Page 319 - Calculus Demystified

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(b) Write Solutions to Exercises

3 2 2 2
sin x cos xdx = sin x(1 − cos x) cos xdx.
Let u = cos x, du =− sin xdx. Then the integral becomes

3 5
2 2 u u
− (1 − u )u du =− + + C.
3 5
Resubstituting x, we obtain the final answer

3 5
3 2 cos x cos x
sin x cos xdx =− + + C.
3 5
2
(c) Let u = tan x, du = sec xdx. Then the integral becomes
4
3 u
u du = + C.
4
Resubstituting x, we obtain the final answer

4
3 2 tan x
tan x sec xdx = + C.
4
(d) Let u = sec x, du = sec x tan x. Then the integral becomes

3
u
2
u du = + C.
3
Resubstituting x, we obtain the final answer
3
sec x
3
tan x sec xdx = + C.
3
5. We do (a), (b), (c), (d).
(a) Use integration by parts twice:

1 1
1
x
x
e sinxdx =sinx·e x − e cosxdx

0 0 0
! "
1 1
x
=[e·sin1−0]− cosxe x − e (−sinx)dx

0 0
1

x
=e·sin1−e·cos1+1− e sinxdx.
0