Solução_Calculo_Stewart_6e

F.
300 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
67. Using the formula for volumes of rotation and the figure, we see that
Volume = d
0 πb2 dy − c
0 πa2 dy − d
c π[g(y)]2 dy = πb 2 d − πa 2 c − d
c π[g(y)]2 dy. Lety = f(x),
which gives dy = f 0 (x) dx and g(y) =x, sothatV = πb 2 d − πa 2 c − π b
a x2 f 0 (x) dx.
Now integrate by parts with u = x 2 ,anddv = f 0 (x) dx ⇒ du =2xdx, v = f(x), and
b
a x2 f 0 (x) dx = x 2 f(x) b
a − b
V = πb 2 d − πa 2 c − π b 2 d − a 2 c −
b
2xf(x) dx a
2xf(x) dx = a b2 f(b) − a 2 f(a) − b
2xf(x) dx,butf(a) =c and f(b) =d ⇒
a
= b
2πx f(x) dx.
a
7.2 Trigonometric Integrals
The symbols
s = and c = indicate the use of the substitutions {u =sinx, du =cosxdx} and {u =cosx, du = − sin xdx}, respectively.
1. sin 3 x cos 2 xdx = sin 2 x cos 2 x sin xdx= (1 − cos 2 x)cos 2 x sin xdx= c (1 − u 2 )u 2 (−du)
= (u 2 − 1)u 2 du = (u 4 − u 2 ) du = 1 5 u5 − 1 3 u3 + C = 1 5 cos5 x − 1 3 cos3 x + C
3.
3π/4
sin 5 x cos 3 xdx = 3π/4
sin 5 x cos 2 x cos xdx= 3π/4
sin 5 x (1 − sin 2 x)cosxdx= s √ 2/2
u 5 (1 − u 2 ) du
π/2 π/2 π/2 1
= √ 2/2
(u 5 − u 7 ) du = 1
1 6 u6 − 1 8 u8√ 2/2
=
1
1/8
− 1/16
6 8
5. Let y = πx,sody = πdxand
sin 2 (πx) cos 5 (πx) dx = 1 π sin 2 y cos 5 ydy= 1 π sin 2 y cos 4 y cos ydy
− 1
− 1
6 8 = −
11
384
= 1 π
= 1 π
sin 2 y (1 − sin 2 y) 2 cos ydy = s 1
π u 2 (1 − u 2 ) 2 du = 1 π (u 2 − 2u 4 + u 6 ) du
1
3 u3 − 2 5 u5 + 1 7 u7 + C = 1
3π sin3 y − 2
5π sin5 y + 1
7π sin7 y + C
= 1
3π sin3 (πx) − 2
5π sin5 (πx)+ 1
7π sin7 (πx)+C
7.
π/2
cos 2 θdθ = π/2
0 0
= 1 2
1
(1 + cos 2θ) dθ [half-angle identity]
2
θ +
1
sin 2θ π/2
= 1 π +0 − (0 + 0) = π 2 0 2 2 4
9.
π
0
sin4 (3t) dt = π
0
= 1 π
4 0
= 1 4
sin 2 (3t) 2 dt = π
1 (1 − cos
0 2 6t)2 dt = 1 π (1 − 2cos6t 4 0 +cos2 6t) dt
3 − 2cos6t + 1 cos 12t 2 2
dt
1 − 2cos6t +
1
2 (1 + cos 12t) dt = 1 4
3
2 t − 1 3
sin 6t +
1
24 sin 12t π
0 = 1 4
3π
2 − 0+0 − (0 − 0+0) = 3π 8
π
0
11. (1 + cos θ) 2 dθ =
(1 + 2 cos θ +cos 2 θ) dθ = θ +2sinθ + 1 2 (1 + cos 2θ) dθ
= θ +2sinθ + 1 θ + 1 sin 2θ + C = 3 θ +2sinθ + 1 sin 2θ + C
2 4 2 4
13.
π/2
sin 2 x cos 2 xdx = π/2 1
(4 0 0 4 sin2 x cos 2 x) dx = π/2 1
(2 sin x cos
0 4 x)2 dx = 1 π/2
4
= 1 4
π/2
0
1
2 (1 − cos 4x) dx = 1 8
π/2
0
(1 − cos 4x) dx = 1 8
sin 2 2xdx
0
x −
1
sin 4x π/2
= 1 π
4 0 8 2
=
π
16