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MATH 105 921 Solutions to Integration Exercises


1)

s2 + 1

s2 1 ds

Solution: Performing polynomial long division, we have that:s2 + 1

s2 1 ds =

(1 +2

s2 1) ds

=

ds +

2

s2 1 ds

= s +

2

s2 1 ds

Using partial fraction on the remaining integral, we get:

2s2 1 =

A

s 1 +B

s + 1=

A(s + 1) + B(s 1)(s + 1)(s 1) =

(A + B)s + (A B)s2 1

Thus, A + B = 0 and A B = 2. Adding the two equations together yields 2A = 2,that is, A = 1, and B = 1. So, we have that:

2

s2 1 ds =

1

s 1 ds

1

s + 1ds

Therefore,

s2 + 1s2 1 ds = s +

2s2 1 ds

= s +

1

s 1 ds

1

s + 1ds

= s + ln |s 1| ln |s + 1| + C

2)

04

x

1 + 2x dx

Solution: Using direct substitution with u = 1 + 2x and du = 2dx, we may write

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x = 12(u 1). Moreover, when x = 4, u = 9, and when x = 0, u = 1. Thus,04

x

1 + 2x dx =

19

1

4(u 1)u du

=

1

9

14

(u3

2 u 12 ) du

= (1

10u

5

2 16

u3

2 ) |19= (

1

10 1

6) ( 243

10 27

6)

=298

15

3)

sin2 x cos2 x dx

Solution: Using half-angle identities sin2 x = 1cos(2x)2

and cos2 x = 1+cos(2x)2

, we get:sin2 x cos2 x dx =

1

4(1 cos(2x))(1 + cos(2x)) dx

=

1

4(1 cos2(2x)) dx

= 14

dx

1

4cos2(2x) dx

=x

4 1

4

cos2(2x) dx

On the remaining integral, we apply the half-angle identity cos2(2x) = 1+cos(4x)2 , andobtain:

cos2(2x) dx =

1 + cos(4x)

2dx =

x

2+

1

8sin(4x) + C

Hence, sin2 x cos2 x dx =

x

4 1

4(

x

2+

1

8sin(4x)) + C =

x

8 1

32sin(4x) + C

4)

sin(

w) dw

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Solution: Using direct substitution with t =

w, and dt = 12w

dw, that is, dw =

2

w dt = 2t dt, we get:

sin(w) dw = 2t sin t dtUsing integration by part method with u = 2t and dv = sin t dt, so du = 2 dt andv = cos t, we get:

2t sin t dt = 2t cos t +

2cos t dt = 2t cos t + 2 sin t + C

Therefore,

sin(

w) dw = 2w cos(w) + 2 sin(w) + C

5)

ln(x)

xdx

Solution: Using direct substitution with u = ln(x) and du = 1x

dx, we get:

ln(x)

xdx =

u du =

u2

2+ C

ln(x)

x

dx =1

2

(ln(x))2 + C

6)

sin t cos(2t) dt

Solution: Recall the double-angle formula that cos(2t) = 2 cos2 t 1, we get:sin t cos(2t) dt =

sin t(2cos2 t 1) dt

=

2sin t cos2

t dt sin t dt = 2sin t cos2 t dt + cos tOn the remaining integral, using direct substitution with u = cos t and du = sin t dt,we have that:

2sin t cos2 t dt =

2u2 du = 2

3u3 + C = 2

3cos3 t + C

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Therefore, sin t cos(2t) dt = 2

3cos3 t + cos t + C

7)

x + 1

4 + x2dx

Solution: Observe that we may split the integral as follows:x + 1

4 + x2dx =

x

4 + x2dx +

1

4 + x2dx

On the first integral on the right hand side, we use direct substitution with u = 4+x2,

and du = 2x dx. We get:x

4 + x2dx =

1

2udu = ln |2u| + C = ln(8 + 2x2) + C

On the second integral on the right hand side, we use inverse trigonometric substitu-

tion with 2 tan t = x (or equivalently, t = arctanx

2

), so 2 sec2 t dt = dx. Thus,

1

4 + x2dx =

1

4 + 4 tan2 t2sec2 t dt =

2sec2 t

4sec2 tdt

=1

2 dt =

t

2 + C =

1

2 arctanx

2

+ C

Therefore,x + 1

4 + x2dx =

x

4 + x2dx +

1

4 + x2dx = ln(8 + 2x2) +

1

2arctan

x2

+ C

8)

sin(tan )

cos2 d

Solution: Using direct substitution with u = tan and du = sec2

d, we get:sin(tan )

cos2 d =

sec2 sin(tan ) d =

sin u du = cos u + C

sin(tan )

cos2 d = cos(tan ) + C

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9)

x

3 2x x2 dx

Solution: Completing the square, we get 3 2x x2 = 4 (x + 1)2. Using directsubstitution with u = x + 1 and du = dx, we get:

x

3 2x x2 dx =

(u 1)

4 u2 du =

u

4 u2 du

4 u2 du

For the first integral on the right hand side, using direct substitution with t = 4u2,and dt = 2u du, we get:

u

4 u2 du =1

2

t dt = 1

3t3

2 + C = 13

(4 u2) 32 + C

For the second integral on the right hand side, using inverse trigonometric substitution

with 2 sin s = u, that is, s = arcsinu2

, and 2 cos s ds = du, we get:4 u2 du =

4 4sin2 s2cos s ds =

4cos2 s ds

=

(2 + 2 cos(2s)) ds (using half-angle formula cos2 s =

1 + cos(2s)

2)

= 2s + sin(2s) + C

= 2s + 2 sin s cos s + C (using double-angle formula sin(2s) = 2 sin s cos s)

= 2 arcsinu

2

+ 2 sin(arcsin

u2

) cos(arcsin

u2

) + C

= 2 arcsinu

2

+ u

4 u2

2

+ C

Therefore,x

3 2x x2 dx =

u

4 u2 du

4 u2 du

= 13

(4 u2) 32 2 arcsinu

2

u

4 u2

2

+ C

x

3 2x x2 dx = 13

(4 (x + 1)2) 32 2 arcsinx + 1

2 (x + 1)4 (x + 1)2

2 10)

3

0

sin3 zcos z dz

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Solution: Using direct substitution with u = sin z, and du = cos z dz, when z = 0,then u = 0, and when z = 3 , u =

32 . We have that:

3

0

sin3 zcos z dz = 3

2

0

u3 du =u4

4 |

3

2

0=

9

64

3

0

sin3 zcos z dz =9

64

11)

1

3x2 + 2x + 1dx

Solution: Completing the square, we get that 3x2 + 2x + 1 = 3x + 13

2

+2

3=

2

3

9

2

x +

1

3

2+ 1

. Using direct substitution with u =

32

x +

1

3

, and du =

32

dx, we get:

1

3x2 + 2x + 1dx =

3

2(92(x +13)

2 + 1)dx =

1

2(u2 + 1)du =

12

arctan u + C

1

3x2 + 2x + 1dx =

12

arctan3

2 x +1

3+ C

12)

1

et + 1dt

Solution: Using direct substitution with u = et + 1 and du = et dt, so dt =1

etdu =

1

u 1 du. Hence, we get:

1

et + 1dt =

1

u(u 1) du

Using partial fraction, we get:

1

u(u 1) =A

u+

B

u 1 =A(u 1) + Bu

u(u 1) =(A + B)s + (A)

u(u 1)

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Thus, A + B = 0 and A = 1. So, A = 1, and B = 1. Thus, we have that:1

u(u 1) du =1

udu +

1

u 1 du

Therefore,1

u(u 1) du =1

udu +

1

u 1 du = ln |u| + ln |u 1| + C

1

et + 1dt = ln |et + 1| + ln |et| + C = ln |et + 1| + t + C

13)

e3a cos(3a) da

Solution: Using direct substitution with t = 3a, and dt = 3 da, we get:e3a cos(3a) da =

1

3et cos t dt

Using integration by parts with u = cos t, du = sin t dt, and dv = et dt, v = et, weget:

1

3et cos t dt =

1

3et cos t +

1

3 et sin t dt

Using integration by parts again on the remaining integral with u1 = sin t, du1 =cos t dt, and dv1 = e

t dt, v1 = et, we get:

1

3

et sin t dt =

1

3sin tet 1

3

et cos t dt

Thus, 1

3et cos t dt =

1

3et cos t +

1

3sin tet 1

3

et cos t dt

13 et cos t dt = 16 et cos t + 16 et sin t + CTherefore,

e3a cos(3a) da =1

6e3a cos(3a) +

1

6e3a sin(3a) + C

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14)

x2

1 + x6dx

Solution: Using direct substitution with u = x3, and du = 3x2 dx, we get:

x2

1 + x6dx =

1

3(1 + u2)du = 1

3arctan u + C = 1

3arctan(x3) + C

15)

1

t(ln t)2dt

Solution: Using direct substitution with u = ln(t) and du = 1t

dt, we get:

1t(ln t)2 dt =

1u2 du = 1u + C

1

t(ln t)2dt = 1

ln t+ C

16)

xe2x

(2x + 1)2dx

Solution: Using integration by parts with u = xe2x, du = (e2x + 2xe2x) dx, anddv = (2x + 1)2 dx, v = 1

2(2x+1), we get:

xe2x

(2x + 1)2dx = xe

2x

2(2x + 1)+

e2x + 2xe2x

2(2x + 1)dx

On the remaining integral, using direct substitution with u = 2x + 1, and du = 2 dx,we get:

e2x + 2xe2x

2(2x + 1)dx =

eu1 + (u 1)eu1

4udu =

1

4eu1 du =

1

4eu1 + C =

1

4e2x + C

Therefore, xe2x

(2x + 1)2dx = xe

2x

2(2x + 1)+

1

4e2x + C =

e2x

4(2x + 1)+ C

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17)

(tan x + cot x)2 dx

Solution:

(tan x + cot x)

2

dx =

(tan2

x + 2 tan x cot x + cot2

x) dx

=

(sec2 x 1 + 2 + csc2 x 1) dx (using identities for tan2 x and c

=

(sec2 x + csc2 x) dx

= tan x cot x + C

18) tet2

sin(t2) dt

Solution: Using direct substitution with x = t2 and dx = 2t dt, we get:tet

2

sin(t2) dt =1

2

ex sin x dx

Using integration by parts with u = sin x, du = cos x dx, and dv = ex dx, v = ex, weget:

1

2ex sin x dx =

1

2ex sin x 1

2 ex cos x dx

Using integration by parts again on the remaining integral with u1 = cos x, du1 = sin x dx, and dv1 = ex dx, v1 = ex, we get:

1

2

ex cos x dx =

1

2ex cos x +

1

2

ex sin x dx

Thus, 1

2ex sin x dx =

1

2ex sin x 1

2ex cos x 1

2

ex sin x dx

12 ex sin x dx = 14 ex sin x 14 ex cos x + CTherefore,

tet2

sin(t2) dt =1

4et

2

sin(t2) 14

et2

cos(t2) + C

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19)

2p 4p2 p dp

Solution: Using partial fraction, we get:

2p 4p(p 1) = Ap + Bp 1 = A(p 1) + Bpp(p 1) = (A + B)p + (A)p(p 1)

Thus, A + B = 2 and A = 4. So, A = 4, and B = 2. We have that:2p 4

p(p 1) dp =

4

pdp

2

p 1 dp

2p 4p(p 1) dp = 4 ln |p| 2 ln |p 1| + C

20)

4

3

1(3x 7)2 dx

Solution: Using direct substitution with u = 3x 7, and du = 3 dx, when x = 3,then u = 2, and when x = 4, u = 5. We have that:4

3

1

(3x 7)2 dx =52

1

3u2du =

13u

|52= 1

15+

1

6=

1

10

4

3

1

(3x

7)2

dx =1

10

21)

t3

(2 t2) 52dt

Solution: Using direct substitution with u = 2 t2, and du = 2t dt, we get:t3

(2 t2) 52dt =

t2

(2 t2) 52(t dt) =

2 u

2u5

2

du

=

(u5

2

+

1

2 u3

2

) du

=2

3u

3

2 u 12 + C

t3

(2 t2) 52dt =

2

3(2 t2) 32 (2 t2) 12 + C

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22)

1

x2

4 x2 dx

Solution: Using inverse trigonometric substitution with x = 2sin y, that is, y =arcsin x2, and dx = 2 cos y dy, we get:

1

x2

4 x2 dx =

2cos y

4sin2 y

4 4sin2 ydy =

2cos y

4sin2 y(2 cos y)dy

=

1

4csc2 y dy = 1

4cot y + C

Therefore, 1

x2

4 x2 dx = 1

4cot(arcsin

x2

) + C =

4 x24x

+ C

23)

y2 1 dy

Solution: Using inverse trigonometric substitution with y = sec u, that is, u =

arccos

1y

, and dy = sec u tan u du, we get:

y2 1 dy =

sec2 u 1(sec u tan u du) =

tan2 u sec u du

= (sec2 u 1)sec u du = sec3 u du sec u duFor the second integral on the right hand side, we have that:

sec u du = ln | sec u + tan u| + C

For the first integral on the right hand side, we use the reduction formula:sec3 u du =

1

2tan u sec u +

1

2

sec u du =

1

2tan u sec u +

1

2ln | sec u + tan u| + C

Observe that since u = arccos1y

, we have that tan u =

y2

1. Therefore,sec3 u du

sec u du =

1

2tan u sec u 1


y2 1 dy = 12

y

y2 1 12

ln |y +

y2 1| + C

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24)

x sin x cos x dx

Solution: Using the double angle identity sin(2x) = 2 sin x cos x, we have that:

x sin x cos x dx = 1

2

x sin(2x) dx

Using direct substitution with t = 2x, and dt = 2 dx, we get:

1

2

x sin(2x) dx =

1

8

t sin t dt

Using integration by parts with u = t, du = dt, and dv = sin t dt, v = cos t, we get:1

8 t sin t dt = 1

8t cos t +

1

8 cos t dt = 1

8t cos t +

1

8sin t + C

Therefore, x sin x cos x dx = 1

4x cos(2x) +

1

8sin(2x) + C

25)

(1 + cos )2 d

Solution:(1 + cos )2 d =

(1 + 2 cos + cos2 ) d

=

d + 2

cos d +

cos2 d

= + 2 sin +

1 + cos(2)

2

d (using half-angle formula)

= + 2 sin +

2+

sin(2)

4+ C

(1 + cos )2 d = 32 + 2 sin + 14 sin(2) + C

26)

1

4x x2 dx

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Solution: Completing the square yields 4x x2 = 4 (x 2)2. Using direct substi-tution with u = x 2, and du = dx, we get:

1

4x x2

dx = 1

4 u2

du

Using inverse trigonometric substitution with u = 2 sin t, that is, t = arcsinu

2

, and

du = 2 cos t dt, we get:1

4 u2 du =

2cos t4 sin2 t

dt =

2cos t

2cos tdt =

dt = t + C

14x x2 dx = arcsin

x 2

2

+ C

27)10

11 + x

1

3

dx

Solution: Using direct substitution with u = 1 + x1

3 , and du =1

3x23 dx, so dx =

3x2

3 du = 3(u 1)2 du. When x = 0, u = 1 and when x = 1, u = 2. We have that:10

1

1 + x1

3

dx =

21

3(u 1)2u

du =

21

(3u 6 + 3u

) du

= (3

2u2 6u + 3ln |u|) |21

= (6 12 + 3 ln2) (32 6 + 3 ln1) = 3

2+ 3ln 2

10

1

1 + x1

3

dx = 32

+ 3ln 2.

28)

1

x3 + xdx

Solution: Using partial fractions, we have:

1

x3 + x=

A

x+

Bx + C

x2 + 1=

A(x2 + 1) + (Bx + C)x

x3 + x=

(A + B)x2 + Cx + A

x3 + x

So, A + B = 0, C = 0 and A = 1. So, B = 1 and we get:1

x3 + xdx =

1

xdx

x

x2 + 1dx = ln |x|

x

x2 + 1dx

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On the remaining integral, using direct substitution with u = x2 + 1 and du = 2x dx,we get:

x

x

2

+ 1

dx = 1

2u

du =1

2

ln

|u

|+ C =

1

2

ln(x2 + 1) + C

Therefore, 1

x3 + xdx = ln |x| 1

2ln(x2 + 1) + C

Remark: This involves partial fractions with non-linear factors, which you are not

required to master in this course!

29) ln(1 + t) dtSolution: Using direct substitution with s = 1 + t, and ds = dt, we have that:

ln(1 + t) dt =

ln s ds

Using integration by parts with u = ln s, du =1

sds, and dv = ds, v = s, we get:

ln s ds = s ln s

s

1

sds = s ln s

ds = s ln s s + C

Therefore, ln(1 + t) dt = (1 + t) ln(1 + t) (1 + t) + C

30)

sin(3x) cos(5x) dx

Solution: Using the trigonometric identity that sin a cos b = 12

(sin(a+b)+sin(a

b)),

we get:sin(3x) cos(5x) dx =

1

2(sin(8x) + sin(2x)) dx = 1

16cos(8x) +

1

4cos(2x) + C

Remark: You are not required to memorize any sum to product or product to sum

trigonometric identities!

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31)

1

k2 6k + 9 dk

Solution: By completing the square, we observe that k2 6k + 9 = (k 3)2. So,using direct substitution with u = k

3, and du = dk, we have that:

1

k2 6k + 9 dk =

1

(k 3)2 dk =

1

u2du = 1

u+ C

1

k2 6k + 9 dk = 1

k 3 + C

32)

1

sec x 1 dx

Solution: Since sec x = 1cos x

, we get:

1

sec x 1 dx =

cos x

1 cos x dx =

1 + 11 cos x

dx = x +

1

1 cos x dx

For the remaining integral, use a direct substitution with t = tanx

2

, so dt =

1

2sec2

x2

dx. We also can compute that sec

x2

=

t2 + 1, cosx

2

=

1t2 + 1

and sinx

2 =t

t2 + 1. So, dx =

2

t2 + 1dt. Using double angle formula, we get:

cos x = cos2x

2

sin2

x2

=

1

t2 + 1 t

2

t2 + 1=

1 t2t2 + 1

So, after the substitution, we get:1

1 cos x dx =

1

1 1t2t2+1

2

t2 + 1

dt =

1

t2dt

= 1t

+ C = cotx

2

+ C

Therefore, 1

sec x 1 dx = x cotx

2

+ C

Remark: This is an extremely challenging question; do not panic if you do not know

how to solve it!

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33)

10

2

ex + 1dx

Solution: Using direct substitution with u = ex + 1, and du = ex dx, that isdx

= 1

u1du

. Whenx

= 0,u

= 2, and whenx

= 1,u

=e1

+ 1. So, we get:10

2

ex + 1dx =

e1+12

2u(u 1) du

Using partial fraction, we get:

2u(u 1) =

A

u+

B

u 1 =A(u 1) + Bu

u(u 1) =(A + B)s + (A)

u(u 1)Thus, A + B = 0 and A = 2. So, A = 2, and B = 2. Thus, we have that:

e

1

+1

2

2u(u 1) du =

e

1

+1

2

2u

du

e1

+1

2

2u 1 du

Therefore,

e1+12

2u(u 1) du = (2ln |u| 2 ln |u 1|) |

e1+12 = (2 ln(e

1 + 1) + 2) (2ln2 0)

2

ex + 1dx = 2 ln(e1 + 1) + 2 2 l n 2.

34)

1

c2 6c + 10 dc

Solution: Completing the square yields c2 6c + 10 = (c 3)2 + 1. So, using directsubstitution with u = c 3, and du = dc, we have that:

1

c2 6c + 10 dc =

1

(c 3)2 + 1 dc =

1

u2 + 1du = arctan u + C

1

c2 6c + 10dc = arctan(c

3) + C

35)

f(x)f(x) dx

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Solution: Using direct substitution with u = f(x), and du = f(x) dx, we get:f(x)f(x) dx =

u du =

1

2u2 + C

f(x)f(x) dx = 12(f(x))2 + C

36)

1

x2 + 4x + 5dx

Solution: Completing the square, we get x2 + 4x + 5 = (x + 2)2 + 1. Using directsubstitution with u = x + 2 and du = dx, we get:

1

x2 + 4x + 5dx =

1(x + 2)2 + 1

dx =

1u2 + 1

du = arctan(u) + C

1

x2 + 4x + 5dx = arctan(x + 2) + C

37)

20

1

(3 + 5x)2dx

Solution: Using direct substitution with u = 3 + 5x, and du = 5 dx, when x = 0,then u = 3, and when x = 2, u = 13. We have that:2

0

1

(3 + 5x)2dx =

133

1

5u2du =

15u

|133 = 1

65+

1

15=

2

39

20

1

(3 + 5x)2dx =

2

39

38) sin(ln u) duSolution: Using direct substitution with t = ln u, that is, u = et, and du = et dt, wehave that:

sin(ln u) du =

et sin t dt

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Using integration by parts twice to compute the integral on the right hand side (seethe solution of question 18 for details), we have that:

et sin t dt =1

2

et sin t

1

2

et cos t + C

Therefore,sin(ln u) du =

1

2elnu sin(ln u) 1

2elnu cos(ln u) + C =

1

2u sin(ln u) 1

2u cos(ln u) + C

39)

r(ln r)2 dr

Solution: Using integration by parts with u = (ln r)2, du =2 ln r

rdr, and dv = r dr,

v =r2

2, we get that:

r(ln r)2 dr =

r2(ln r)2

2

r ln r dr

Using integration by parts again on the remaining integral with u1 = ln r, du1 =1

rdr,

and dv1 = r dr, v1 =r2

2, we get that:

r ln r dr =

r2 ln r

2

r

2dr =

r2 ln r

2 r

2

4+ C

Therefore, r(ln r)2 dr =

r2(ln r)2

2 r

2 ln r

2+

r2

4+ C

40)

1x3 x dx

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Solution: Using partial fraction, we get:

1

x3 x =A

x+

B

x + 1+

C

x 1 =A(x2 1) + B(x2 x) + C(x2 + x)

x3 x= (A + B + C)x

2

+ (C B)x + (A)x3 x

Thus, A + B + C = 0, C B = 0 and A = 1. Therefore, A = 1, and B + C = 1,which gives C = 12 and B = 12 . So,

1

x3 x dx =1

xdx

1

2(x + 1)dx +

1

2(x 1) dx

1

x3 x dx = ln |x| 1

2ln |x + 1| + 1

2ln |x 1| + C

Remark: This involves partial fractions with 3 distinct roots in the denominator, which

you are not required to master in this course!

41)

sec3 u du

Solution: We use the reduction formula:sec3 u du =

1

2tan u sec u +

1

2

sec u du =

1

2tan u sec u +

1


42)

x2 2x 8

x 1 dx

Solution: Observe that x2 2x 8 = (x 1)2 9. Using direct substitution witht = x 1, and dt = dx, we get:

x2 2x 8x 1 dx =

t2 9

tdt

Using inverse trigonometric substitution with t = 3sec y, and dt = 3 sec y tan y dy, we

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get:

t2 9

tdt =

9sec2 y 9

3sec y3sec y tan y dy =

3tan2 y dy

=

3(sec2 y 1) dy = 3tan y 3y + C

x2 2x 8x 1 dx = 3 tan(arccos

3

t

) 3 arccos

3

t

+ C =

t2 9 3 arccos

3

t

+

=

(x 1)2 9 3 arccos

3

x 1

+ C

43)

r2 1r

dr

Solution: Using inverse trigonometric substitution with sec s = r, that is, s =

arccos

1

r

, and sec s tan s ds = dr, we get:

r2 1

rdr =

sec2 s 1

sec ssec tan s ds =

tan2 s ds

=

(sec2 s 1) ds = tan s s + C

r2

1

r dr = tan(arccos1

r

) arccos1r+ C = r2 1 arccos1r+ C

44)

(et

2

+ 16)tet2

dt

Solution: Using direct substitution with u = et2

and du = 2tet2

dt, we get:

(et2 + 16)tet

2

dt = 1

2(u + 16) du =

1

4u2 + 8u + C

(et2

+ 16)tet2

dt =1

4e2t

2

+ 8et2

+ C

45)

y ln y dy

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Solution: Using integration by parts with u = ln y, du =1

ydy and dv =

y dy,

v =2

3y3

2 , we get:

y ln y dy = 2

3y3

2 ln y

23

y1

2 dy = 23

y3

2 ln y 49

y3

2 + C

46)

cos

1 + sin2 d

Solution: Using direct substitution with u = sin , and du = cos d, we have that:

cos 1 + sin2 d =

11 + u2 du = arctan u + C

cos

1 + sin2 d = arctan(sin ) + C

47)

1

x2

x2 + 4dx

Solution: Using inverse trigonometric substitution with 2 tan u = x, and 2 sec u du =

dx, we get:1

x2

x2 + 4dx =

1

4tan2 u

4tan2 u + 42sec2 u du =

2sec2 u

8tan2 u sec udu

=

cos2 u

4cos u sin2 udu =

1

4cot u csc u du

= 14

csc u + C = 14

csc(arctanx

2

) + C

1

x2

x2 + 4dx =

x2 + 4

4x+ C

48)

tet

2

dt

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Solution: Using direct substitution with u = t2, and du = 2t dt, we have that:tet

2

dt =

1

2eu du =

1

2eu + C

tet2 dt = 12 et2 + C

49)

cos(t) cos(sin(t)) dt

Solution: Using direct substitution with u = sin(t) dt, and du = cos(t) dt, wehave that:

cos(t) cos(sin(t)) dt =

1 cos u du = 1 sin u + C

cos(t) cos(sin(t)) dt =1

sin(cos(t)) + C

50)

4

0

sin5(x) dx

Solution: Using direct substitution with u = cos x, and du =

sin x dx, when x = 0,

then u = 1, and when x = 4

, u = 12

. We have that:

4

0

sin5(x) dx =

4

0

(sin2(x))2 sin x dx =

4

0

(1 cos2 x)2 sin x dx

=

12

1

(1 u2)2 du = 1

2

1

(1 + 2u2 u4) du

= (u + 23

u3 15

u5) |12

1

= (1

2 +1

32 1

202) (1 +2

3 1

5) = 43

602 +8

15

4

0

sin5(x) dx = 4360

2+

8

15

P 22 f 22

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