Solução_Calculo_Stewart_6e

F.
√ 3
x
√
x2 + x +1 dx = tan θ − √ 1
2 2 3
√
3
sec θ 2 sec2 θdθ
2
= √
3
2 tan θ − 1 2
TX.10
sec θdθ= √
3
tan θ sec θdθ− 1
2 2
sec θdθ
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 307
= √ 3
2 sec θ − 1 2 ln |sec θ +tanθ| + C 1
= √ x 2 + x +1− 1 2 ln 2 √3
√
x2 + x +1+ 2 √
3
x +
1
2
+ C1
= √ x 2 + x +1− 1 2 ln 2 √3
√
x2 + x +1+ x + 1 2
+ C1
= √ x 2 + x +1− 1 2 ln 2 √
3
− 1 2 ln √ x 2 + x +1+x + 1 2
+ C1
= √ x 2 + x +1− 1 2 ln√ x 2 + x +1+x + 1 2
+ C, whereC = C1 − 1 2 ln 2 √
3
27. x 2 +2x =(x 2 +2x +1)− 1=(x +1) 2 − 1. Letx +1=1secθ,
so dx =secθ tan θdθand √ x 2 +2x =tanθ. Then
√
x2 +2xdx= tan θ (sec θ tan θdθ)= tan 2 θ sec θdθ
= (sec 2 θ − 1) sec θdθ= sec 3 θdθ− sec θdθ
= 1 sec θ tan θ + 1 ln |sec θ +tanθ| − ln |sec θ +tanθ| + C
2 2
= 1 sec θ tan θ − 1 ln |sec θ +tanθ| + C = 1 (x +1)√ 2 2 2
x 2 +2x − 1 ln √ 2 x +1+ x2 +2x + C
29. Let u = x 2 , du =2xdx.Then
x
√
1 − x4 dx = √ 1 − u 2 1
2 du = 1 2
= 1 2
where u =sinθ, du =cosθdθ,
cos θ · cos θdθ
and √ 1 − u 2 =cosθ
1 (1 + cos 2θ) dθ = 1 θ + 1 sin 2θ + C = 1 θ + 1 sin θ cos θ + C
2 4 8 4 4
= 1 4 sin−1 u + 1 4 u √ 1 − u 2 + C = 1 4 sin−1 (x 2 )+ 1 4 x2 √ 1 − x 4 + C
31. (a) Let x = a tan θ,where− π 2