Solução_Calculo_Stewart_6e

F.
15.
cos 5
α
√ dα =
sin α
TX.10
cos 4
α
1 − sin 2 α 2
√ cos αdα= √ cos αdα s (1 − u 2 ) 2
= √ du
sin α sin α u
SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 301
1 − 2u 2 + u 4
=
du = (u −1/2 − 2u 3/2 + u 7/2 ) du =2u 1/2 − 4
u 1/2 5 u5/2 + 2 9 u9/2 + C
17.
= 2
45 u1/2 (45 − 18u 2 +5u 4 )+C = 2 45
√
sin α (45 − 18 sin 2 α +5sin 4 α)+C
sin
cos 2 x tan 3 3
x
xdx =
cos x dx =
c (1 − u 2
)(−du) −1
=
u
u + u du
= − ln |u| + 1 2 u2 + C = 1 2 cos2 x − ln |cos x| + C
19.
cos x +sin2x
sin x
cos x +2sinx cos x cos x
dx =
dx =
sin x
sin x dx +
=ln|u| +2sinx + C =ln|sin x| +2sinx + C
2cosxdx= s 1
du +2sinx
u
Or: Use the formula cot xdx=ln|sin x| + C.
21. Let u =tanx, du =sec 2 xdx.Then sec 2 x tan xdx= udu= 1 2 u2 + C = 1 2 tan2 x + C.
Or: Let v =secx, dv =secx tan xdx.Then sec 2 x tan xdx= vdv = 1 2 v2 + C = 1 2 sec2 x + C.
23. tan 2 xdx= (sec 2 x − 1) dx =tanx − x + C
25. sec 6 tdt = sec 4 t · sec 2 tdt= (tan 2 t +1) 2 sec 2 tdt= (u 2 +1) 2 du
[u =tant, du =sec 2 tdt]
= (u 4 +2u 2 +1)du = 1 5 u5 + 2 3 u3 + u + C = 1 5 tan5 t + 2 3 tan3 t +tant + C
27.
π/3
0
tan 5 x sec 4 xdx = π/3
0
tan 5 x (tan 2 x +1)sec 2 xdx= √ 3
0
u 5 (u 2 +1)du [u =tanx, du =sec 2 xdx]
= √ 3
(u 7 + u 5 ) du = 1
0 8 u8 + 1 6 u6√ 3
= 81 + 27 = 81 + 9 = 81 + 36 = 117
0 8 6 8 2 8 8 8
Alternate solution:
π/3
0
tan 5 x sec 4 xdx = π/3
0
tan 4 x sec 3 x sec x tan xdx= π/3
0
(sec 2 x − 1) 2 sec 3 x sec x tan xdx
= 2
1 (u2 − 1) 2 u 3 du [u =secx, du =secx tan xdx] = 2
1 (u4 − 2u 2 +1)u 3 du
= 2
1 (u7 − 2u 5 + u 3 ) du = 1
8 u8 − 1 3 u6 + 1 4 u4 2
1 = 32 − 64 3 +4 − 1
8 − 1 3 + 1 4
29. tan 3 x sec xdx = tan 2 x sec x tan xdx= (sec 2 x − 1) sec x tan xdx
=
117
8
= (u 2 − 1) du [u =secx, du =secx tan xdx] = 1 3 u3 − u + C = 1 3 sec3 x − sec x + C
31. tan 5 xdx = (sec 2 x − 1) 2 tan xdx= sec 4 x tan xdx− 2 sec 2 x tan xdx+ tan xdx
= sec 3 x sec x tan xdx− 2 tan x sec 2 xdx+ tan xdx
= 1 4 sec4 x − tan 2 x +ln|sec x| + C [or 1 4 sec4 x − sec 2 x +ln|sec x| + C ]
33.
tan 3 θ
cos 4 θ dθ =
tan 3 θ sec 4 θdθ=
tan 3 θ · (tan 2 θ +1)· sec 2 θdθ
= u 3 (u 2 +1)du [u =tanθ, du =sec 2 θdθ]
= (u 5 + u 3 ) du = 1 6 u6 + 1 4 u4 + C = 1 6 tan6 θ + 1 4 tan4 θ + C