# solution exam3 review

```Lenarz
Math 262
Exam #3 Review
October 30, 2012
Name:
Directions: Answer the following questions on a separate piece of paper. You may NOT
use a calculator. If you do not show your work, you will receive no credit.
1. Evaluate the following integrals.
Z
1
√
(a)
dx
x2 16 − x2
Solution:
Use the trig substitution x = 4 sin θ so that dx = 4 cos θ dθ and
√
2
16 − x = 4 cos θ. Then we have
Z
Z
1
1
√
(4 cos θ) dθ
dx =
2
2
2
(4 sin θ) (4 cos θ)
x 16 − x
Z
1
1
dθ
=
2
16
Z sin θ
1
=
csc2 θ dθ
16
1
= − cot θ + C
16 1 cos θ
= −
+C
16 sin θ
!
√
16−x2
1
4
+C
= −
x
16
4
√
16 − x2
= −
+C
16x
Z
(b)
x3
1
dx
+x
Solution: Do a partial fraction decomposition:
1
1
A Bx + C
=
=
+ 2
x3 + x
x(x2 + 1)
x
x +1
So we have
1 = A(x2 + 1) + (Bx + C)x = (A + B)x2 + Cx + A
and hence
A+B =0
C=0
A=1
Math 262
October 30, 2012
Page 2
Thus A = 1, B = −1 and C = 0 and we have
Z Z
1
1
x
dx =
−
dx
x3 + x
x x2 + 1
Z
Z
1
x
=
dx −
dx
x
x2 + 1
Z
Z
1
1
1
=
dx −
du
x
2
u
1
= ln |x| − ln |u| + C
2
1
= ln |x| − ln(x2 + 1) + C
2
where the substitution u = x2 + 1 (so that du = 2x dx) was used in the third
line.
Z
(c)
x2
x
dx
+ 6x + 13
Solution: Complete the square in the denominator:
Z
Z
x
x
dx =
dx
2
x + 6x + 13
(x + 3)2 + 4
Z
u−3
=
du
u2 + 4
Z
Z
u
1
=
du
−
3
du
u2 + 4
u2 + 4
Z
Z
1
1
1
=
dw − 3
du
2
2
w
u + 22
1
1
−1 u
ln |w| − 3
tan
=
+C
2
2
2
1
3
2
−1 u
=
ln(u + 4) − tan
+C
2
2
2
3
x+3
1
2
−1
+C
=
ln(x + 6x + 13) − tan
2
2
2
where the substitution u = x + 3 (so du = dx) was made in the second line, and
the substitution w = u2 + 4 (so that dw = 2u du) was made in the fourth line.
Math 262
Z
(d)
October 30, 2012
√
Page 3
x
dx
4 − x2
Solution: Let u = 4 − x2 so that du = −2x dx. Then we have
Z
Z
x
1
√
dx = −
u−1/2 du
2
2
4−x
1
= − (2u1/2 ) + C
2
√
= − 4 − x2 + C
Z
(e)
x2
7
dx
− 6x + 18
Solution: Complete the square in the denominator.
Z
Z
7
7
dx =
dx
2
x − 6x + 18
(x − 3)2 + 9
Z
1
= 7
du
2
u + 32
1
−1 u
tan
+C
= 7
3
3
7
x−3
−1
=
tan
+C
3
3
where the substitution u = x − 3 (so that du = dx) was made in the second line.
Z
(f)
√
x sin
√
x dx
√
Solution: Let w = x so that w2 = x and 2w dw = dx. Then we have
Z
Z
√
√
x sin x dx =
(w sin w)(2w) dw
Z
=
2w2 sin w dw
Now use integration by parts with
u = 2w2
dv = sin w dw
so that
du = 4w dw
v = − cos w
Math 262
October 30, 2012
and we have
Z
2
Page 4
Z
2
2w sin w dw = (2w )(− cos w) − (− cos w)(4w) dw
Z
2
= −2w cos w + 4w cos w dw
Use integration by parts again with
u = 4w
dv = cos w dw
so that
du = 4 dw
v = sin w
and we have
2
−2w cos w +
Z
Z
2
4w cos w dw = −2w cos w + (4w)(sin w) − (sin w)(4) dw
Z
2
= −2w cos w + 4w sin w − 4 sin w dw
= −2w2 cos w + 4w sin w + 4 cos w + C
√
√
√
√
√
= −2( x)2 cos x + 4 x sin x + 4 cos x + C
√
√
√
= (−2x + 4) cos x + 4 x sin x + C
Z
(g)
x3/2 ln x dx
Solution: Use integration by parts with
u = ln x
so that
du =
1
dx
x
dv = x3/2 dx
2
v = x5/2
5
Then we have
Z Z
2 5/2
2 5/2
1
3/2
x ln x dx = (ln x)
x
−
x
dx
5
5
x
Z
2 5/2
2
=
x ln x −
x3/2 dx
5
5
2 5/2
2 2 5/2
=
x ln x −
x
+C
5
5 5
2 5/2
4
=
x ln x − x5/2 + C
5
25
Math 262
Z
(h)
October 30, 2012
Page 5
x
√
dx
3
x−1
√
√
Solution: Let u = 3 x − 1 so that u + 1 = 3 x and (u + 1)3 = x. This means
that 3(u + 1)2 du = dx and we have
Z
Z
x
(u + 1)3
√
dx
=
(3(u + 1)2 ) du
3
u
x−1
Z
(u + 1)5
=
du
u
Z 1
4
3
2
u + 5u + 10u + 10u + 5 +
= 3
du
u
1 5 5 4 10 3
2
= 3
u + u + u + 5u + 5u + ln |u| + C
5
4
3
√
√
3 √
15 √
=
( 3 x − 1)5 + ( 3 x − 1)4 + 10( 3 x − 1)3 + 15( 3 x − 1)2
5
4 √
√
+15( 3 x − 1) + 3 ln | 3 x − 1| + C
Z
(i)
sin4 x cos3 x dx
Solution: Let u = sin x so that du = cos x dx. Then we have
Z
Z
4
3
sin x cos x dx =
sin4 x cos2 x cos x dx
Z
=
sin4 x(1 − sin2 x) cos x dx
Z
=
u4 (1 − u2 ) du
Z
=
(u4 − u6 ) du
1 5 1 7
u − u +C
5
7
1 5
1
=
sin x − sin7 x + C
5
7
=
```