Math 142 Homework # 6 (Solutions)
1. Use substitution x = 2 tan θ, −π/2 < θ < π/2, to evaluate
Z
1
√
dx.
2
x x2 + 4
Solution. We have
p
p
x2 + 4 = 4 tan2 θ + 4 =
2
2
=
,
| cos θ|
cos θ
because cos θ > 0 for −π/2 < θ < π/2. Therefore
Z
Z
1
1
1
cos θ
√
dx =
2 dθ = − 4 sin θ + C.
2
2
4
sin
θ
x x +4
Now we need to express the above result in terms of x. Because cos θ > 0, x has the same sign as sin θ:
x2
+4
x
sin θ = √
.
2
x +4
sin2 θ =
x2
This gives the final answer:
√
1
x2 + 4
+ C.
dx = −
4x
x2 x2 + 4
Z
2
2–9. Evaluate the following integrals 2.
dx
2
x −1
Solution.
Z
Z
Z
2
1
1
dx =
dx −
dx = ln |x − 1| − ln |x + 1| + C.
x2 − 1
x−1
x+1
Z
1
3.
dx
2
x (x + 1)
Solution.
Z
Z
Z
1
x
1
1
dx =
dx −
dx = ln |x| − ln x2 + 1 + C.
2
2
x (x + 1)
x
x +1
2
Z
7x − 35
4.
dx
x2 − 5x − 6
Solution.
Z
Z
Z
7x − 35
6
1
dx
=
dx
+
dx = 6 ln |x + 1| + ln |x − 6| + C.
2
x − 5x − 6
x+1
x−6
Z
x+7
5.
dx
2x2 + x + 1
Solution.
Z
Z
Z
x+7
1
1
4x + 1
27
dx
=
dx
+
7 dx
2x2 + x + 1
4
2x2 + x + 1
8
(x + 41 )2 + 16
1
27
4x + 1
2
√
= ln 2x + x + 1 + √ arctan
+ C.
4
2 7
7
Z
6.
e2t sin(3t) dt
Z
√
Solution. Integration by parts:
Z
Z
1
2
4
e2t sin(3t) dt = − e2t cos(3t) + e2t sin(3t) −
e2t sin(3t) dt.
3
9
9
Therefore
Z
2 2t
3
e sin(3t) − e2t cos(3t) + C.
13
13
e2t sin(3t) dt =
1
Z
1 + x2 e−x dx
7.
0
Solution. Integration by parts:
Z 1
1
6
1 + x2 e−x dx = − x2 + 2x + 3 e−x 0 = 3 −
e
0
√
Z
8. √
π
θ3 cos θ2 dθ
π/2
Solution. Substitution u = θ2 and then integration by parts:
√
Z
π
1
θ3 cos θ2 dθ =
√
2
π/2
Z
Z
π
u cos u du =
π/2
π
1
π 1
=− − .
(cos u + u sin u)
2
4
2
π/2
2
1
dt
t2 − 1
2
√
Solution. Substitution u = t2 − 1:
9.
√
t3
√
√
2
Z
√
1
√
dt =
3
t t2 − 1
2
Z
1
3
1
(u2
2
+ 1)
du
Then we can use integration by parts:
Z
Z
1
(u2
2
+ 1)
du =
1
du −
u2 + 1
u2
Z
du = arctan u −
1
2
Z
u·
2u
2 du
+ 1)
+ 1)
Z
u
1
u
1
1
− 2
+
du = arctan u +
+ C.
= arctan u −
2
u +1
u2 + 1
2
2 (u2 + 1)
(u2
Therefore
Z
2
√
2
2
(u2
√
1
π
1
3
√
dt =
− +
.
3
2
24
4
8
t t −1
10, 11.
Z Use integration by parts to
Z prove the following reduction formulas:
n
n−1
10. (ln x) dx = x(ln x)
− n (ln x)n−1 dx
Solution. The above reduction formula is wrong, so it’s impossible to prove it. The correct formula is:
Z
Z
n
n
(ln x) dx = x(ln x) − n (ln x)n−1 dx.
Z
11.
n x
n x
x e dx = x e − n
Z
xn−1 ex dx.
Solution. Integration by parts: u = x2 , du = nxn−1 ; dv = ex , v = ex .
12, 13. Use either the Table of Integrals or Maple to find the following integrals. If you use the Table of
Integrals, please indicate all the formulas you are using. If you use Maple, write down all Maple commands
used.Z
e2x
√
dx
12.
2 + ex
Solution.
Z
√
e2x
2 x
√
dx
=
(e
−
4)
ex + 2 + C.
3
2 + ex
Using
Z the2 Maple command: int(exp(2*x)/sqrt(2+exp(x)),x);.
x +x+5
√
dx
13.
x2 + 1
Solution.
Z 2
p
1 p
9
x +x+5
√
dx = x x2 + 1 + arcsinh x + x2 + 1 + C.
2
2
x2 + 1
Using the Maple command: int((x^2+x+5)/sqrt(x^2+1),x);.
14*. Approximate the following definite integral to six decimal places using the Simpson Rule.
Z 1
sin(x2 ) dx.
0
(You will have to estimate the fourth(!) derivative of sin(x2 ). Use the fact that −1 6 sin(x2 ) 6 1)
Solution. Let f (x) = sin(x2 ). We will use the following error estimate:
|Error| 6
K(1 − 0)5
,
180n4
where K is a number such that f (4) (x) 6 K on [0, 1]. We have:
f (4) (x) = 16x4 − 12 sin(x2 ) − 48x2 cos(x2 ).
Now using the facts that | sin(x2 )| 6 1, | cos(x2 )| 6 1, |16x4 − 12| 6 12, and |a + b| 6 |a| + |b|, we get
|f (4) (x)| 6 12 + 48 = 60.
We can take K = 60. Now we need the error to be less than 0.000001. Therefore:
60
< 0.000001
180n4
1
1000000
n4 >
=
3 · 0.000001
3
n > 24.028.
Since n must be even we take n = 26. Now it is just a matter of applying Simpson’s Rule:
Z 1
1
2
1
f (0) + 4f
+ 2f
+ · · · + f (1) ≈ 0.310268
sin(x2 ) dx ≈
26
26
26
0