9. I sin4 xcos2 xdx.

Solution. I sin

2

2

2

xdx

1 cos2x 2

1 cos2x

dx

x cos

2

2

1

1 2 cos 2x cos2

2x 1 cos 2x dx

8

1

1 cos 2x cos2

2x cos3 2x dx

1 x

sin 4x

sin

3

2x

C.

8

8

8 2

6

10.

Find tgm x dx.

Solution. Keep in mind that tg x = sec2x and that tg2x = sec2x 1. The

steps are few: tgm x dx tgm 2 x tg2

xdx tgm 2

x sec2 x 1 dx

tgm 2 x sec2 xdx tgm 2 xdx

t tg x

tm 2dt tgm 2 xdx

= tgm 1 x tgm 2 xdx.

m 1

Repeated application of this recursion eventually produces tg xdx

or

tg0

xdx. Both are easily computed: tg xdx ln

cos x

C;

tg0

xdx x C.

11.

Obtain a recursion formula for secn xdx.

Solution. Write secnx as secn –2x sec2x and use integration by parts:

secn 2 x sec2

xdx

u secn 2 x du n 2 secn 2 x tg xdx

dv sec2 xdx

v tg x

secn 2 tg x tg x n 2 secn 2

x tg xdx secn 2 x tg x

n 2 secn 2

x sec2

x 1 dx secn 2

x tg x n 2

secn

xdx

n 2 secn 2 xdx.

Collecting secn x dx, we obtain

n 1

secn xdx secn 2

x tg x n 2

secn 2 xdx and therefore

secn xdx secn 2

x tg x

n 2

secn 2 dx .

n 1

n 1

Solution. sec xdx

dx

cos x dx

cos x dx

cos x

2

2

cos

x

1 sin

x

t sin x;

dt

1 ln

1

t

C.

dt cos x dx.

2

t

1 t

2

1

13. 32 sin 6 x cos4 xdx .

Solution. Multiplying the integrand we get

32 sin 6 x cos4 x 2(2 sin x cos x)4

sin 2 x sin 4 2x(1 cos 2x)

sin 4 2x sin 4 2x cos 2x

1

(1 cos 4x)2

sin 4 2x cos 2x

4

32 sin 6

x cos4 xdx

1

dx 1 cos 4xdx

1

dx 1

cos 8xdx

4

2

8

8

sin 4

2x cos 2xdx

x

1

cos 4xd (4x)

x

1

cos 8xd(8x)

4

8

8

64

1 sin 4 2xd(sin 2x)

3x

1 sin 4x

1

sin 8x

1

sin 5 2x C .

8

64

10

2

8

14.

dx

.

(sin x cos x)

2

(t 1) 2

dt

dt

2

dt

ln

t 1

2

C

2

t 1

2

t

1

(t 1)

(t 1)

ln

sin x 1

2

C.

sin x 1

16. tg3 x sec5 x dx.

and that tgx = sec2x 1.

Solution. Recall d(secx) = tgx secx dx

Let t = secx and dt = secx tgx dx. Then tg3 x sec5 x dx

tg2 x sec4 x tg x sec x dx sec2 x 1 sec4 x tg x sec x dx

t 2 1 t 4dt t

6 t4 dt

t7

t5

C sec7 x sec5 x

C .

7

5

7

5

t

4 t 2 dt

t5

t3

C

tg5 x

tg3 x

C.

5

5

3

3

18.

dx

.

sin

3

x cos x

Solution. Integrand

R(sin x, cos x)

1

is even for both sin x

sin 3 x cos x

R( sin x, cos x)

1

1

R(sin x, cos x) .

( sin x)3 ( cos x)

sin 3 x cos x

dt

dx

and 1

tg2 x

1

. Consequently,

cos2

cos2 x

x

dx

dx

(1 tg2 x)d tg x

sin

3

x cos x

tg

3

x cos

4

x

tg

3

x

d tg x

d tg x

1 ctg2

x ln

tg x

C .

3

tg

x

tg x

2

19. I

sin

2xdx

.

4

4

x

cos

x sin

Solution. The substitutions tg x

= t transform the integral to

4

1

4

t4

2t

dt

t tg x;cos

x 1 t2 2

;sin x 1 t2 2

;

I

1 t2

1 t2

2t

dt

1

t

4

sin 2x

;dx

.

1 t2 2

1 t2 2

1 t2

1 t2

2tdt

u t2

du

arctg u C arctg(tg2

x) C.

4

du 2tdt

1 u

2

1 t

21.

dx

.

5 3cos x

x

Solution. The substitution tg

t transforms the integral into

2

dx

2dt

2

dt

5 3 cos x

(1 t

5 3 1 t

2

5 5t

2

3

3t

2

2 )

2

1 t

dt

1 arctg(2t) C

1 arctg(2 tg

x

) C .

1

4t

2

2

2

2

22.

2 sin x

dx .

2 cos x

x

Solution. Let

tg

t .

2

Then

x 2 arctg t , dx

2

dt , sin x

2t

;

cos x

1 t

2

.

1

t

2

1 t 2

1 t

2

Consequently,

2t

2