Math1024 Answer to Homework 4
Exercise 3.5.10 (3)
When p 6= −2, −1, then
Z
Z
p
(x − 1)(x + 1) dx = [(x + 1)p+1 − 2(x + 1)p ]dx
=
1
2
(x + 1)p+2 −
(x + 1)p+1 + C.
p+2
p+1
When p = −2, −1, we have
Z
2
+ C,
(x − 1)(x + 1)−2 dx = log |x + 1| +
x+1
Z
(x − 1)(x + 1)−1 dx = x − 2 log |x + 1| + C.
Exercise 3.5.11 (2)
Z
Z
x 2
(x + a ) dx =
=
=
=
=
(x2 + 2xax + a2x )dx
Z
1 3
a2x
x +
+ 2 xax dx
3
2 log a
Z
1 3
a2x
2
x +
+
xdax
3
2 log a log a
Z
1 3
a2x
2
2
x
x +
+
xa −
ax dx
3
2 log a log a
log a
1 3
a2x
2
2
x +
+
xax −
ax + C.
3
2 log a log a
(log a)2
Exercise 3.5.11 (3)
Z
xex dx
=−
(x + 1)2
Z
Z
1
1
1
x
xe d
=−
xe +
d(xex )
x+1
x+1
x+1
Z
1
1
1
ex
x
xe +
(x + 1)ex dx = −
xex + ex + C =
+ C.
=−
x+1
x+1
x+1
x+1
x
Exercise 3.5.14
Using integration by parts, we know
Z 1
Z 1
Z 1
Z 1
1
1
a
n
n x
n
x
x
n−1
x a dx =
x da =
(a −
a nx dx) =
−
xn−1 ax dx
log
a
log
a
log
a
log
a
0
0
0
0
1
Denote bn =
R1
0
xn ax dx, Then
bn =
a
n
−
bn−1
log a log a
Thus we have,
n−1
X
n · · · (n − k + 1)
n!
a
+
bn =
(−1)k
a + (−1)n
b0 ,
k+1
log a k=1
(log a)
(log a)n
n≥1
where b0 = a/ log a − 1/ log a.
Exercise 3.5.16 (3)
Z
Z
Z log(x + b)
log(x + a) log(x + b) +
dx = log(x + a)d(log |x + b|) +
dx
x+b
x+a
x+a
Z
Z
log(x + b)
log(x + b)
= log(x + a) log(x + b) −
dx +
dx
x+a
x+a
= log(x + a) log(x + b) + C.
Exercise 3.5.16 (6)
Z log(x +
√
1 + x2 )
2
x
1+ √
2 Z
√
√
2
√1 + x dx
dx = x log(x + 1 + x2 ) − 2x[log(x + 1 + x2 )]
x + 1 + x2
√
Z
2
√
x log(x + 1 + x2 )
√
= x log(x + 1 + x2 ) − 2
dx
2
1
+
x
Z
2
√
√
√
2
= x log(x + 1 + x ) − 2 log(x + 1 + x2 )d 1 + x2
2
√
√
√
= x log(x + 1 + x2 ) − 2 1 + x2 log(x + 1 + x2 ) + 2x + C.
Exercise 3.5.17
(1) From integration by parts,
Z b
b Z b
2
0
0
2
((x + A) + B)df (x) = f (x)((x + A) + B) −
2(x + A)f 0 (x)dx
a
a
a
Z b
b
0
2
= f (x)((x + A) + B) − 2(x + A)f (x) + 2
f (x)dx
a
Then by removing terms, we have the stated result.
(2)First, we find A such that
(b + A)f (b) − (a + A)f (a) = f (a)
2
b−a
b−a
+ f (b)
,
2
2
a
we deduce A = −(a + b)/2.
Then we take B = −(b − a)2 /4 to make
f 0 (x)((x + A)2 + B)|ba = 0.
With these values, we get
Z b
Z
f (a) + f (b)
1 b
f (x)dx =
(b − a) +
((x + A)2 + B)f 00 (x)dx.
2
2
a
a
Direct computation shows that
Z b
Z b
(b − a)3
2
|(x + A) + B|dx = −
[(x + A)2 + B]dx =
6
a
a
(3) From the result of (2),
Z b
Z
f (a) + f (b)
1 b
f (x)dx −
((x + A)2 + B)f 00 (x)dx|
|
(b − a)| = |
2
2 a
a
Z b
1
00
≤
sup f (x)
|(x + A)2 + B|dx
2 x∈(a,b)
a
(b − a)3
= sup f (x)
12
x∈(a,b)
00
Thus we can get the error formula for the trapezoidal rule in Theorem 3.3.1 by applying
this error bound to n sub intervals with the same interval length (b − a)/n.
Exercise 3.5.23 (1)
Z
x
2
√
Z
a2
−
x2 dx
3
1
− xd(a2 − x2 ) 2
3
Z
3
1 2
1 2
2 32
= − (a − x ) +
(a − x2 ) 2 dx
3
3
√
3
1
1
3a4
x
= − (a2 − x2 ) 2 − x(2x2 − 5a2 ) a2 − x2 +
arcsin + C.
3
8
8
a
=
Exercise 3.5.23 (2)
Z
x
√
Z
a2
−
x2 dx
3
1
− d(a2 − x2 ) 2
3
3
1
= − (a2 − x2 ) 2 + C.
3
=
Exercise 3.5.26 (1)
3
Let y = 2x − 1. Then x = (y + 1)/2 and
Z
2
10
1
(y + 1)2
+ 1)y 10 dy
4
2
1
1 12 1 11 5 10
=
( y + y + y )dy
2
4
2
4
1 13
1 12
5 11
=
y +
y +
y + C.
104
192
352
Z
(x + 1)(2x − 1) dx =
(
Z
Exercise 3.5.27 (2)
Z
bx + c
dx =
x2 + a2
Z
Z
b 1
c
2
d(x + 1) +
dx
2
2
2x +1
x + a2
b
c
x
= log(x2 + 1) + arctan + C.
2
a
a
x3
√
dx =
3
x 2 + a2
Z
Exercise 3.5.27 (6)
Z
2
3 2 2
x d(x + a2 ) 3
4
Z
5
3 2 2
3
3
2 32
= x (x + a ) −
d(x2 + a2 ) 3
4
4
5
2
5
3
9
= x2 (x2 + a2 ) 3 − x2 (x2 + a2 ) 3 + C.
4
20
Exercise 3.5.28 (6)
Z
Z
ey sin ydy
sin(log x)dx =y=log x
1
= ey (sin y − cos y) + C
2
1
= x(sin(log x) − cos(log x)) + C.
2
Exercise 3.5.28 (9)
Z
√
sin 2x a +
cos2
xdx = −
Z √
a + cos2 xd(a + cos2 x)
3
2
= − (a + cos2 x) 2 + C.
3
4
Exercise 3.5.29 (9)
Z
2
Z
2
Z
2
2
Z
y d(cos y) = y cos y − 2 y cos ydy = y cos y − 2 yd(sin y)
Z
2
= y cos y − 2y sin y + 2 sin ydy = y 2 cos y − 2y sin y − 2 cos y + C
√
= x((arccos x)2 − 2) − 2 1 − x2 arccos x + C.
(arccos x) dx =
Exercise
Z √ 3.5.30
Z (2)
Z
Z
√
√
x
y
y
y
e dx = e 2ydy = 2 yde = 2ye − 2 ey dy = 2yey − 2ey + C = 2( x − 1)e x + C.
Exercise 3.5.30 (9)
√
2ydy
, and
Let y = ex + a. Then x = log(y 2 − a), dx = 2
y −a
Z
Z Z
√
2a
2ydy
x
=
2+ 2
e + adx = y 2
dy
y −a
y −a

√ y − a
√


√ + C,
2y + a log if a > 0



y
+
a

√
y
+ C, if a < 0
= 2y + 2 −a arctan √

−a


2a


2y −
+ C,
if a = 0
y
√ x
√ 
e + a − a
√ x
√


2 e + a + a log √ x
√ + C, if a > 0


e +√a + a 
√ x
√
ex + a
=
√
2
+ C, if a < 0
e
+
a
+
2
−a
arctan



−a

 √ x
2 e + C,
if a = 0
 √
√
√
√
√

2 ex + a + ax − 2 a log( ex + a + a) + C, if a > 0


r

√
√
ex
= 2 ex + a + 2 −a arctan − − 1 + C,
if a < 0

a


1
2e 2 x + C,
if a = 0
Exercise 3.5.33 (3)
Z
Z
2f (x)
f (x) 0
2 f (x)dx = 2f (x) df (x) =
+ C.
log 2
Exercise 3.5.34 (2)
5
Let x := π − y, then
Z
I=
π
0
Z
(π − y)f (sin(π − y))d(π − y)
xf (sin x)dx =
0
π
π
Z
(π − y)f (sin y)dy
Z π
Z π
yf (sin y)dy
f (sin y)dy −
=π
0
Z0 π
f (sin y)dy − I
=π
=
0
0
so
π
Z
π
I=
2
f (sin y)dy
0
Exercise 3.5.35
Because φ(x) = x1 is not differentiable at 0 ∈ (−1, 1).
Without change of variable,
Z 1
π
dx
= arctan x|1−1 = .
2
2
−1 1 + x
But if you insist on using the change of variable y = x1 , then
Z
− y12
1
−1
1+
1
y2
Z
1
dy = −
−1
dy
π
=
−
,
1 + y2
2
which is a wrong answer.
Exercise 3.5.36
Z
b
Z
f (x + t)dx =
a
b
f (x + t)d(x + t)
a
Z
b+t
=
f (y)dy
a+t
Z b+t
Z
f (y)dy −
=
0
then
d
dt
Z
a+t
f (y)dy
0
b
f (x + t)dx = f (b + t) − f (a + t)
a
Exercise 3.5.38
6
If f (x) is even, then f (−x) = f (x).
a
Z
Z
0
f (x)dx =
−a
a
Z
f (x)dx +
f (x)dx
Z a
Z 0
f (−x)d(−x) +
f (x)dx
=−
−a
0
Z a
Z 0
f (x)dx
f (y)dy +
=−
0
a
Z a
Z a
f (x)dx.
f (y)dy +
=
−a
0
0
0
If f (x) is odd, then f (−x) = −f (x).
Z
a
Z
0
a
f (x)dx
Z a
f (x)dx
−f (x)d(−x) +
=
0
−a
Z 0
Z a
=
f (−x)d(−x) +
f (x)dx
−a
0
Z 0
Z a
=
f (y)dy +
f (x)dx
a
0
Z a
Z a
f (x)dx = 0.
f (y)dy +
=−
f (x)dx +
f (x)dx =
−a
Z
−a
Z 0
0
0
0
Exercise 3.5.40 (1)
7
If p 6= −1, −2,
Z
Z
√ p
(1 + x) dx = (1 + y)p dy 2
Z
= 2y(1 + y)p dy
Z
(1 + y)p+1
= 2yd
p+1
Z
(1 + y)p+1
2y(1 + y)p+1
−2
d(y + 1)
=
p+1
p+1
2
p+1
1
p+1
=
(1 + y)
y−
+C
p+1
p+2
p+2
2
(1 + y)p+1 [(p + 1)y − 1] + C
=
(p + 1)(p + 2)
√
√
2
=
(1 + x)p+1 [(p + 1) x − 1] + C
(p + 1)(p + 2)
If p = −2,
Z
(1 +
√
Z
p
x) dx =
(1 + y)−2 dy 2
Z
2y(1 + y)−2 dy
Z
= − 2yd(1 + y)−1
Z
2y
1
=−
+2
d(y + 1)
1+y
1+y
2y
+ 2 log |1 + y| + C
=−
1+y
√
√
2 x
√ + 2 log |1 + x| + C
=−
1+ x
=
8
If p = −1,
Z
(1 +
√
Z
p
(1 + y)−1 dy 2
x) dx =
Z
2y(1 + y)−1 dy
Z
1+y−1
dy
=2
1+y
Z 1
dy
=2
1−
1+y
= 2y − 2 log |1 + y| + C
=
Exercise 3.5.42 (5)
Recall that
csc2 x = 1 + cot2 x.
So
Z
6
Z
4
cot x csc xdx = −
Z
=−
cot6 x csc2 xd(cotx )
cot6 x(1 + cot2 x)d(cotx )
1
1
= − cot7 x − cot9 x + C.
7
9
Exercise 3.5.43 (6)
Recall that a sin x + b cos x = sin(x + φ), where
cos φ = √
a
b
, sin φ = √
.
2
2
+b
a + b2
a2
Then
Z
Z
a
dx
dx
=√
2
2
a sin x + b cos x
a + b Z sin(x + φ)
a
dx
=√
x+φ
2
2
2 sin 2 cos x+φ
a +b
2
Z
a
dx
=√
2 tan x+φ
cos2 x+φ
a2 + b 2
2
2
Z
x+φ
d tan 2
a
=√
tan x+φ
a2 + b 2
2
a
x
+ φ =√
log tan
+C
2 a2 + b 2
9
Exercise 3.5.43 (7)
Z
1 + sin x
dx =
1 + cos x
Z
Z
=
Z
=
Z
=
Z
=
+ sin2 x2 + 2 cos x2 sin x2
dx
2 cos2 x2
x
1 1
2 x
− tan
+ tan
dx
2 2
2
2
1 1
x
x 1
2 x
− tan
+ tan
cos2
dx
2 2
2
2
2 cos2 x2
x
1
1
1 1
2 x
− tan
+ tan
dx
x
2
2 2
2
2
1 + tan 2 cos2 x2
x
x
1
x
d tan
1 − tan2 + 2 tan
x
2
2
2 1 + tan 2
2
cos2
x
2
1 − t2 + 2t
dt
1 + t2
Z
2
2t
+
=
dt
−1 +
1 + t2 1 + t2
= −t + 2 arctan t + log |1 + t2 | + C
x x
= − tan + x + log 1 + tan2 + C
2
2
Z
=
Exercise 3.5.45
We want to find A, B, C, such that
Z 1
B
C
A sin x
−
−
+ const.
dx =
n
n−1
n−2
(a + b cos x)
(a + b cos x)
(a + b cos x)
(a + b cos x)n−1
Differentiate both side, then
0
1
B
C
A sin x
−
−
=
.
(a + b cos x)n (a + b cos x)n−1 (a + b cos x)n−2
(a + b cos x)n−1
Transform all trigonometric function into cos x, we have
[(2 − n)Ab + Cb2 ] cos2 x + (Aa + Bb + 2abC) cos x + Ba − 1 + (n − 1)Ab + Ca2 = 0.
Let y := cos x, then
[(2 − n)Ab + Cb2 ]y 2 + (Aa + Bb + 2abC)y + Ba − 1 + (n − 1)Ab + Ca2 = 0,
therefore,

2

 b(2 − n)A + b C = 0
aA + bB + 2abC = 0


aB + b(n − 1)A + a2 C = 1.
10
If b 6= 0, then you will find that n 6= 1 and
A=
b2
(3 − 2n)ab
(n − 2)b
,B = 2
,C = 2
.
2
2
2
(b − a )(n − 1)
(b − a )(n − 1)
(b − a2 )(n − 1)
If b = 0, then a 6= 0, since |a| 6= |b|. So A = 0 and aB + a2 C = 1. That is, for all B and
C satisfying aB + a2 C = 1, we still have the required equation. You can directly check this
situation by letting b = 0 in the required equation.
Exercise 3.5.47 (3)
Z
cos x+a
− x−a
2
2
dx
sin x−a
2 cos x+a
2
2
Z
cos x+a
cos x−a
+ sin x+a
sin x−a
1
2
2
2
2
dx
x−a
cos a
sin
2 cos x+a
2
2
Z cos x−a
sin x+a
1
2
2
+
dx
x+a
2 cos a
sin x−a
cos
2
2
Z
Z
cos x−a
sin x+a
1
2
2
dx +
dx
2 cos a
sin x−a
cos x+a
2
2
sin x−a
1
2 log +C
cos a
cos x+a
2
dx
1
=
sin x − sin a
cos a
=
=
=
=
Z
Exercise 3.5.48 (9)
Z
3
2
(x(x + 1)) dx =
Z √
x2 + x
3
dx
3
Z p
1
2
(2x + 1) − 1) dx
=
2
Z p
1
=
( y 2 − 1)3 dy
16
Z p
1 p 2
3
3
= y( y − 1) −
y 2 − 1dy 2
16
64
1 p
3
= y( y 2 − 1)3 −
log |y 2 − 1| + C
16
64
p
1
3
= (2x + 1)( (2x + 1)2 − 1)3 −
log |(2x + 1)2 − 1| + C
16
64
3
1
3
= (2x + 1)(x2 + x) 2 −
log(4x2 + 4x) + C
2
64
3
1
3
= (2x + 1)(x2 + x) 2 −
log(x2 + x) + C
2
64
11
Exercise 3.5.49 (2)
Z
√
arctan xdx =
Z
arctan tdt2
Z
2
= t arctan t − t2 d(arctan t)
Z 2
t +1−1
2
dt
= t arctan t −
t2 + 1
Z 1
2
1− 2
= t arctan t −
dt
t +1
= t2 arctan t − t + arctan t + C
√
√
√
= ( x + 1) arctan x − x + C
12