Solutions to Math 142 Homework # 5
π/3
Z
tan θ
dθ
cos θ
1.
0
Solution. Substitution u = cos θ, du = − sin θ dθ; if x = 0, u = 1; if x =
Z
π/3
0
Z
tan θ
dθ =
cos θ
Z
π/3
0
sin θ
dθ = −
cos2 θ
1/2
Z
1
1
du =
u2
Z
1
1/2
1
du =
u2
π
3,
u = 12 :
−
1
1 = −1 + 2 = 1.
u 1/2
x5 cos x3 dx
2.
Solution. Substitution u = x3 , du = 3x2 dx and then integration by parts:
Z
Z
Z
1
1
1 3
5
3
x cos x dx =
u cos u du =
u sin u − sin u du =
x sin x3 + cos x3 + C.
3
3
3
Z
xe2x dx
3.
Solution. Integration by parts:
Z
Z
1
1
1 2x 1
2x
e2x dx = xe2x − e2x + C.
xe dx = xe −
2
2
2
4
Z
(ln x)2 dx
4.
Solution. Integration by parts twice:
Z
Z
2
Z
1
2
1 · (ln x) dx = x(ln x) − x · 2(ln x) dx = x(ln x) − 2 1 · ln x dx
x
Z
2
= x(ln x) − 2 x ln x − x · 1x dx = x(ln x)2 − 2x ln x + 2x + C.
2
(ln x) dx =
1/2
Z
5.
Z
2
sin−1 x dx
0
Solution. Integration by parts f = arcsin x, f 0 =
1/2
Z
−1
1 · sin
x dx =
0
4
Z
6.
1/2
x arcsin x|0
1
4
√
Z
Z
−
√
t, du =
0
1
2u dt;
2
t ln t dt =
1
7.
1/2
g 0 = 1, g = x:
√
1/2
x
1 π p
π
3
2
√
dx = · + 1 − x =
+
− 1.
2
2 6
12
2
0
1−x
√
t ln t dt
Solution. Substitution u =
Z
Z
√ 1
,
1−x2
1
u ln u2 · 2u du = 4
if t = 1, u = 1; if t = 4, u = 2; then integration by parts:
Z
1
2
2
Z
4 3
4 2 2
u ln u −
u du
3
3 1
1
2
32
4 3 32
32 4
32
28
=
ln 2 − u =
ln 2 −
+ =
ln 2 − .
3
9 1
3
9
9
3
9
u2 ln u du =
x−9
dx
(x + 5)(x − 2)
Solution.
x−9
2
1
=
−
,
(x + 5)(x − 2)
x+5 x−2
so
Z
3
Z
8.
2
1
dx
x2 − 1
Solution.
1
1
1
1
=
−
,
(x + 1)(x − 1)
2 x−1 x+1
so
Z
2
Z
9.
3
3
1
1
1
1
1
1
1
1
dx = ln |x − 1| − ln |x + 1| = ln 2 − ln 4 + ln 3 = ln 3 − ln 2.
2
x −1
2
2
2
2
2
2
2
2
1
√
dx
x− x+2
Solution. Substitution u =
Z
x−9
dx = 2 ln |x + 5| − ln |x − 2| + C.
(x + 5)(x − 2)
1
√
dx =
x− x+2
Z
√
x + 2, du =
√1
2 x+2
2u
du =
u2 − u − 2
Z
dx =
1
2u
dx, then partial fractions:
Z
Z
2u
2
1
4
1
du =
du +
du
(u + 1)(u − 2)
3
u+1
3
u−2
4 √
2 √
= ln x + 2 + 1 + ln x + 2 − 2 + C.
3
3
10. Here is a neat calculus proof that zero is equal to one:
R
Proof that 0 = 1. Let us apply the integration by parts formula to the integral
Z
Z
1
1
dx =
· 1 dx
x
x
1
x
dx:
Let f (x) = x1 and g 0 (x) = 1. Then f 0 (x) = − x12 and g(x) = x. Therefore
Z
Z 0
Z Z
Z
Z
1
1
1
1
1
1
1
· 1 dx =
· (x)0 dx = · x −
· x dx = 1 −
− 2 · x dx = 1 +
·
x
dx
=
1
+
dx
x
x
x
x
x
x2
x
We have obtained the following equality
Z
Now subtracting
R
1
x
1
dx = 1 +
x
Z
1
dx
x
dx from both sides we get
0=1
This proves that zero is equal to one.
But zero is not equal to one! There must be some mistake in the above proof. Find this mistake (if it exists). Or
maybe zero is equal to one after all?
Solution. The mistake is in the very last step - when we subtract indefinite integrals we need to pay
attention to constants:
Z
Z
1
1
dx −
dx = C - a constant that is not necessarily equal to zero.
x
x
Therefore the subtraction should be performed in the following way:
Z
Z
1
1
dx = 1 +
dx
x
x
Z
Z
Z
Z
1
1
1
1
dx −
dx = 1 +
dx −
dx
x
x
x
x
C = 1 + D.
The above equality will no longer imply that 0 = 1.