TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS
241
Fig. 29-4
29.24
29.25
29.26
29.27
29.28
29.29
is present, let
Since
Then
and (Fig. 29-5)
appears, let
Since
Then, using Problem 29.1,
is present, let
Since
and
Then, using Problem 29.2,
Then
md use Problem 29.27
By completing the square, x
2 - 6x + 13 = (x - 3)
2 + 4. Let x-3 = 2tanft
dx = 2 sec
2 9 d6, x
1 -
By Problem 29.1, we have
Fig. 29-5
x = 2 sin 6, dx = 2 cos 6 d0,
= 2 cos ft
x
2 + 9
x = 3 tan e,
dx = 3 sec
2 6 dO,
x
2 + 9 = 9 sec
2 0.
sin « cos e) + C =
4 cos 9.
x = 3 sin ft <& = j cos 9 d0,
jVVi-*
2
d*.
Let
x = sin ft d^ = cos 6 dO,
J sin
2 9 cos 0cos9 d0 = J sin
2 0 cos
2 e d0 =
g[0 - sin e cos 0(1 - 2 sin
2 0)] + C = i[
sin ~' x-x
J e
3
* Vl - e
2 ' dx.
Let Jt = lnu,
6x + 13 = 4sec
2 0 (see Fig. 29-6). Then