TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS
243
29.37
29.38
29.39
29.40
29.41
Fig. 29-10
Refer to Fig. 29-10.
Then
Find
By completing the square,
Find
Then
(Fig. 29-9),
Then
Fig. 29-9
Fig. 29-8
Find
Find
Refer to Fig. 29-8.
Let
Find
First, use integration by parts.
For the latter integral, use a trigonometric substitution. Let
Then
Then
Hence, the answer is
Problem 29.2).
Let x = a sin 6, dx = a cos 6 dO.
= a(J" esc 0 dO - J sin 0 dO) = a(ln |csc 0 - cot 0| + cos 0) + C =
Let « = sin ' x, dv = x dx, du =
J A: sin * x dx.
x
2 -4x = (x-2)
2 -4.
So 4* -x
2 = 4- (x -2)
2 .
Let
u = x-2,
du = 2 cos 0 dO.
Let M = 2 sin e
rfw = dx.
x = 2 sin 6,
dx = 2 cos 6 d6. So
dx = I cos x dx + $ cos x sin x <& = sin * + | sin2 x + C.
— cos x( I + sin x) = cos AC + cos x sin *. Thus,
x=sin0, dx=cos0d0.