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

Sin-1( )45 = θ and Sin-1( )3

5 = β.

252 (θ – β)

*recall the identity sin(θ – β) = sinθcosβ – cosθsinβ. Draw

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

36. 23 Tan

-1 x + 1 3 + c

37. ─12 cosy –

14 cos2y +

y2 +

12 siny + c

38. (p3 + 5 )4.3ln4

12.9 + 3ln4 + c

39. 12 ln(2 + tan2x) + c

40. ─12

(Arccos2x)5

5 + c

41. cos32x

3 ─ cos52x

5 + c

42. 14 [

32 x -

54 sin2x +

14 sin4x +

sin32x6 ]