32 STEP 2. Determine Your Test Readiness
Chapter 10
28.
∫
1 −x^2
x^2
dx=
∫ (
1
x^2
−
x^2
x^2
)
dx
=
∫ (
1
x^2
− 1
)
dx
=
∫
(x−^2 −1)dx=
x−^1
− 1
−x+C
=−
1
x
−x+C
KEY IDEA
You can check the answer by
differentiating your result.
- Letu=ex+1;du=exdx.
f(x)=
∫
ex
ex+ 1
dx=
∫
1
u
du
=ln|u|+C=ln|ex+ 1 |+C
f(0)=ln|e^0 + 1 |+C=ln (2)+C
Sincef(0)=ln 2⇒ln (2)+C
=ln 2⇒C= 0.
Thus, f(x)=ln (ex+1) andf(ln 2)
=ln (eln 2+1)=ln (2+1)
=ln 3.
- See Figure DS-10.
To find the points of intersection, set
sin 2x=
1
2
⇒ 2 x=sin−^1
(
1
2
)
⇒ 2 x=
π
6
or 2x=
5 π
6
⇒x=
π
12
orx=
5 π
12
.
Volume of solid
=π
∫ 5 π/ 12
π/ 12
[
(sin 2x)^2 −
(
1
2
) 2 ]
dx.
Using your calculator, you obtain:
Volume of solid≈(0.478306)π
≈ 1. 50264 ≈ 1. 503.
Figure DS-10
31.
∫ 5
2
1
x^2 + 2 x− 3
dx=
∫ 5
2
1
(x+3)(x−1)
dx
Use a partial fraction decomposition with
A
(x+3)
+
B
(x−1)
=
1
(x+3)(x−1)
, which
givesA=
− 1
4
andB=
1
4
. Then the integral
becomes
=
∫ 5
2
− 1 / 4
(x+3)
dx+
∫ 5
2
1 / 4
(x−1)
dx
=
− 1
4
∫ 5
2
1
(x+3)
dx+
1
4
∫ 5
2
1
(x−1)
dx
=
− 1
4
ln
∣∣
x+ 3
∣∣
+
1
4
ln
∣∣
x− 1
∣∣]^5
2
=
1
4
(
ln
∣∣
∣∣x−^1
x+ 3
∣∣
∣∣
)] 5
2
=
1
4
[(
ln
∣∣
∣∣^4
8
∣∣
∣∣
)
−
(
ln
∣∣
∣∣^1
5
∣∣
∣∣
)]
=
1
4
ln
(
5
2
)
- Integrate
∫
x^2 cosxdxby parts withu=x^2 ,
du= 2 xdx,dv=cosxdx, andv=sinx.
The integral becomes
=x^2 sinx−
∫
sinx(2x)dx
=x^2 sinx− 2
∫
xsinxdx.
Use parts again for the remaining integral,
lettingu=x,du=dx,dv=sinxdx, and
v=−cosx. The integral
=x^2 sinx− 2
[
−xcosx−
∫
(−cosx)dx
]