MATH 101 HOMEWORK 5 – SOLUTIONS
1. Find the area of the finite planar region which lies inside the circle x2 + y 2 = 4 and
outside of the circle x2 − 6x + y 2 = 0.
We first find the x-coordinate of the points of intersection: if x2 +y 2 = 4 and x2 −6x+y 2 = 0
then 4 − 6x = 0, x = 2/3. Thus the area in question is equal to
Z
2/3
2
p
−2
We compute
Z
4−
x2 dx
−2
2/3 p
6x − x2 dx.
0
√
R√
a2 − x2 dx by substituting x = a sin θ, dx = a cos θ dθ, a2 − x2 = a cos θ:
Z p
Z
Z
2
2
a − x dx = a cos θ · a cos θ dθ = a2 cos2 θ dθ
Z
a2
a2
a2
(cos(2θ) + 1) dθ =
sin(2θ) + θ + C
2
4
2
√
a2
a2
x a 2 − x2
a2
x
=
sin θ cos θ + θ + C =
+
sin−1 ( ) + C.
2
2
2
2
a
=
We use this to evaluate the above integrals:
Z
2/3
−2
1 2
= ·
2 3
√
p
x 4 − x2
x 2/3
2
+ 2 sin−1 ( )
4 − x dx =
2
2 −2
r
√
4
4 2
1
−1 1
−1
+ 2 sin−1 ( ) + π,
4 − + 2 sin ( ) − 0 − 2 sin (−1) =
9
3
9
3
and, substituting u = x − 3,
Z
2/3
0
7
=−
6
r
p
Z
6x − x2 dx =
−7/3
−3
p
√
−7/3
9
u
9 − u2
−1 x 2
+ sin ( )
9 − u du =
2
2
3 −3
√
72
9
7
14 2 9
7
9
9
−1
−1
9 − 2 + sin (− ) − 0 − sin (−1) = −
+ sin−1 (− ) + π.
9
2
9
2
9
2
9
4
1
Plug this into the above formula for the area, and after some more simplifications we get
2
28√2
9
−1
+ 2 sin
9
5 1
−1 7
( ) + sin ( ) − π .
3
2
9
4
2. Evaluate the integrals:
(a) We substitute x = 2 tan θ, dx = 2 sec2 θ dθ, x2 + 4 = 4 sec2 θ. Then
Z
x2
dx =
(x2 + 4)3
Z
4 tan2 θ
1
· 2 sec2 θ dθ =
2
3
(4 sec θ)
8
Z
tan2 θ
dθ
sec4 θ
Z
sin2 θ/ cos2 θ
1
dθ =
sin2 θ cos2 θ dθ
1/ cos2 θ
8
Z
Z
1
sin(4θ)
1
θ
2
=
−
+ C.
sin (2θ) dθ =
(1 − cos(4θ)) dθ =
32
64
64
258
√
√
We now plug in θ = tan−1 (x/2), sin θ = x/ 4 + x2 , cos θ = 2/ 4 + x2 , and
1
=
8
Z
4x 8
sin(4θ) = 2 sin(2θ) cos(2θ) = 2 sin θ cos θ(2 cos θ − 1) = 2
−1 ,
x + 4 x2 + 4
2
and get the answer
1
8x
x
x tan−1 ( ) − 2
+ C.
+
64
2
(x + 4)2
x2 + 4
(b) We want to write
1
1
A
B
C
D
E
=
=
+
+
+
+
.
x(x2 − 1)(x2 − 4)
x(x − 1)(x + 1)(x − 2)(x + 2)
x x−1 x+1 x−2 x+2
The constants can be evaluated as follows:
1
1
1
1
A=
= , B=
=− ,
(x − 1)(x + 1)(x − 2)(x + 2) x=0
4
x(x + 1)(x − 2)(x + 2) x=1
6
C=
Hence
Z
x(x2
1
1
1
1
,
=− , D=
=
x(x − 1)(x − 2)(x + 2) x=−1
6
x(x − 1)(x + 1)(x + 2) x=2 24
1
1
E=
=
.
x(x − 1)(x + 1)(x − 2) x=−2
24
1
1
1
dx
= ln |x| − (ln |x − 1| + ln |x + 1|) + (ln |x − 2| + ln |x + 2|)
2
− 1)(x − 4)
4
6
24
2
=
1
1
1
ln |x| − ln |x2 − 1| +
ln |x2 − 4|.
4
6
24
(c) Again, we want to use partial fractions:
4x(x + 1)
A
B
Cx + D
dx =
+
,
+ 2
2
2
2
(x − 2) (x + 1)
x − 2 (x − 2)
x +1
4x(x + 1) = A(x − 2)(x2 + 1) + B(x2 + 1) + (Cx + D)(x − 2)2 ,
4x2 + 4x = Ax3 − 2Ax2 + Ax − 2A + Bx2 + B + Cx3 − 4Cx2 + 4Cx + Dx2 − 4Dx + 4D.
This gives a system of equations for A, B, C, D:
0 = A + C,
4 = −2A + B − 4C + D,
4 = A + 4C − 4D,
0 = −2A + B + 4D.
4
4
We solve this and get (after a calculation) A = 25
, B = 120
= 24
, C = − 25
, D = − 28
.
25
5
25
Thus
Z
Z
Z
Z
4x(x + 1)
4
dx
24
dx
4x − 28
1
dx =
+
dx
+
2
2
2
(x − 2) (x + 1)
25
x−2
5
(x − 2)
25
x2 + 1
=
4
28
24
2
ln |x − 2| −
+
ln(x2 + 1) −
tan−1 (x) + C.
25
5(x − 2) 25
25
Z
2 − sin x
dx is convergent. Justify your answer. – This
2−x
0
problem was cancelled – it will be included in HW # 5.
2
3. Decide whether the integral
3