = -16t2 + 10t to get
dsdt /the velocity
dsdt = -32t + 10
When t = 3.4. v = -32(3.4) + 10 v = -98.8 ft/s
*Again, it is negative kasi opposite siya nung initial direction
DEFINITE INTEGRAL
PROPERTIES:
1. b
af(x)dx = -
b
af(x)dx
- interchanging the limits changes the sign of the integral
2. b
af(x)dx =
c
af(x)dx +
b
cf(x)dx
- The interval of integration may be broken down to any number of subintervals and the integration performed over each interval separately
3. b
af(x)dx =
c
af(t)dt +
b
cf(z)dz
- The definite integral of an integrand is independent of the variable of integrations
EXAMPLES:
1. 2
-1
5x
2 +
13 x –
12 dx
5x3
3 + 16 x
2 –
12 x ]
2
−1
s = vot + at
2
v = dsdt
a = g = dvdt
*substitute 2 and -1 sa mga x. Subtract the lower number from the upper number.
53 (2
3 – (-1)
3) +
16 (2
2 – (-1)
2) -
12 (2+1)
53 (8 + 1) +
16 (4 – 1) -
12 (2+1) =
282 = 14
2. 1
0
x3 + 1
x + 1 dx
1
0
(x + 1)(x2 - 2x + 1)
x + 1
1
0 (x
2 – 2x + 1)dx
x3
2 ─ x2 + x ]
1
0
13 (1
3 – 0) – (1
2 – 0) + (1 – 0) =
56
THE GENERAL POWER FORMULA
∫undu
if n ≠ -1:
∫undu =
un + 1
n + 1 + c
if n = -1:
∫undu = lnu + c
1. ∫ (x + 1)2dx
u = x + 1 du = dx
∫ u2du
u3
3 + c
(x + 1)3
3 + c
13 (x
3 + 3x
2 + 3x + 1) + c
13 x
3 + x
2 + x +
13 + c
2. ∫ dx
(2x - 7)4
∫ (2x – 7)-4
dx u = 2x – 7 du = 2dx
du2 = dx
*trinanspose yung 2 sa other side para maging equal yung value ng du sa original formula. Pero yung 2 na trinanspose aka yung 12 , gagawin mong constant. so if you like, hide
12 , du = dx. So it
still follows the original formula na ∫und. Pag hindi ‘to nagets explain ko sa other examples. =)))
12 ∫ u
-4du
12 ▪
u-3
-3 + c
─1
6(2x - 7)3 + c
3. ∫ tdt
4t2 + 9
∫ (4t2 + 9)
-1/2tdt
u = 4t2 + 9
du = 8tdt
du8 = tdt
n = -12
* diba sa orig na formula it’s (4t2 + 9)-1/2tdt so diba u = (4t2 + 9)-1/2
tapos after that yung tdt. trinanspose natin yung 8 to the other side to follow the general formula na undu. Diba nakuha nating du nung una is 8tdt. Para maging tdt lang which is yung nasa original formula, linipat yung 8. Pero gagawin siyang constant or “preparation” sa integration.
18 ∫u
-1/2du
18 ▪
u1/2
12
+ c
14 4t
2 + 9 + c
4. ∫ e
2tdt
e4t
+ 22t
+ 1
∫ e
2tdt
(e2t
+ 1)2
∫ (e2t
+ 1)-2
e2t
dt n = -2 u = e2t + 1 du = 2e2tdt
du2 = e2tdt
12 ∫ u
-2du
12 ▪
u-1
-1 + c
─ 1
2e2t
+ 1 + c
5. ∫ y
1/3
(y4/3
+ 9)2 dy
∫ (y4/3
+ 9)-2
y1/3
dy
u = y4/3
+ 9
du = 43 y
1/3dy
34 du = y
1/3dy
34 ∫ u
-2du
34 ▪
u-1
-1 + c
─ 3
4(y4/3
+ 9) + c
6. ∫ (1 + 2e3x
)e3x
dx
u = 1 + 2e3x
du = 6e
3xdx
du6 = e
3xdx
16 ∫ u
1du
16 ▪
u2
2 + c
(1 + 2e3x
)2
12 + c
7. ∫ x
3/4 + 9
x1/4 dx
n = 12
u = x3/4
+ 9
du = 34 x
-1/4dx
43 du =
dxx
1/4
43 ∫ u
1/2du
43 ▪
23 u
2/3 + c
89 (x
3/4 + 9)
2/3 + c
8. ∫ (6cos2x + sin
2x)
1/2sinxcosx dx
n = 12
u = 6cos2x + sin
2x
du = 6[2cosx ▪ d(cosx)
dx ] + 2cosx ▪ d(sinx)
dx
du = 6 [2cosx(-sinx)]dx + 2sinxcosxdx du = -12sinxcosxdx + 2sinxcosxdx du = -10sinxcosxdx
─du10 = sinxcosxdx
─1
10 ∫ u1/2
du
─1
10 ▪ 23 u
3/2 + c
─1
15 (6cos2x + sin
2x)
3/2 + c
9. 10
8
1
4 x - 1 -3
dx
n = -3
u = 14 x – 1
du = 14 dx
4du = dx 4 u
-3du
4u-2
-2
─ 2u
2 = ─ 2
(14 x - 1)
2 ]
10
8
─ 2
(14 (10) - 1)
2 +
2
(14 (8) - 1)
2 =
109
EXAMPLES
when∫duu or u
-1du = lnu + c :
1. ∫ sec5θtan5θ3 + 2sec5θ dθ
u = 3 + 2sec5θ du = 2(5)sec5θtan5θ dθ du10 = sec5θtan5θ dθ
110 ∫
duu
110 lnu + c
110 ln(3 + 2sec5θ) + c
2. ∫ dx
x + x
*factor x + x for it to be x ( x + 1)
∫ dx
x ( x + 1)
u = x + 1
du = dx
2 x
2du = dx
x
2∫ duu = 2lnu + c
2ln( x + 1) + c
3. ∫ secxdx
∫ secxdx ▪ secx + tanxsecx + tanx
∫ secxtanx + sec
2x
secx + tanx dx
u = secx + tanx du = (secxtanx + sec
2x)dx
∫ duu = ln(secx + tanx) + c
4. ln2
0
exdx
1 + 3ex
u = 1 + 3ex
du = 3exdx
du3 = e
xdx
*change the limits. To do that, substitute sa limits sa mga x sa
equation ng u which is 1 + 3ex. when x = ln2, eln2 = 2. So 1 + 3(2) = 7. And when x = 0, it’s going to be e0 = 1. So 1 + 3(1) = 4.
x ln2 0
u 7 4
13 ∫
duu
13 ln(1 + 3e
x)]
7
4
13 [ln7 – ln4]
13 ln
74
5. ∫ x
3 - 2x + 5x - 3 dx
*when the degree/exponent of the numerator is higher than the denominator, divide.
∫ (x2 + 3x + 7 +
26x - 3 )dx
x3
3 + 3x
2
2 + 7x + 26∫ (x – 3)-1
dx
x3
3 + 3x
2
2 + 7x + 26ln(x – 3)
6. -2
-3
y + 2y
2 + 4y dy
u = y2 + 4y
du = (2y + 4)dy du2 = (y + 2)dy
*change limits x -3 -2
u -3 -4
12
-2
-3 duu =
12 ln |u| ]
−4
−3
12 [ ln|-4| - ln|-3|]
12 ln
43
EXPONENTIAL FUNCTION
∫ audu =
1lna a
u + c
∫ eudu = e
u + c
TRIGONOMETRIC FUNCTIONS
1. ∫ sin u du = – cos u + c
2. ∫ cos u du = sin u + c
3. ∫ tan u du = ln sec u + c
= – ln cos u + c
4. ∫ cot u du = ln sin u + c
= – ln csc u + c
5. ∫ sec u du = ln(sec u + tan u ) + c
6. ∫ csc u du = ln(csc u – cot u) + c
7. ∫ sec2u du = tan u + c
8. ∫ csc2u du = –cot u + c
9. ∫ sec u tan u du = sec u + c
10. ∫ csc u cot u du = –csc u + c
EXAMPLES:
1. ∫ sin4xdx u = 4x du = 4dx du4 = dx
14 ∫ sin u du
14 (-cos u du) + c
─ 14 cos4x + c
2. ∫ tan x
x dx
u = x
du = 1
2 x dx
2du = dx
x
2∫ tan u du = 2lnsec u + c
2ln(sec x ) + c
3. ∫ e2x
cos e2x
dx u = e2x
du = 2e2x dx
2du = e2x dx
12 ∫ cos u du
12 sin e
2x + c
TRIGONOMETRIC TRANSFORMATIONS
I. ∫ sinm
x cosnx dx
where m or n is a positive odd integer tools: change the one w/ odd powers sin
2x = 1 – cos
2x
cos2x = 1 – sin
2x
Ex:
∫ sin52x cos
42x dx
y = 2x dy = 2dx dy2 = dx
12 ∫ sin
5y cos
4y dy
12 ∫ sin
4y cos
4y siny dy
12 ∫ (sin
2)
2 cos
4y siny dy
12 ∫ (1 – cos
2y)
2 cos
4y siny dy
12 ∫ (1 – 2cos
2y + cos
4y) cos
4y siny dy
12 ∫ (cos
4y – 2cos
6y + cos
8y) siny dy
*integrate each term. so their n’s sa un would be 4, 6, and 8 respectively. u = cosy du = -siny dy
So we’ll be using the form ∫un =
un+1
n + 1 for each term. And
substitute 2x to y na ulit.
12 ∫ (u
4 – 2u
6 + u
8) siny dy
–12
cos
52x
5 – 2cos
72x
7 – cos
92x
9 + c
–12 cos
52x
1
5 – 2cos
22x
7 – cos
42x
9 + c
II. ∫ secm
x tannx dx or ∫ csc
mx cot
nx dx
a. Where m is positive even integer tools: sec
2x = 1 + tan
2x
csc2x = 1 + cot
2x
Ex:
∫ tan412 x sec
412 x dx
y = 12 x
2dy = dx
2∫ tan4y sec
4y dy
2∫ tan4y sec
2y sec
2y dy
2∫ tan4y (1 + tan
yx) sec
2y dy
2∫ (tan4y + tan
6y) sec
2y dy
u = tany du = sec
2y
2∫ (u4 + u
6)du = 2
tan
512 x
5 + tan
712 x
7 + c
b. Where n is a positive odd integer
tools: tan2x = sec
2x – 1
cot2x = csc
2x – 1
Ex:
∫ tan53x sec
33x dx
y = 3x dy3 = dx
13 ∫ tan
5y sec
3y dy
13 ∫ tan
4y sec
2y tany secy dy
13 ∫ (sec
2y – 1)
2 sec
2y tany secy dy
13 ∫ (sec
4y – 2sec
2y + 1) sec
2y tany secy dy
13 ∫ (sec
6y – 2sec
4y + sec
2y) tany secy dy
u = secy du = tany secy dy
13 ∫ (u
6 – 2u
4 + u
2)du
13
sec
73x
7 – 2sec
53x
5 + sec
33x
3 + c
∫
1 + sinθ
cosθ 2dθ
∫ 1 + 2sinθ + sin
2θ
cos2θ dθ
∫ 1
cos2θ dθ + ∫
2sinθcos
2θ dθ + ∫
sin2θ
cos2θ dθ
∫ sec2θ dθ + 2∫ secθ ▪
sinθcosθ dθ + ∫ tan
2θ dθ
2tan2θ + 2∫ secθ tanθ dθ + ∫ (sec
2θ – 1) dθ
2tan2θ + 2secθ + ∫ sec
2θ dθ – ∫dθ
2tan2θ + 2secθ + tanθ – θ + c
III. ∫ tannx dx or cot
nx dx
where n is an integer tools: tan
2x = sec
2x – 1
cot2x = csc
2x – 1
a. n is a positive even integer EX:
∫ tan6x dx
∫ tan4x ▪ tan
2x dx
∫ tan4x (sec
2x – 1) dx
∫ (tan4x sec
2x – tan
4x) dx
*step by step nating i-solve each part, okay? So we’ll start
with ∫tan4x sec
2x
∫ tan4x sec
2x dx
u = tanx du = sec
2x dx
∫ u4du
u5
5 = tan
5x
5
*next is –∫ tan4x dx
–∫ tan4x dx
–∫ tan2x ▪ tan
2x dx
–∫ tan2x(sec
2x – 1) dx
–∫ (tan2x sec
2x – tan
2x) dx
*it’s possible na to integrate tan2x sec
2x. Use ∫u
ndu. And
distribute the negative sign so magiging positive yung tan
2x.
–tan
3x
3 + ∫ tan2x dx
–tan
3x
3 + ∫ (sec2x – 1) dx
–tan
3x
3 + ∫ sec2x dx – ∫dx
–tan
3x
3 + tan x – x
*combine na the two parts. So the final answer would be:
tan5x
5 – tan
3x
3 + tanx – x + c
b. n is a positive odd integer EX:
∫ tan5x dx
∫ tan3x ▪ tan
2x dx
∫ tan3x(sec
2x – 1) dx
∫ tan3x sec
2x dx – ∫ tan
3x dx
*pwede na ma-integrate yung first term using undu so
we’ll focus on the second term which is tan3xdx
tan
4x
4 – ∫ tan2x ▪ tanx dx
tan4x
4 – ∫ tanx(sec2x – 1) dx
tan4x
4 – ∫ (tanx sec2x – tanx) dx
tan4x
4 – ∫ (tanx sec2x) dx – ∫ tanx dx
tan4x
4 – tan
2x
2 – ln(secx) + c
IV. ∫ sinm
x cosnx dx
where m & n are positive even integers
tools: sinx cosx = 12 sin2x
sin2x =
12 (1 – cos2x)
cos2x =
12 (1 + cos2x)
Ex:
∫ sin23x cos
23x dx
y = 3x dy = 3dx dy3 = dx
13 ∫ sin
2y cos
2y dx
13 ∫ (siny cosy)
2 dx
13 ∫ (
14 sin
22y)dy
1
12 ∫ sin22ydy
112 ∫
12 (1 – cos4y)dy
124 ∫ dy –
124 ∫ cos4ydy
124 [y –
14 sin4y] + c
124 [3x –
14 sin12x] + c
∫ sin2x cos
4x dx
∫ sin2x cos
2x cos
2x dx
∫ (sinx cosx)2 cos
2x dx
∫ (12 sin2x)
2cos
2x dx
14 ∫ sin
22x cos
2x dx
18 ∫ (1 – cos4x) cos
2x dx
18 ∫ cos
2x dx –
18 ∫ cos4x cos
2x dx
18 ∫
12 (1 + cos2x) dx –
18 ∫ cos4x ▪
12 (1 + cos2x) dx
116 ∫ (1 + cos2x) dx –
116 ∫ (cos4x + cos2x cos4x) dx
116 (x +
12 sin2x) –
116 ∫ cos4x dx –
116 ∫ (cos2x cos4x) dx
116 (x +
12 sin2x) –
116 ▪
14 sin4x –
116 ∫ (cos2x(1 – 2sin
22x) dx
116 (x +
12 sin2x) –
164 sin4x –
116 ∫ (cos2x – 2sin
22x cos2x) dx
116 (x +
12 sin2x) –
164 sin4x –
116 ∫ cos2x dx –
116 ▪ 2∫ sin
22x
*use u = sin2x, du = 2cos2x dx cos2x dx
116 (x +
12 sin2x) –
164 sin4x –
116 ▪
12 sin2x –
18 ▪
12 ∫ u
2du
116 x +
132 sin2x –
164 sin4x –
132 sin2x –
116 ▪
u3
3 + c
116 x +
132 sin2x –
164 sin4x –
132 sin2x –
116 ▪
u3
3 + c
116 x +
132 sin2x –
164 sin4x –
132 sin2x –
148 sin
32x + c
V.∫ sin ax sin bx dx
∫ sinm
x cosnx dx
∫ sinm
x cosnx dx
tools: sinα sinβ = 12 *cos(α ─ β) – cos(α + β)]
cosα cosβ = 12 *cos(α ─ β) + cos(α + β)+
sinα cosβ = 12 *sin(α ─ β) + sin(α + β)+
EX:
1. ∫ sin4x sin7x dx
12 ∫ [cos(4x – 7x) – cos(4x + 7x)] dx
12 ∫ [cos(–3x) – cos(11x)] dx
12 ∫ (cos3x – cos11x) dx
12 [
13 sin3x –
111 sin11x] + c
2. ∫ cos7x sin4x dx *let α = 4x and β = 7x
12 ∫ [sin(4x – 7x) + sin(4x + 7x)] dx
12 ∫ [sin(–3x) + sin(11x)] dx
12 ∫ (–sin3x + sin11x) dx
12 [
13 cos3x –
111 cos11x] + c
3. π4
0 cosx cos3x dx
12 ∫ [cos(x – 3x) + cos(x + 3x)] dx
12 ∫ (cos2x + cos4x) dx
12 [
12 sin2x +
14 sin4x]
π4
0
12 [
1
2 sin2(π4 ) +
14 sin4(
π4 ) –
1
2 sin2(0) + 14 sin4(0) ]
12 [
12 (1) +
14 (0)] – 0 =
14
4. π3
0 sinx sin2x sin3x dx
12 ∫ sinx[cos(2x – 3x) – cos(2x + 3x)] dx
12 ∫ sinx[cosx – cos5x] dx
12 ∫ (sinx cosx – sinx cos5x) dx
*for sinx cosx, u = sinx, du = cosxdx. So the formula you’ll use would be ∫undu.
12 ▪
sin2x
2 – 12 ∫
12 [sin(x – 5x) + sin(x + 5x)] dx
sin2x
4 – 14 ∫ (-sin4x + sin6x) dx
sin2x
4 + 14 ∫ sin4x dx –
14 ∫ sin6x dx
sin2x
4 + 1
16 cos4x – 1
24 cos6x ]
π3
0
= 9
32
INVERSE TRIGONOMETRIC FUNCTIONS
1. ∫ du
a2 ─ u
2 = Sin-1
ua + c
2. ∫ du
a2 + u
2 = 1a Tan
-1ua + c
3. ∫ du
u u2 ─ a
2 = 1a Sec
-1 ua + c
Examples:
1. ∫ dx
25 + 64x2
∫ dx
(5)2 + (8x)
2 a = 5 u = 8x du8 = dx
18 ∫
dua
2 + u
2
18 ▪
15 Tan
-1 8x5 + c
140 Tan
-1 8x5 + c
2. ∫ dx
9 ─ 4x2
∫ dx
(3)2 ─ (2x)
2 a = 3 u =2x
du2 = dx
12 ∫
du
a2 ─ u
2
12 Sin
-12x3 + c
3. ∫ sec
2x dx
50 ─ sec2x
∫ sec
2x dx
50 ─ (1 + tan2x)
∫ sec
2x dx
49 ─ tan2x
∫ sec
2x dx
(7)2 ─ (tanx)
2 a = 7 u =tanx du = sec2x dx
Sin-1
tanx
7 + c
4. ∫ dx
21 - 4x + x2
*add and subtract 4 para maging perfect square yung x2 – 4x
∫ dx
21 + 4 ─ (x2 ─ 4x + 4)
*nakalagay sa equation, + 4 sa pareho, kasi yung second na 4, negative siya pag dinitribute yung nega
∫ dx
25 ─ (x ─ 2)2 a = 5 u = x – 2 du = dx
Sin-1
x ─ 2
5 + c
5. ∫ dx
5 - 2x - 3x2
*to get c in ax2 + bx + c, get the value of b2
4a . so in this
formula c = 22
4(3) = 13
∫ dx
5 + 13 ─ (3x
2 + 2x +
13 )
∫ dx
4
3
2 ─ ( 3 x +
1
3 )
2
a = 4
3 u = 3 x +
1
3
∫ du
a2 ─ u
2 du
3 = dx
1
3 Sin
-1
3x + 1
4 + c
6. ∫ dx
5x2 - 4x + 2
∫ dx
(5x2 - 4x +
45 ) + (2 -
45 )
∫ dx
( 5 x ─ 2
5 )
2 +
6
5
2 a =
2
5 u = 5 x ─
2
5
∫ du
a2 + u
2 du
5 = dx
1
5 ▪
5
6 Tan
-1
5x - 2
5
6
5
+ c
1
6 Tan
-15x - 2
6 + c
7. ∫ dx
x 9x2 - 25
*multiply the whole equation to 33 para maging 3x
yung x na nasa baba.
∫ 3dx
3x (3x)2 - (5)
2 a = 5 u = 3x
du3 = dx
∫ du
u u2 ─ a
2
15 Sec
-13x5 + c
8. 1
0 (x + 1)dx
x2 + 1
∫ x
x2 + 1 dx + ∫
dxx
1 + 1
2
*For x
x2 + 1 :
u = x2 + 1 du2 = xdx
∫ duu + ∫
dxx
1 + 1
2
*For dx
x1 + 12 :
u = x a = 1 du = dx
12 ∫
duu + ∫
duu
2 + a
2
12 ln(x
2 + 1) + Tan
-1x]
1
0
12 [(ln1
2 + 1) – (ln0
2 +1)] + [Tan
-11 – Tan
-10]
12 [ln2 – ln1] + [
π4 - 0]
12 ln2 +
π4
9. e
1
dyy(1 + ln
2y)
∫ dyy(1
2 + (lny)
2) a = 1 u = lny du =
dyy
∫ dua
2 + u
2
Tan-1
(lny) ] e
1
Tan-1
(lne) – Tan-1
(ln1) Tan
-11 – Tan
-10
π4
10. 4
2
dy
y y ─ 1
∫ dy
y ( y )2 ─ 1
2
*Diba u = y . So yung y sa labas, ihiwalay mo para
maging
1
y
y . Para yung form maging
du
u u2 ─ a2
u = y 2du = dy
y
2∫ du
u u2 ─ a
2
2 Sec-1
y ] 4
2
2 [Sec-1
4 ─ Sec-1
2 ]
2 [ π3 ─
π4 ] =
π12
11. 6
2
(5x - 2)dx x
2 - 4x + 20
*let u = x2 – 4x + 2z and du = (2x – 4)dx *divide the numerator by the derivative of the denominator. Then follow this:
∫52 (2x - 4) + 8
x2 - 4x + 20
52∫ (2x - 4)dx
x2 - 4x + 200 + 8∫ dx
x2 - 4x + 200
u = x2 – 4x + 200
du = 2x – 4
52∫du
u + 8∫ dx(x
2 - 4x + 4) + (20 - 4)
52 ln(x
2 – 4x + 20) + 8∫ dx
(x - 2)2 + (4)
2
a = 4 u = x – 2 du = dx
52 ln(x
2 – 4x + 20) + 8∫ du
a2 + u
2
52 ln(x
2 – 4x + 20) + 8 ▪
14 Tan
-1
x - 2
4 ] 6
2
52 [ln32 – ln16] + 2 [Tan
-11 – Tan
-10]
52 ln
3216 + 2 ▪
π4
52 ln2 +
π2
12. 2
1
(3x + 1)dx
3 + 2x - x2 u = 3 + 2x – x2 du = (2 – 2x)dx
*divide the numerator by du. Yung ginawa sa previous numbahhh
∫─
32 (-2x + 2) + 4
3 + 2x - x2
dx
─32 ∫ -2x + 2
3 + 2x - x2 dx + 4∫ dx
3 + 2x - x2
u = 3 + 2x – x2 du = (2 – 2x)dx
─32 ∫u
-1/2du + 4∫ dx
(3 + 1) - (x2 - 2x + 1)
─32 ▪
2u1/2
1 + 4∫ dx
(2)2 - (x
- 1)
2
– 3 3 + 2x - x2 + 4∫
du
a2 ─ u
2
– 3 3 + 2x - x2 + 4 Sin
-1
x - 1
2 ] 2
1
*─3 3 – (-6)] + 4[ π6 ─ 0+
6 - 3 3 + 2π3
13. 2
1
dx
(x + 1) 2x(x + 2)
∫ dx
(x + 1)( 2 ) x(x + 2) *linabas lang yung 2
1
2 ∫ dx
(x + 1) x2 + 2x + 1 - 1
1
2 ∫ dx
(x + 1) (x + 1)2 - 1
2 u = x + 1 du = dx a = 1
1
2 ∫ du
u u2 ─ a
2
1
2 Sec
-1(x + 1) ]
2
1
1
2 [Sec
-13 – Sec
-12]
1
2 [Sec
-13 –
π3 ]
ADDITIONAL FORMULAS:
1. ∫ u2 ± a
2 du =
12 { u u
2±a
2 ± a
2 ln |u + u
2±a
2 |} + c
2. ∫ du
u2 ± a
2 = ln|u + u
2±a
2 |} + c
3. ∫ a2 ─ u
2 du =
12 { u a
2 ─ u
2 + a
2 Sin
-1
u
a } + c
4. ∫ duu
2 - a
2 = 1
2a ln | u - au + a | + c
5. ∫ dua
2 - u
2 = 1
2a ln | u + au - a | + c
EXAMPLES:
1. ∫ xdx
9x4 - 1
∫ xdx
(3x2)
2 - 1
2 u = 3x2
du6 = xdx a = 1
16 ∫ du
u2 ± a
2
16 ln|3x
2 + 9x
4 - 1 | + c
2. ∫ dx
x2 + x + 1
∫ dx
x2 + x +
14 + (1 -
14 )
∫ dx
(x + 12 )2 + (
3 2 )2
a = x + 12 du = dx a =
3 2
∫ du
u2 ± a
2
ln|(x + 12 + x
2 + x + 1 | + c
3. ∫ dxx
2 - 3x - 10
∫ dx
(x2 - 3x +
94 ) - (10 +
94 )
∫ dx
(x -
32 )
2 -
494
u = x - 32 du = dx a =
72
∫ duu
2 - a
2
17 ln |
(x - 32 ) -
72
(x - 32 ) +
72
| + c
17 ln |
x - 5x + 2 | + c
4. 1
0
dx2 - x
2
∫ dx
( 2 )2 - x
2 u = x du = dx a = 2
∫ dua
2 - u
2
1
2 2 ln |
x + 2
x - 2 | ]
1
0
1
2 2 [ ln|1 + 2
1 - 2 |─ ln|
0+ 2
0 - 2 |]
1
2 2 [ ln|1 + 2
1 - 2 |─ |ln1| ]
1
2 2 ln|1 + 2
1 - 2 |
5. 4
325 - x
2 dx a = 5 u = x du = dx
∫ a2 ─ u
2 du
12 { x 25 - x
2 + 25 Sin
-1
x
5 } ] 4
3
12 { [12 + 25 Sin
-1
4
5 ] – [12 + 25 Sin-1
3
5
252 [ Sin
-1
4
5 – Sin-1
3
5 ]
*diba yung notation na “Sin-1(x)” means ANGLE yung
value niya? Like, “Sin-1(1)” = 90 or π2 . So let’s represent
ka ng triangle of each angle. Since sin function yung 45 at
35 , ile-label mo yung numbers na yan sa
oppositehypotenuse . So
the triangles would look like this:
sin(θ – β) = sinθcosβ – cosθsinβ
sin(θ – β) =
4
5
4
5 –
3
5
3
5
sin(θ – β) = 7
25 *note that we’re only getting (θ – β)
(θ – β) = Sin-1
7
25
252 Sin
-1
7
25
6. π4
π12
cos2x
sin22x -
116
dx a = 14 u = sin2x
du2 = cos2xdx
12 ∫ du
u2 - a
2
12 ▪
1
2(14 )
ln | sin2x -
14
sin2x + 14
| ]
π4
π
12
ln|
sin2
π
4 - 14
sin2
π
4 + 14
|─ ln|
sin2
π
12 - 14
sin2
π
12 + 14
|
ln|
sin
π
2 - 14
sin
π
2 + 14
|─ ln|
sin
π
6 - 14
sin
π
6 + 14
|
ln
34
54
─
14
34
ln
3
5 ─ ln
1
3
ln95
7. ∫ 6y + 19y
2 - 6y - 3 dy u = 9y2 – 6y – 3 du = (18y – 6)dy
*divide the numerator by du. Yung method na ginawa before
∫13 (18y - 6) + 3
9y2 - 6y - 3 dy
13 ∫ (18y - 6)dy
9y2 - 6y - 3 + 3∫ dy
(9y2 - 6y + 1) - (3 + 1)
13 ∫du
u + 3∫ duu
2 - a
2 u = 3y + 1 du3 = dy a = 2
13 ln(9y
2 – 6y – 3) +
34 ln|
3y - 33y + 1 | + c
8. ∫ 2x - 3
x2 + x + 2
dx u = x2 + x + 2 du = (2x + 1)dx
*divide the numerator by du. Yung method na ginawa before
∫ (2x - 1) - 4
x2 + x + 2
dx
∫ (2x - 1)
x2 + x + 2
dx ─ 4∫ dx
(x2 + x +
14 ) + (2 -
14 )
∫u-1/2
du ─ 4∫ du
u2 - a
2
2(x2 + x + 2)
1/2 – 4 ln|(x +
12 ) + x
2 + x + 2 |} + c
2 x2 + x + 2 ─ 4 ln|(x +
12 ) + x
2 + x + 2 |} + c
9. e^3
e^2
lnxx(ln
4x - 1) dx u = ln2x
du2 =
lnxx dx a = 1
12 ∫ du
u2 - a
2
12 ▪
12 ln |
ln2x - 1
ln2x + 1 |]
e^3
𝑒^2
14
ln | ln
2e
3 - 1
ln2e
3 + 1 |─ ln |
ln2e
2 - 1
ln2e
2+ 1 |
14 [ ln|
33 - 1
33 + 1 |─ ln |
22 - 1
22 + 1 |]
14 [ ln
45 ─ ln
35 ] =
14 ln
43
10. ∫ 1 + 1x dx
∫ x + 1
( x )2 dx
∫ 1
x ( x )
2 + 1
2 dx u = x 2du =
dx
x a = 1
2 ∫ u2 ± a
2 du
12 { x ▪ x + 1 + ln | x + x + 1 |} + c
HYPERBOLIC FUNCTIONS
1. ∫ sinh u du = cosh u + c
2. ∫ cosh u du = sinh u + c
3. ∫ tanh u du = ln |cosh u | +c
4. ∫ coth u du = ln |sinh u | +c
5. ∫ sech2
u du = tanh u + c
6. ∫ csch2u du = –coth u + c
7. ∫ sech u tanh u du = –sech u + c
8. ∫ csch u coth u du = –csch u + c
EXAMPLE:
∫(sech 1 - t )(tanh 1 - t )
1 - t dx
u = 1 - t ─2du = dt
1 - t
─2∫sech u ▪ tanh u ▪ du
─2(–sech u) + c
2sech 1 - t + c
IMPROPER INTEGRALS I. Integrals with infinite limits in the integrand
*in other words, isa or both a and b sa formula
na b
af(x)dx, infinity.
∞
af(x)dx = limb
∞
b
af(x)dx
b
-∞f(x)dx = lima
-∞ b
af(x)dx
∞
-∞f(x)dx = lima
-∞ and b
∞ b
af(x)dx
NOTE:
∞∞ &
00 = ‘pag ganyan yung situation, dun sa
equation/s kung sa’n naka substitute yung “b” or “a”, derive both the numerator and the denominator. Then you may start dividing
1∞ = 0
EXAMPLES:
1. ∞
1
2dyy(y + 16)
limb
∞ b
1
2dyy
2 + 16y + 64 - 64
2 b
1
dy(y + 8)
2 - (8)
2 a = 8 u = y + 8 du = dy
∫ duu
2 - a
2
2 ▪ 1
2(8) ln | y + 8 - 8y + 8 + 8 |
2 ▪ 1
2(8) ln | y
y + 16 |] b
1
18
ln| b
b + 16 |─ ln | 1
1 + 16 | 18
ln| ∞∞ |─ ln |
117 |
*so diba infinity over infinity, so bawal yun. Babalik tayo sa equation before this. Yung may b over b + 16. Derive that.
18
ln| 11 |─ ln |
117 |
*recall that ln1 = 0
─ 18 ln
117
*recall that lnab = lna - lnb
─ 18 [ln1 – ln17]
─ 18 [–ln17] =
18 [ln17]
2. ∞
0xe
-x^2dx
limb
∞ b
1 xe
-x^2dx u = ─x2 ─
du2 = xdx
─12
b
1e
udu
─12
1
ex^2 ]
b
0
─12
1e
b^2 ─ 1e
0
─12
1∞ ─
11
─12 ▪ ─1 =
12
3. ∞
2
dxx
2 + 1
limb
∞ b
1
duu
2 + a
2
12a ln |
u - au + a | =
12 ln|
x - 1x + 1 |]
b
2
12
ln| b - 1b + 1 |─ ln |
2 - 12 + 1 |
*so diba infinity over infinity, so derive the numerator and the denominator
12
ln| 11 |─ ln |
13 |
12
─ ln | 13 |
─ 12 ln
13 = ─
12 [ln1 – ln3] =
12 ln3
II. Integrals with infinite discontinuities in the integrand *in other words, isa or both a and b sa formula
na b
af(x)dx, pag sinubstitute sa f(x)dx,
UNDEFINED yung lalabas. a) If f(x) increases numerically without limit as x a, then
n
mf(x)dx = lima
m+
n
af(x)dx
a) If f(x) increases numerically without limit as x b, then
n
mf(x)dx = limb
n-
b
mf(x)dx
a) If f(x) increases numerically without limit as x c, a < c < b , (kumbaga yung point of discontinuity,
hindi given pero nasa gitna siya ng a and b) then,
b
af(x)dx =
c
af(x)dx +
b
cf(x)dx
= limn
c- n
af(x)dx + limm
c+
b
mf(x)dx
EXAMPLES:
1. 2
0
dx
x(2 - x)
*pag sinubstite both 0 & 2, magiging undefined yung sagot so ii-integrate both limits
lima
0 and b
2 b
a
dx
1 - (x2 - 2x + 1)
b
a
du
a2 - u
2
Sin-1
(x – 1) ] b
𝑎
Sin-1
(b – 1) – Sin-1
(a – 1) Sin
-1(1) – Sin
-1(-1)
π2 ─
─
π2 = π
* ─90° yung Sin-1(-1) instead of 180 kasi pag negative yung value tas Arcsin yung hinahanap, clockwise mo siya babasahin
2. 2
0
dx(x - 1)
2/3
*If you substitute 0 & 2, the value will not be undefined. But if you substitute 1, it will be undefined. So you’ll apply the a < c < b rule.
1
0
dx(x - 1)
2/3 + 2
1
dx(x - 1)
2/3
limb
1- b
0(x – 2)
-2/3dx + lima
1+
2
a(x – 2)
-2/3dx
3(x – 1)1/3]
b
0
+ 3(x – 1)1/3]
2
𝑎
3[ (b – 1)1/3
– (0 – 1)1/3
] + 3[ (2 – 1)1/3
– (a – 1)1/3
] 3[ (1 – 1)
1/3 – (– 1)
1/3 ] + 3[ (1)
1/3 – (1 – 1)
1/3 ]
3 + 3 = 6
EXERCERISES A. 1. A curve is such that y’’’ = 72x + 6
a. The curvature at any point (x,) on the curve: y’’= ______ b. The slop at any point (x,y) on the curve: y’ = _______ c. The general equation of the curve: y = _______
If the curve has a critical point at (0,1) and the curve also passes through (1,3):
d. The values of the constants of integration: c1 = ____ c2 = ____ c3 = ____
e. At x = 1, y = _____ y’ = _____
2. A stone that was tossed upward with a velocity of 16ft/sec from the top of a 96-ft high tower falls to the ground under the influence of gravity only (g = 32ft
2/sec). Determine the
equations of the motion of the stone as functions of time (Show the evaluation of the constants of integration):
a. acceleration: a(t) = __________ b. velocity: v(t) = __________ c. displacement: s(t) = ___________
Based on the equations above, determine: d. time the stone takes before it hits the ground: t = ____s e. its velocity as it hits the ground: v = _____ft/s
3. Determine the area bounded by the curve y = x2 – 3x + 2
and the x-axis, from x = 0 to x = 2: a. A(x) = ___________________ b. the intersections of the curve with the x-axis:
x1 = ____ x2 = ____ c. from x = 0 to x = x1 : c = ____ A = ____ d. from x = x1 to x = 2 : c = ____ A = ____ e. total area from x = 0 to x = 2: AT = ______
4. Find the equation of the curve for which dydx =
(lnx)2
x if the
curve passes through (1,2).
B. Evaluate:
1. ∫
2 x +
1
x
2 dx
2. ∫ (x - 2)(5x + 1)
x dx
3. ∫ x3 + 2x
2 + x dx
4. ∫ 3x
3 + 3x - 5x - 2 dx
5. ∫ (x - e
-2x)
x2 + e
-2x dx
6. ∫ (x + x
2)
26
(1 + 2x)-1 dx
7. ∫ (1 - 2x
2)(3 + lnx - x
2)
-1
x dx
8. ∫ xdx
(3π)4x^2 dx
9. ∫ 3xex^2
sin2ex^2
dx
10. ∫ (1 + e
x^2)
x-1
(x2 + e
x^2) dx
11. ∫
1
3 x-2/3
+ x
x2 - 1
(x1/3
+ x2 - 1 )
-2 dx
12. ∫ (x
-1 + cosx)
cos2(sinx + lnx) dx
13. ∫
2x ─
1cos
2x csc
2(x
2 – tanx) dx
14. ∫ sinh
2(x
2 - cosx)(2x + sinx)
sech(x2 - cosx) dx
15. ∫ x
3 + x
2 - 3 x
2 + 5
x + 1 dx
16. ∫ 2x
(3x + 2)2 dx
17. ∫ (2x-3
+ 3x2 + x
-1)
2 dx
18. ∫ e
(2x + 1)
e(-5x - 2) dx
19. ∫ dx
x(1 + x2)
20. ∫ x + 6
(x + 2)2 dx
21. ∫ 7 - lnx
x(3 + lnx) dx
22. 3
1
xdx
4x - 3 dx
23. ∫ cose
3x
e-3x dx
24. ∫ cos3xsin
33x dx
25. ∫ cotx
lnsinx dx
26. ∫ csc
2y coty
1 + csc2y dy
27. ∫ (2x + 1) 4x2 + 4x - 3 dx
28. ∫ sin2t 4 - cos2t dx
29. ∫ x cotx2 cscx
2 dx
30. ∫ e
2t
1 + 6e2t
+ 9e4t dx
31. ∫ cos(tanx
3)
x-2
cos2x
3 dx
32. ∫ x cosx2 (4
sinx^2) dx
33. ∫ (1 + 6x
2/3)
3 x + 6x5/3 dx
34. ∫ (4 - tanx)
cos2x 4 - tan
2x
dx
35. ∫ (sinx + tanx)2 dx
36. ∫ dx
x + 1 (x + 10) dx
37. ∫ [ sin3y2 + cos
y2 ] cos
y2 dx
38. ∫ p2(p
3 + 5 )(p
3 + 5 )
2.3 + ln4 dx
39. ∫ sinx
cos3x(2 + tan
2x) dx
40. ∫ (1 - 4x
2)
-1/2
(Arccos2x)-4 dx
41. ∫ sin32x(1 + cos4x) dx
42. ∫ sin6x dx
ANSWERS:
A. 1. a. 36x
2 + 6x + c1
b. 12x3 + 3x
2 + c1x + c2
c. 3x4 + x
3 +
c1x2
2 + c2x + c3
d. c1 = -4 c2 = 0 c3 = 1 e. y = 3 y’ = 11
2. a. -32
b. -32t + 16
c. -16t2 + 16t
d. 3 sec
e. -80ft/s
3. a. x
3
3 ─ 3x
2
2 + 2x + c
b. x1 = 1 x2 = 2
c. c = 0 A = 56
d. c = 56 A =
16
e. AT = 1 s.u.
4. y = (lnx
3)
3 + 2
B.
1. 2x2 + 4x + lnx + c
2. 2x5/2
– 6x3/2
– 4x1/2
+ c
3. 2x
5/2
5 + 2x
3/2
3 + c
4. x3 + 3x
2 + 15x + 25ln(x – 2) + c
5. 12 ln(x
2 + e
-2x) + c
6. (x + x
2)
27
27 + c
7. ln(3 + lnx – x2) + c
8. ─18
(3π)-4x^2
ln3π + c
9. ─34 cos2e
x^2 + c
10. 12 ln(x
2 + e
x^2) + c
11. 13 (x
1/3 + x
2 - 1 )
3 + c
12. tan(sinx + lnx) + x 13. ─cot(x
2 – tanx) + c
14. sinh
3(x
2 - cosx)
3 + c
15. x
3
3 ─ 3x + 8ln(x + 1) + c
16. 29 [ ln(3x + 2) +
23x + 2 ] + c
17. ─ 4
5x5 +
9x5
5 ─ 1x + 12lnx ─
43x
3 + 3x2 + c
18. 17 e
7x + 3 + c
19. lnx ─ 12 ln(1 + x
2) + c
20. ln(x + 2) ─ 4
x + 2
21. 7ln(3 + lnx) ─ (3 + lnx) + 3ln(3 + lnx) + c
22. 116
23. 13 sin(e
3x) + c
24. ─ 16 ▪
1cos
23x + c
25. ln(ln sinx) + c
26. 12 ln(1 + csc
2y) + c
27. 16 (4x
2 + 4x – 3)
3/2 + c
28. 4 - cos2t
3 + c
29. ─12 cscx
2 + c
30. ln(1 + 3e2t
) + c
31. 13 sin(tanx
3)
32. 4
sinx^2
2ln4 + c
33. ln(x1/3
+ 2x) + c
34. 4 Sin-1tanx
4 ─ 4 - tan2x + c
35. 12 [x -
12 sin2x] + tanx – x + 2ln(secx + tanx) – 2sinx + c