Solutions
Calculus II - Fall 2013 - Written homework 7
due Thursday, October 31, 2013
Show all work.
x3
Z
1. Integrate (give your answer in terms of x):
3/2
(1 − x2 )
dx
Trig substitution with x = sin θ and dx = cos θd θ gives a new integral of the form
Z
sin3 θ
dθ
cos2 θ
This becomes a trigonometric integral with an odd power of sin(x), which means that our strategy is to
isolate a power of sin(x) and convert the remaining terms to cos(x).
Z
1 − cos2 θ
sin θ
dθ
cos2 θ
Substituting u = cos θ, in which case du = − sin θ dθ, we have
Z
Z
1
1
1 − u2
1 − 2 du = u + + C
du
=
−
u2
u
u
Now we have to back-substitute twice:
1
+ cos θ + C
cos θ
1
+ cos arcsin x + C
cos arcsin x
√
Lastly, we can use triangles to determine that cos arcsin x = 1 − x2 :
p
1
2 − x2
+ 1 − x2 + C = √
+C
2
1−x
1 − x2
Z
x2
√
2. Integrate (give your answer in terms of x):
dx
16x2 − 4
=√
Keeping in mind that 16x2 − 4 = (4x)2 − (2)2 , we make the substitution 4x = 2 sec θ. Then dx =
1
2 sec θ tan θ dθ, and the integral becomes:
Z 1
1
2
4 sec θ · 2 sec θ tan θ
√
dθ
4 sec2 θ − 4
Z
1
=
sec3 θ dθ
16
We know from past experience that one can use integration by parts twice to calculate this integral:
1
(sec θ tan θ + ln | sec θ + tan θ|) + C
32
1
Again, we use a triangle to determine values for θ = arcsec 2x:
=
p
p
1 2x · 4x2 − 1 + ln 2x + 4x2 − 1 + C
32
Z
3. Integrate, using partial fraction decomposition:
x
dx
(x + 1)(x − 3)
There are two linear factors with multiplicity one, which means our decomposition will have two elements:
x
A
B
=
+
(x + 1)(x − 3)
x+1 x−3
A(x − 3) + B(x + 1)
=
(x + 1)(x − 3)
x = Ax − 3A + Bx + B
= (A + B)x − 3A + B
This gives us two equations in two unknowns:
−3A + B = 0
A+B =1
A + 3A = 1
B = 3A
3
B=
4
4A = 1
A=
1
4
It remains to evaluate the integral
Z
Z
1
dx
3
dx
1
3
+
= ln |x + 1| + ln |x − 3| + C
4
x+1 4
x−3
4
4
Z
4. Integrate, using partial fraction decomposition:
x3
x3 − 1
dx
− 2x2 + x
Since the numerator has the same degree as the denominator, we must do some long-division first.
The steps below follow the intuitive method mentioned in class; if you’d like to see the traditional
long-division technique, you can check wolfram alpha or another online source.
x3 − 1
x3 −2x2 + x+2x2 − x − 1
x3 − 2x2 + x
2x2 − x − 1
2x2 − x − 1
=
=
+
=
1
+
x3 − 2x2 + x
x3 − 2x2 + x
x3 − 2x2 + x x3 − 2x2 + x
x3 − 2x2 + x
We know the antiderivative of 1, so we can come back to that part and decompose the rational function
on the right. Factor the denominator as x(x − 1)2 , which gives two linear factors, one which has
2
multiplicity one, and one with multiplicity two. This means that our decomposition will have three
summands.
A
B
C
2x2 − x − 1
= +
+
x3 − 2x2 + x
x
x − 1 (x − 1)2
A(x2 − 2x + 1) + B(x)(x − 1) + Cx
=
x3 − 2x2 + x
2
2
2x − x − 1 = Ax − 2Ax + A + Bx2 − Bx + Cx
= (A + B)x2 + (−2A − B + C)x + A
This gives three equations in three unknowns:
−2A − B + C = −1
2 − 3 + C = −1
A+B =2
−1 + B = 2
B=3
C=0
This gives us the integral
Z
−
dx
+3
x
Z
dx
= − ln |x| + 3 ln |x − 1| + C
x−1
3
A = −1