Solução_Calculo_Stewart_6e

F.
324 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
81. The function y =2xe x2 does have an elementary antiderivative, so we’ll use this fact to help evaluate the integral.
(2x 2 +1)e x2 dx = 2x 2 e x2 dx + e x2 dx = x
2xe x2 dx + e x2 dx
= xe x2 − e x2 dx + e x2 dx
u = x,
du = dx
dv= 2xe x2 dx,
v= e x2
= xe x2 + C
7.6 Integration Using Tables and Computer Algebra Systems
Keep in mind that there are several ways to approach many of these exercises, and different methods can lead to different forms of the answer.
1. We could make the substitution u = √ 2 x to obtain the radical √ 7 − u 2 andthenuseFormula33witha = √ 7.
Alternatively, we will factor √
7
2 out of the radical and use a = . 2
√
7 − 2x
2
dx = √ ⎡
⎤
7
2
2
− x2
dx = 33 √ 2⎣− 1
7
x 2 x 2 2
x
− x2 − sin −1 x
⎦ + C
= − 1 x
7
2
√
7 − 2x2 − √ 2sin −1
2
7 x
+ C
3. Let u = πx ⇒ du = πdx,so
sec 3 (πx) dx = 1 π sec 3 udu= 71 1 1 sec u tan u + 1 ln |sec u +tanu| + C
π 2 2
5.
1
0
= 1
1
sec πx tan πx + ln |sec πx +tanπx| + C
2π 2π
2x cos −1 xdx=2
91 2x 2 − 1
cos −1 x − x √ 1
1 − x 2
=2 1
4
4
· 0 − 0 4
− − 1 · π − 0 4 2
=2
π
8 =
π
4
0
7. Let u = πx,sothatdu = πdx.Then
tan 3 (πx) dx = tan 3 u 1
du
π
= 1 π tan 3 udu 69
= 1 1
π 2 tan2 u +ln|cos u| + C
= 1
2π tan2 (πx)+ 1 π
ln |cos (πx)| + C
9. Let u =2x and a =3.Thendu =2dx and
1
dx
x √ 2 4x 2 +9 = du
2
u 2 =2
√
u2 + a 2
11.
0
−1
4
√
4x2 +9
= −2
9 · 2x
√
du
u √ 28 a2 + u
= −2
2
+ C
2 a 2 + u 2 a 2 u
√
4x2 +9
+ C = − + C
9x
0
t 2 e −t dt =
97 1
−1 t2 e −t − 2 0
0
0
te −t dt = e +2 te −t dt 96 1
= e +2
−1
−1 −1
−1
(−1) (−t − 1) 2 e−t −1
= e +2 −e 0 +0 = e − 2
tan 3 (1/z)
13.
dz
z 2
u =1/z,
du = −dz/z 2
= −
tan 3 udu 69 = − 1 2 tan2 u − ln |cos u| + C
= − 1 tan2
1
2 z − ln cos
1 + C
z