Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 5.5 THE SUBSTITUTION RULE ¤ 251
9. Let u =3x − 2. Thendu =3dx and dx = 1 3 du,so (3x − 2) 20 dx = u 20 1
3 du = 1 3 · 1
21 u21 + C = 1
63 (3x − 2)21 + C.
11. Let u =2x + x 2 .Thendu =(2+2x) dx =2(1+x) dx and (x +1)dx = 1 du, so
2
(x +1) √u
2x + x 2 dx =
1 du = 1 u 3/2
2
2 3/2 + C = 1
3 2x + x
2 3/2
+ C.
Or: Letu = √ 2x + x 2 .Thenu 2 =2x + x 2 ⇒ 2udu =(2+2x) dx ⇒ udu =(1+x) dx, so
(x +1)
√
2x + x2 dx = u · udu= u 2 du = 1 3 u3 + C = 1 3 (2x + x2 ) 3/2 + C.
13. Let u =5− 3x. Thendu = −3 dx and dx = − 1 3
du, so
dx 1
5 − 3x =
−
1
u
du = − 1 ln |u| + C = − 1 ln |5 − 3x| + C.
3 3 3
15. Let u = πt. Thendu = πdtand dt = 1 du,so sin πt dt = sin u 1
du = 1 (− cos u)+C = − 1 cos πt + C.
π π π π
17. Let u =3ax + bx 3 .Thendu =(3a +3bx 2 ) dx =3(a + bx 2 ) dx, so
a + bx 2
√ dx = 3ax + bx
3
1 du 3
u = 1
1/2 3
u −1/2 du = 1 3 · 2u2 + C = 2 3
3ax + bx3 + C.
19. Let u =lnx. Thendu = dx (ln x)
2
x ,so dx = u 2 du = 1 3
x
u3 + C = 1 (ln 3 x)3 + C.
21. Let u = √ t.Thendu = dt
2 √ t and √ 1 √
cos t
dt =2du,so √ dt = cos u (2 du) =2sinu + C =2sin √ t + Ċ.
t t
23. Let u =sinθ. Thendu =cosθdθ,so cos θ sin 6 θdθ= u 6 du = 1 7 u7 + C = 1 7 sin7 θ + C.
25. Let u =1+e x .Thendu = e x dx,so e x√ 1+e x dx = √ udu= 2 3 u3/2 + C = 2 3 (1 + ex ) 3/2 + C.
Or: Let u = √ 1+e x .Thenu 2 =1+e x and 2udu = e x dx, so
e
x √ 1+e x dx = u · 2udu= 2 3 u3 + C = 2 3 (1 + ex ) 3/2 + C.
27. Let u =1+z 3 .Thendu =3z 2 dz and z 2 dz = 1 3 du,so
z 2
3√ dz = 1+z
3
u −1/3 1
3 du = 1 3 · 3
2 u2/3 + C = 1 2 (1 + z3 ) 2/3 + C.
29. Let u =tanx. Thendu =sec 2 xdx,so e tan x sec 2 xdx= e u du = e u + C = e tan x + C.
31. Let u =sinx. Thendu =cosxdx,so
[or −csc x + C ].
cos x
sin 2 x dx =
1
u 2 du =
u −2 du = u−1
−1 + C = − 1 u + C = − 1
sin x + C
33. Let u =cotx. Thendu = − csc 2 xdxand csc 2 xdx= −du,so
√cot
x csc 2 xdx=
√u
(−du) =−
u 3/2
3/2 + C = − 2 3 (cot x)3/2 + C.