Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 7.2 TRIGONOMETRIC INTEGRALS ¤ 303
57. A =
=2
π/4
−π/4
π/4
0
=1− 0=1
π/4
(cos 2 x − sin 2 x) dx =
−π/4
cos 2xdx
π/4
1
cos 2xdx=2
2 sin 2x = sin 2x π/4
0
0
59. It seems from the graph that 2π
0
cos 3 xdx=0, since the area below the
x-axis and above the graph looks about equal to the area above the axis and
below the graph. By Example 1, the integral is sin x − 1 3 sin3 x 2π
0
=0.
Note that due to symmetry, the integral of any odd power of sin x or cos x
between limits which differ by 2nπ (n any integer) is 0.
61. Using disks, V = π
π π/2 sin2 xdx= π π 1
(1 − cos 2x) dx = π 1
x − 1 sin 2x π
= π π
− 0 − π +0 = π2
π/2 2 2 4 π/2 2 4 4
63. Using washers,
V = π/4
π (1 − sin x) 2 − (1 − cos x) 2 dx
0
= π π/4
0 (1 − 2sinx +sin 2 x) − (1 − 2cosx +cos 2 x) dx
= π π/4
(2 cos x − 2sinx +sin 2 x − cos 2 x) dx
0
= π π/4
(2 cos x − 2sinx − cos 2x) dx = π 2sinx +2cosx − 1 sin 2x π/4
0 2 0
= π √ 2+ √ √
2 − 1 2 − (0 + 2 − 0) = π 2 2 −
5
2
65. s = f(t) = t
sin ωu 0 cos2 ωu du. Lety =cosωu ⇒ dy = −ω sin ωu du. Then
s = − 1 cos ωt
y 2 dy = − 1 1 y3 cos ωt
= 1 (1 − ω 1 ω 3 3ω cos3 ωt).
67. Just note that the integrand is odd [f(−x) =−f(x)].
69.
Or: If m 6= n, calculate
π
−π
1
π
1
sin mx cos nx dx = [sin(m − n)x +sin(m + n)x] dx = 1
cos(m − n)x
− −
2
−π
2 m − n
If m = n,thenthefirst term in each set of brackets is zero.
π cos mx cos nx dx = π 1
[cos(m − n)x +cos(m + n)x] dx.
−π −π 2
If m 6= n, this is equal to 1 2
sin(m − n)x
m − n
+
π
sin(m + n)x
=0.
m + n
−π
If m = n,weget π 1
[1 + cos(m + n)x] dx = π
1
x π sin(m + n)x
+ −π 2 2
= π +0=π.
−π
2(m + n)
−π
π
cos(m + n)x
=0
m + n
−π