Solução_Calculo_Stewart_6e

F.
21.
dx
√
x2 − 4x =
dx
(x2 − 4x +4)− 4 =
2secθ tan θdθ
=
2tanθ
TX.10
dx
(x − 2) 2 − 2 2
x − 2=2secθ,
dx =2secθ tan θdθ
CHAPTER 7 REVIEW ¤ 345
= sec θdθ=ln|sec θ +tanθ| + C 1
=ln
x − 2
√ x2 − 4x
+
2 2 + C1
=ln x − 2+ √ x 2 − 4x + C,whereC = C 1 − ln 2
23. Let x =tanθ,sothatdx =sec 2 θdθ.Then
25.
27.
dx
x √ x 2 +1 =
sec 2
θdθ
tan θ sec θ =
sec θ
tan θ dθ
= csc θdθ=ln|csc θ − cot θ| + C
√ =ln
x2 +1
− 1 √ x x + C =ln x2 +1− 1
x + C
3x 3 − x 2 +6x − 4
(x 2 +1)(x 2 +2) = Ax + B
x 2 +1 + Cx + D
x 2 +2
⇒ 3x 3 − x 2 +6x − 4=(Ax + B) x 2 +2 +(Cx + D) x 2 +1 .
Equating the coefficients gives A + C =3, B + D = −1, 2A + C =6,and2B + D = −4
A =3, C =0, B = −3,andD =2.Now
3x 3 − x 2 +6x − 4
(x 2 +1)(x 2 +2) dx =3 x − 1
x 2 +1 dx +2
dx
x 2 +2 = 3 2 ln x 2 +1 − 3tan −1 x + √ x√2
2tan −1 + C.
π/2
cos 3 x sin 2xdx= π/2
cos 3 x (2 sin x cos x) dx = π/2
2cos 4 x sin xdx= − 2 0 0 0 5 cos5 x π/2
= 2 0 5
29. The product of an odd function and an even function is an odd function, so f(x) =x 5 sec x is an odd function.
By Theorem 5.5.7(b), 1
−1 x5 sec xdx=0.
31. Let u = √ e x − 1. Thenu 2 = e x − 1 and 2udu= e x dx. Also,e x +8=u 2 +9.Thus,
ln 10
0
e x√ e x − 1
e x +8
3
u · 2udu 3
dx =
0 u 2 +9 =2 0
u 2 3
u 2 +9 du =2
0
⇒
1 − 9
du
u 2 +9
=2 u − 9 u
3
3 tan−1 =2 (3 − 3tan −1 1) − 0
=2
3
0
3 − 3 · π
4
=6− 3π 2
33. Let x =2sinθ ⇒ 4 − x 2 3/2 =(2cosθ) 3 , dx =2cosθdθ,so
x 2
(4 − x 2 ) 3/2 dx = 4sin 2 θ
8cos 3 θ 2cosθdθ=
=tanθ − θ + C =
tan 2 θdθ=
sec 2 θ − 1 dθ
x
x
√ − 4 − x
2 sin−1 + C
2