Page 224 - Calculus Workbook For Dummies

208 Part IV: Integration and Infinite Series

7. You want to solve for # e x sin x dx, so bring them both to the left side and solve.

=
2 e # x sin x dx e x sin x - e x cosx + C
e # x sin x dx = e x sin x - e x cosx + C
2 2
3
g # 3 sin cos x dx = 3 sin / 4 3 x - 10 sin 10 /3 x + C
3
x
4
1. Split off one cosx: # 3 sin cos x cosx dx
2
x
2. Convert the even number of cosines into sines with the Pythagorean Identity.
= # 3 sin x 1 - sin xi cosx dx = # 3 sin cosx dx - # sin x cosx dx
/ 7 3
2
x
_
3 / 4 3 3 10 /3
3. Integrate with u-substitution with u = sin x: = sin x - sin x + C
4 10
/ π 6
*h # cos sin t dt = π
2
4
t
96
0
1 + cos x 1 - cos x
2
2
2
2
1. Convert to odd powers of cosine with trig identities cos x = and sin x = .
2 2
/ π 6 2
2
2
1 + cos t 1 - cos t
= # c m dt
2 2
0
2. Simplify and FOIL.
/ π 6 / π 6 / π 6 / π 6 / π 6
1 1 1 1 1
2
2
2
2
2
3
= # _ 1 - cos t 1 + cos t dt = # 1 dt + # cos tdt- # cos tdt- # cos tdt
2 ^ i
2 h
8 8 8 8 8
0 0 0 0 0
3. Integrate. The first and second are simple; for the third, you use the same trig identity again;
the fourth is handled like you handled problem 7. Here’s what you should get:
/ π 6 / π 6 / π 6 / π 6 / π 6 / π 6
= # 1 dt + # cos tdt- 1 # 1 dt - 1 # cos tdt- # cos tdt+ # sin t 2 cos tdt
1
1
1
1
2
2
2
4
2
8 8 16 16 8 8
0 0 0 0 0 0
/ π 6 / π 6 / π 6
1
1 # 1 # cos tdt+ # 2
2
= dt - 4 sin t 2 cos tdt
16 16 8
0 0 0
/ π 6 / π 6 / π 6
1 1 1 3
= t E - sin t 4 E + sin t 2 E
16 64 48
0 0 0
π 3 3
= - +
96 128 128
π
=
96
*i # sec x tan x dx = 1 sec x - 1 sec x + C
3
3
5
3
5 3
1. Split off sec tan x: = # sec x tan x sec tan x dx
2
2
x
x
2. Use the Pythagorean Identity to convert the even number of tangents into secants.
= # sec x _ sec x - 1i sec tan x dx
2
x
2
= # sec x sec tan x dx - # sec x sec tan x dx
4
2
x
x
1 5 1 3
3. Integrate with u-substitution: = sec x - sec x + C
5 3