Solução_Calculo_Stewart_6e

F.
302 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
35. Let u = x, dv =secx tan xdx ⇒ du = dx, v =secx. Then
x sec x tan xdx= x sec x −
sec xdx= x sec x − ln |sec x +tanx| + C.
37.
π/2
π/6 cot2 xdx= π/2
π/6 (csc2 x − 1) dx = − cot x − x π/2
= √ √
0 − π π/6 2 − − 3 −
π
6 = 3 −
π
3
39. cot 3 α csc 3 αdα = cot 2 α csc 2 α · csc α cot αdα= (csc 2 α − 1) csc 2 α · csc α cot αdα
= (u 2 − 1)u 2 · (−du) [u =cscα, du = − csc α cot αdα]
= (u 2 − u 4 ) du = 1 3 u3 − 1 5 u5 + C = 1 3 csc3 α − 1 5 csc5 α + C
41. I = csc xdx =
csc x (csc x − cot x)
csc x − cot x
dx =
− csc x cot x +csc 2 x
csc x − cot x
du =(− csc x cot x +csc 2 x) dx. ThenI = du/u =ln|u| =ln|csc x − cot x| + C.
dx. Letu =cscx − cot x
⇒
43.
2a
sin 8x cos 5xdx = 1
[sin(8x − 5x)+sin(8x +5x)] dx = 1
2 2 sin 3xdx+
1
2 sin 13xdx
= − 1 6
1
cos 3x − cos 13x + C
26
45.
47.
2b
sin 5θ sin θdθ = 1
[cos(5θ − θ) − cos(5θ + θ)] dθ = 1
2 2 cos 4θdθ−
1
2 cos 6θdθ=
1
sin 4θ − 1 sin 6θ + C
8 12
1 − tan 2 x
sec 2 x
cos
dx = 2 x − sin 2 x
dx =
cos 2xdx= 1 sin 2x + C
2
49. Let u =tan(t 2 ) ⇒ du =2t sec 2 (t 2 ) dt. Then t sec 2 (t 2 )tan 4 (t 2 ) dt = u 4 1
2 du = 1 10 u5 + C = 1 10 tan5 (t 2 )+C.
In Exercises 51–54, let f(x) denote the integrand and F (x) its antiderivative (with C =0).
51. Let u = x 2 ,sothatdu =2xdx.Then
x sin 2 (x 2 ) dx = sin 2 u 1
du
2
= 1 1
2 2
(1 − cos 2u) du
= 1 4 u −
1
sin 2u + C = 1 u − 1 1 · 2sinu cos u + C
2 4 4 2
= 1 4 x2 − 1 4 sin(x2 )cos(x 2 )+C
We see from the graph that this is reasonable, since F increases where f is positive and F decreases where f is negative.
Note also that f is an odd function and F is an even function.
53. sin 3x sin 6xdx = 1
[cos(3x − 6x) − cos(3x +6x)] dx
2
(cos 3x − cos 9x) dx
= 1 2
= 1 6
1
sin 3x − sin 9x + C
18
Notice that f(x) =0whenever F has a horizontal tangent.
55. f ave = 1 π
2π −π sin2 x cos 3 xdx= 1 π
2π −π sin2 x (1 − sin 2 x)cosxdx
= 1
2π
=0
0
0 u2 (1 − u 2 ) du
[where u =sinx]