Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 305
9. Let x =4tanθ,where− π 2 0, we don’t need the absolute value.)
11. Let 2x =sinθ,where− π ≤ θ ≤ π . Thenx = 1 sin θ, dx = 1 cos θdθ,
2 2 2 2
and √ 1 − 4x 2 = 1 − (2x) 2 =cosθ.
√
1 − 4x2 dx = cos θ 1
cos θ
dθ = 1 2 4 (1 + cos 2θ) dθ
= 1 4 θ +
1
sin 2θ + C = 1 (θ +sinθ cos θ)+C
2 4
= 1 4
sin −1 (2x)+2x √ 1 − 4x 2 + C
13. Let x =3secθ,where0 ≤ θ< π 2 or π ≤ θ< 3π 2 .Then
dx =3secθ tan θdθand √ x 2 − 9=3tanθ,so
√
x2 − 9
dx =
x 3
= 1 3
3tanθ
27 sec 3 θ 3secθ tan θdθ= 1 3
sin 2 θdθ= 1 3
= 1 6 sec−1 x
3
tan 2 θ
sec 2 θ dθ
1 (1 − cos 2θ) dθ = 1 θ − 1 sin 2θ + C = 1 θ − 1 2 6 12 6 6
sin θ cos θ + C
− 1 √
x2 − 9 3
6 x x + C = 1 x
√
x2 − 9
6 sec−1 − + C
3 2x 2
15. Let x = a sin θ, dx = a cos θdθ, x =0 ⇒ θ =0and x = a ⇒ θ = π 2 .Then
√ a
0 x2 a 2 − x 2 dx = π/2
a 2 sin 2 θ (a cos θ) a cos θdθ= a 4 π/2
sin 2 θ cos 2 θdθ= a 4 π/2
1 (2 sin θ cos 0 0 0 2 θ)2 dθ
θ − 1 π/2
4 sin 4θ
= a4
4
π/2
0
sin 2 2θdθ= a4
4
= a4 π
8 2 − 0 − 0 = π 16 a4
17. Let u = x 2 − 7,sodu =2xdx.Then
π/2
0
1
2
x
√
x2 − 7 dx = 1
2
(1 − cos 4θ) dθ =
a4
8
1
√ u
du = 1 2 · 2 √ u + C = x 2 − 7+C.
0