Solução_Calculo_Stewart_6e

F.
252 ¤ CHAPTER 5 INTEGRALS
TX.10
sin 2x
sin x cos x
35.
1+cos 2 x dx =2 dx =2I. Letu =cosx. Thendu = − sin xdx,so
1+cos 2 x
udu
2I = −2
1+u = −2 · 1 ln(1 + 2 2 u2 )+C = − ln(1 + u 2 )+C = − ln(1 + cos 2 x)+C.
37.
Or: Let u =1+cos 2 x.
cos x
cot xdx=
sin x dx. Letu =sinx. Thendu =cosxdx,so
39. Let u =secx. Thendu =secx tan xdx,so
sec 3 x tan xdx= sec 2 x (sec x tan x) dx = u 2 du = 1 3 u3 + C = 1 3 sec3 x + C.
41. Let u =sin −1 x.Thendu =
1
√ dx,so 1 − x
2
1
cot xdx= du =ln|u| + C =ln|sin x| + C.
u
dx
√
1 − x2 sin −1 x = 1
u du =ln|u| + C =ln sin −1 x + C.
43. Let u =1+x 2 .Thendu =2xdx,so
1+x
1+x 2 dx =
1
1+x dx + 2
x
1+x dx 2 =tan−1 x +
1
2 du
u
=tan−1 x + 1 2 ln|u| + C
=tan −1 x + 1 2 ln 1+x 2 + C =tan −1 x + 1 2 ln 1+x 2 + C [since 1+x 2 > 0].
45. Let u = x +2.Thendu = dx,so
x
u − 2
4√ dx = x +2
4√ du = (u 3/4 − 2u −1/4 ) du = 4 7 u7/4 − 2 · 4
3 u3/4 + C
u
= 4 7 (x +2)7/4 − 8 3 (x +2)3/4 + C
In Exercises 47–50, let f(x) denote the integrand and F (x) its antiderivative (with C =0).
47. f(x) =x(x 2 − 1) 3 . u = x 2 − 1 ⇒ du =2xdx,so
x(x 2 − 1) 3 dx = u 3 1
2 du = 1 8 u4 + C = 1 8 (x2 − 1) 4 + C
Where f is positive (negative), F is increasing (decreasing). Where f
changes from negative to positive (positive to negative), F has a local
minimum (maximum).
49. f(x) =sin 3 x cos x. u =sinx ⇒ du =cosxdx,so
sin 3 x cos xdx= u 3 du = 1 4 u4 + C = 1 4 sin4 x + C
Note that at x = π 2
, f changes from positive to negative and F has a local
maximum. Also, both f and F are periodic with period π,soatx =0and
at x = π, f changes from negative to positive and F has local minima.
51. Let u = x − 1, sodu = dx. Whenx =0, u = −1; whenx =2, u =1. Thus, 2
0 (x − 1)25 dx = 1
−1 u25 du =0by
Theorem 7(b), since f(u) =u 25 is an odd function.