CHAPTER 29
Trigonometric Integrands
and Substitutions
29.1
29.2
29.3
29.4
29.5
29.6
29.7
29.8
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Find J cos
2 ax dx.
Find / sin
2 ax dx.
Using Problem 29.1, J sin
2 ax dx =
In Problems 29.3-29.16, find the indicated antiderivative.
Now,
you show that this answer agrees with Problem 28.36?
Hence, the entire answer is
So we get
J sin x cos
2 x dx.
Let M = cos x, du = —sin x dx.Then J sin x cos2 dx = - J u2 du = - i«3 + C = - \ cos3 x + C.
J sin
4 x cos
5 x dx.
Since the power of cos* is odd, let « = sin;e, du = cos x dx. Then Jsin
4 A: cos
5 x dx = J sin
4 x(l —
sm2x)2cosxdx = $u\l-u2)2 du = J «4(1 - 2«2 + u4) du = J (u4 - 2u6 + «8) du = ^w5 - §w7 + \u" + C =
w5(i - |«2 + |a4) + C = sin5 x($ - f sin2 * + | sin4 x) + C.
J cos
6 x dx.
Also, in
let u = sin 2x, du=2 cos 2* djt.
(1 + 3 cos 2x + 3 cos
2 2x + cos
3 2*) dx =
$cos
6 xdx = J (cos
2 x)
3 <& =
J(l-sin
2 2jt)cos2;edx,
[Can
J cos
4 x sin
2 x dx.
J cos
4 x sin
2 x dx =
cos 2x - cos 2x - cos 2*) dx = I (x + | sin 2x - J cos 2x dx - J cos 2x dx). Now, J cos 2x dx = \ (x +
\ sin 2x cos 2x) by Problem 29.1. Also, J cos3 2x dx = / (1 - sin2 2x) cos 2x dx = / cos 2x dx -
J sin2 2* cos 2x dx = | sin 2x — \ sin3 IK. Hence, we get \ [x + \ sin 2x — \ (x + | sin 2x cos 2x) + \ sin 2x -
g sin3 2x] + C = s [(x/2) + sin 2x - \ sin 2x cos 2x - g sin3 2*] + C.
Let * = 2«, <& = 2 d«. Then
J tan
4 x dx.
J tan
4 x dx = / tan
2 * (sec
2 * - 1) dx = J tan
2 x sec
2 x <& - J tan
2 x dx = j tan
3 x - J (sec
2 x - 1) dx
= 3 tan
3 x - tan x + x + C