Math 121: Homework 3 solutions
1. (a)
ln(ln x )
dx,
I=
x
R
Let u = ln x, du = dx
.
So
I
=
ln udu. Let U = ln u, dV = du, dU =
x
V = u.
Z
I = u ln u −
Z
du
u
and
du = u ln u − u + C
= (ln x )(ln(ln x )) − ln x + C.
(b)
Z
(sin−1 x )2 dx.
R
Let x = sin θ, dx = cos θdθ. So I = θ 2 cos θdθ. Let U = θ 2 , dV = cos θdθ,
dU = 2θdθ and V = sin θ.
I=
2
I = θ sin θ − 2
Z
θ sin θdθ,
let U = θ, dV = sin θdθ, dU = dθ and V = − cos θ.
2
I = θ sin θ − 2(−θ cos θ +
Z
cos θdθ )
= θ 2 sin θ + 2θ cospθ − 2 sin θ + C
= x (sin−1 x )2 + 2 1 − x2 (sin−1 x ) − 2x + C.
(c)
I=
Z
xe x cos xdx,
let U = x, dV = e x cos xdx, dU = dx and V = 12 e x (sin x + cos x ).
1 x
I =
xe (sin x + cos x ) −
2
1 x
=
xe (sin x + cos x ) −
2
1 x
=
xe (sin x + cos x ) −
2
1
e x (sin x + cos x )dx
2
1 x
e (sin x − cos x + sin x + cos x ) + C
4
1 x
e sin x + C.
2
Z
(d)
x3 + 1
37x + 85
= x−7+
2
( x + 4)( x + 3)
x + 7x + 12
37x + 85
A
B
=
+
,
( x + 4)( x + 3)
x+4 x+3
1
solving to have A = 63 and B = −26. Now we have
Z
x3 + 1
dx =
12 + 7x + x2
=
Z
(x − 7 +
63
26
−
)dx
x+4 x+3
x2
− 7x + 63 ln | x + 4| − 26 ln | x + 3| + C.
2
(e)
dt
(t − 1)(t2 − 1)2
Z
dt
=
( t − 1)3 ( t + 1)2
Z
du
=
,
u3 ( u + 2)2
Z
I =
with setting u = t − 1. Then
1
u3 ( u + 2)2
so we have A =
3
16 ,
=
A
B
C
D
E
+ 2+ 3+
+
u
u + 2 ( u + 2)2
u
u
3
B = − 41 , C = 14 , D = − 16
and E = − 18 .
1
3
du 1 du 1 du
3
du
du
−
−
+
−
2
3
16
u
4
4
16 u + 2 8 (u + 2)2
u
u
1
1
1
3
3
ln |t − 1| +
−
ln |t + 1| +
+ C.
−
=
2
16
4( t − 1) 8( t − 1)
16
8( t + 1)
Z
I =
Z
Z
Z
Z
(f)
Z
dx
dx
= x
=
2x
x
e − 4e + 4
( e − 2)2
Z
du
,
u ( u − 2)2
by setting u = e x .
1
A
B
C
A(u2 − 4u + 4) + B(u2 − 2u) + Cu
=
+
+
=
,
u
u − 2 ( u − 2)2
u ( u − 2)2
u ( u − 2)2
to have A = 41 , B = − 14 and C = 12 .
Z
du
1 du 1
du
1
du
=
−
+
2
4
u
4 u − 2 2 ( u − 2)2
u ( u − 2)
1
1
1
1
=
ln |u| − ln |u − 2| −
+C
4
4
2 ( u − 2)
1
x 1
=
− ln |e x − 2| −
+ C.
x
4 4
2( e − 2)
Z
Z
2
Z
(g)
dx
sec θ tan θdθ
cos3 θdθ
I=
=
=
x2 ( x2 − 1)3/2
sec2 θ tan3 θ
sin2 θ
by setting x = sec θ, so dx = sec θ tan θdθ. Then let u = sin θ, du = cos θdθ,
then
Z
Z
Z
1 − sin2 θ
cos θdθ
sin2 θ
Z
1 − u2
1
=
du = − − u + C
2
u
u
1
= −(
+ sin θ ) + C
sin θ
√
x
x2 − 1
= −( √
+
) + C.
x
x2 − 1
Z
I =
(h) Let x = tan θ, dx = sec2 θdθ.
dx
x (1 + x2 )3/2
Z
Z
cos2 θdθ
sec2 θdθ
=
=
sin θ
tan θ sec3 θ
Z
Z
Z
2
cos θ sin θdθ
du
u2 du
=
= u+
,
=−
2
2
2
1−u
u −1
sin θ
with u = cos θ, du = − sin θdθ. We have
I =
Z
u2
1
1
1
1
= (
−
).
2 u−1 u+1
−1
Thus
u−1
1
ln |
|+C
2
u+1
1
cos θ − 1
= cos θ + ln |
|+C
2
cos θ + 1
√ 1
−1
2
1
1
= √
+ ln | 11+x
|+C
√
+
1
1 + x2 2
2
√1+x
1
1
1 + x2 − 1
+ ln( √
= √
) + C.
1 + x2 2
1 + x2 + 1
I = u+
(i) Let x = tan 2θ , dθ =
Z
2
dx,
1+ x 2
dθ
=
1 + sin θ + cos θ
=
cos θ =
1− x 2
,
1+ x 2
sin θ =
2x
.
1+ x 2
So
( 1+2x2 )dx
Z
2
x
1 + ( 11−
) + ( 1+2xx2 )
+ x2
Z
dx
θ
= ln |1 + x | + C = ln |1 + tan | + C.
1+x
2
3
2. We set u = ln x, dx = eu du.
Z
I =
2
4
x (ln x ) dx =
Z
u4 e3u du
= ( a4 u4 + a3 u3 + a2 u2 + a1 u + a0 )e3u + C.
We have
dI
= (4a4 u3 + 3a3 u2 + 2a2 u + a1 )e3u + 3( a4 u4 + a3 u3 + a2 u2 + a1 u + a0 )e3u
du
= u4 e3u .
So 3a4 = 1, 3a3 + 4a4 = 0, 3a2 + 3a3 = 0, 3a1 + 2a2 = 0, and 3a0 + a1 = 0. Thus
8
8
a4 = 31 , a3 = − 94 , a2 = 49 , a1 = − 27
and a0 = 81
. We now have
4
4
8
8
1
I = ( u4 − u3 + u2 − u + )e3u + C
3
9
9
27
81
I = x3 (
(ln x )4 4(ln x )3 4(ln x )2 8 ln x
8
−
+
−
+ ) + C.
3
9
9
27
81
3.
A
B
C
D
Ex + F
x5 + x3 + 1
=
+
+
+
+ 2
.
2
3
2
3
x − 1 ( x − 1)
x+1 x +x+1
( x − 1)( x − 1)( x − 1)
( x − 1)
R
2
2
4. Suppose that I = e− x dx = P( x )e− x + C, where P is a polynomial having degree
m ≥ 0. Then we must have
2
2
dI
= ( P0 ( x ) − 2xP( x ))e−x = e−x .
dx
It follows that P0 ( x ) − 2xP( x ) = 1. The left side of this equation must be a polynomial of degree m + 1 ≥ 1 because 2xP( x ) has degree m + 1 and P0 ( x ) only has
degree m − 1. But the right side of the equation is a polynomial of degree 0. This
contradiction shows that no such polynomial P( x ) exists.
Since
d
dx Er f ( x )
=
2
√2 e− x
π
by the Fundamental Theorem of Calculus, we have
Z
√
e
− x2
dx =
π
Er f ( x ) + C.
2
Let us try the form
J=
Z
2
Er f ( x )dx = P( x ) Er f ( x ) + Q( x )e− x + C,
where P and Q are polynomials to be determined. Then
2
dJ
2
= P0 ( x ) Er f ( x ) + ( √ P( x ) + Q0 ( x ) − 2xQ( x ))e−x = Er f ( x ).
dx
π
4
Hence we must have P0 ( x ) = 1 and √2π P( x ) + Q0 ( x ) − 2xQ( x ) = 0. The first of these
DEs says that P( x ) = x + k, without loss of generality we can take the constant k to
be zero. The second DE says that
2x
Q0 ( x ) − 2xQ( x ) = − √ .
π
The right side has degree 1 and so must the left side. Thus Q must have degree zero.
Hence Q0 ( x ) = 0 and Q( x ) = √1π . Therefore
J=
Z
2
1
Er f ( x )dx = xEr f ( x ) + √ e− x + C.
π
5.
In =
Z π/2
0
x n sin xdx,
let U = x n , dV = sin xdx, dU = nx n−1 dx, V = − cos x.
I = −x
n
cos x |0π/2
+n
Z π/2
x n−1 cos xdx.
0
Let U = x n−1 , dV = cos xdx, dU = (n − 1) x n−2 dx, V = sin x.
I = n[ x
n −1
sin x |0π/2
− ( n − 1)
Z π/2
π
= n( )n−1 − n(n − 1) In−2 ,
2
I0 =
Z π/2
0
0
x n−2 sin xdx ]
( n ≥ 2).
sin xdx = 1.
I6 = 6(π/2)5 − 6(5){4(π/2)3 − 4(3)[2(π/2) − 2(1) I0 ]} =
3 5
π − 15π 3 + 360π − 720.
16
Set U = sinn−1 x, dV = sin dx, dU = (n − 1) sinn−2 x cos xdx, V = − cos x,
In =
Z
sinn xdx
= − sin
nIn
In
n −1
x cos x + (n − 1)
Z
sinn−2 x cos2 xdx
= − sinn−1 x cos x + (n − 1)( In−2 − In )
= − sinn−1 x cos x + (n − 1) In−2
1
n−1
= − sinn−1 x cos x +
In−2 .
n
n
5
Note that I0 = x + C and I1 = − cos x + C, hence
5
1
I6 = − sin5 x cos x + I4
6
6
1
5 1
3
= − sin5 x cos x + (− sin3 x cos x + I2 )
6
6 4
4
5
5
1
1
sin3 cos x + (− sin x cos x + 1/2I0 )
= − sin5 x cos x −
6
24
8 2
1
5
5
5
= − sin5 x cos x −
sin3 cos x −
sin x cos x + x + C
6
24
16
16
5x
− cos x (sin5 x/6 + 5 sin3 x/24 + 5 sin x/16) + C.
=
16
1
I7 = − sin6 x cos x +
7
1
= − sin6 x cos x +
7
1
= − sin6 x cos x −
7
1
= − sin6 x cos x −
7
6
I5
7
6 1
4
(− sin4 x cos x + I3 )
7 5
5
6
24 1
2
4
sin x cos x + (− sin2 x cos x + I1 )
35
35 3
3
6
8
16
sin4 x cos x −
sin2 x cos x − x + C
35
35
35
16
= − cos x (sin6 x/7 + 6 sin4 x/35 + 8 sin2 x/35 + ) + C.
35
6