MATH 142
MIDTERM 1 SOLUTIONS
September 19, 2013
1. (5 points) Compute the definite integral
Z π/8
cos2 (8x) dx.
0
SOLUTION: We use the half-angle formula
1
cos2 (8x) = (1 + cos(16x)).
2
Substituting into the integral gives
Z π/8
Z
2
cos (8x) dx =
π/8
1
(1 + cos(16x)) dx
2
0
π/8
x sin(16x) = +
2
32 0
π
= .
16
0
2. (5 points) Evaluate the indefinite integral
Z
2
x3 ex dx.
SOLUTION: We use integration by parts with
2
u = x2 , dv = xex dx.
To integrate dv and get v we use a substitution w = x2 . With this, dw = 2xdx. So
Z
Z
1
1 2
1
x2
ew dw = ew = ex .
xe dx =
2
2
2
So we have
1 2
du = 2xdx , v = ex .
2
Then,
Z
2
x2 e x
2
x e dx =
− xex dx.
2
We just computed the last integral so we have
Z
2
x2 ex
1 2
3 x2
x e dx =
− ex + C.
2
2
Z
3 x2
1
3. (5 points) Use partial fractions to expand the following rational function
2x2 − x − 1
.
(x − 1)2 (x2 + x + 1)
SOLUTION: We have x − 1 repeated plus an irreducible quadratic term x2 + x + 1
so partial fractions tell us that
2x2 − x − 1
A
B
Cx + D
=
+
+ 2
2
2
2
(x − 1) (x + x + 1)
x − 1 (x − 1)
x +x+1
for some constants A, B, C, and D. We find a common denominator to get
2x2 − x − 1
A(x − 1)(x2 + x + 1) + B(x2 + x + 1) + (Cx + D)(x − 1)2
=
.
(x − 1)2 (x2 + x + 1)
(x − 1)2 (x2 + x + 1)
We need to have
2x2 − x − 1 = A(x − 1)(x2 + x + 1) + B(x2 + x + 1) + (Cx + D)(x − 1)2 .
Plugging x = 1 in gives
B = 0.
To get the rest, we need to expand. Expanding the numerator on the right hand side
gives
(A + C)x3 + (−2C + D)x2 + (C − 2D)x + (−A + D).
So we need
A+C
−2C + D
C − 2D
−A + D
=0
=2
= −1
= −1.
Which gives A = 1, C = −1, and D = 0. Thus,
1
x
x2 − x − 1
=
−
.
(x − 1)2 (x2 + x + 1)
x − 1 x2 + x + 1
4. (5 points) Evaluate the following definite integral
Z 1
1
dx.
2
3/2
0 (x + 4)
SOLUTION: We use trigonometric substitution with x = 2 tan θ. We convert everything into the θ variables. We have
dx = d(2 tan θ) = 2 sec2 θdθ.
2
When x = 0, then θ = 0. When x = 1, then θ = tan−1 (1/2) which we do not have a
nice formula for immediately. Thus,
Z 1
Z tan−1 (1/2)
1
2 sec2 θ dθ
dx
=
.
2
3/2
(4 tan2 θ + 4)3/2
0 (x + 4)
0
We substitute the identity
4 tan2 θ + 4 = 4 sec2 θ
to get
Z tan−1 (1/2)
0
2 sec2 θ dθ
=
(4 tan2 θ + 4)3/2
tan−1 (1/2)
Z
0
1
=
4
Z
2 sec2 θ dθ
8 sec3 θ
tan−1 (1/2)
0
tan−1 (1/2)
sin (tan−1 (1/2))
1
=
cos θ dθ = sin θ
.
4
4
0
We turn to the right triangle
p
(1/2)2 + 12 =
1/2
√
5
2
1
Thus,
1/2
1
sin tan−1 (1/2) = √
=√ .
5/2
5
And,
1
Z
0
(x2
1
1
dx = √ .
3/2
+ 4)
4 5
5. (5 points) Evaluate the following indefinite integral
Z
4 dx
.
2
x − 2x − 3
SOLUTION: We factor x2 − 2x − 3 = (x − 3)(x + 1) and use partial fractions. Since
we have distinct linear factors, we know that
x2
4
A
B
=
+
− 2x − 3
x+1 x−3
for constants A and B. Finding a common denominator gives
x2
4
A(x − 3) + B(x + 1)
=
.
− 2x − 3
x2 − 2x − 3
3
We need to solve for A and B so that
4 = A(x − 3) + B(x + 1).
Plugging x = 3 gives B = 1. Plugging in x = −1 gives A = −1. Thus,
Z
Z
x − 3
1
1
4
+ C.
dx = −
+
dx = ln x2 − 2x − 3
x+1 x−3
x + 1
6. (5 points) Compute the following definite integral
Z
e2
1
ln x
√ dx.
x
SOLUTION: We use integration by parts with
1
u = ln x , dv = √ dx.
x
We then have
√
1
dx , v = 2 x.
x
Using the formula from integration by parts, we have
e2 Z e2
Z e2
√
√ 1
ln x
√ dx = 2 x ln x −
2 x · dx
x
x
1
1
1
e2 Z e2
√
1
= 2 x ln x −
2 √ dx
x
1
1
e2
√
√ = 2 x ln x − 4 x du =
1
= 4e ln e − 4e + 4
= 4.
7. (5 points) Evaluate the indefinite integral
Z
x2
√
dx.
9 − x2
SOLUTION: Since we have
√
9 − x2 we do a trigonometric substitution with
x = 3 sin θ.
We convert dx to θ’s
dx = d(3 sin θ) = 3 cos θdθ.
4
Substituting in gives
Z
x2
√
dx =
9 − x2
Z
27 sin2 θ cos θ
p
dθ.
9 − 9 sin2 θ
We use the identity
9 − 9 sin2 θ = 9 cos2 θ
to get
Z
27 sin2 θ cos θ
p
dθ =
9 − 9 sin2 θ
Z
27 sin2 θ cos θ
dθ =
3 cos θ
Z
9 sin2 θ dθ.
We next use the half-angle formula
1
sin2 θ = (1 − cos(2θ))
2
to get
Z
9
9 sin θ dθ =
2
2
Z
9
9
(1 − cos(2θ)) dθ = θ − sin(2θ) + C.
2
4
Now we need to convert back to x’s. We have x = 3 sin θ so θ = sin−1 ( x3 ). Using the
identity,
sin(2θ) = 2 cos(θ) sin(θ)
we just have to express cos θ in terms of x. We use the identity
r
p
x2
2
cos θ = 1 − sin θ = 1 − .
9
This gives
9
9
9 −1 x 1 √
θ − sin(2θ) + C = sin
− x 9 − x2 + C.
2
4
2
3
2
8. (5 points) What is the value of the integral
Z 1
3t(2t − 1)6 dt ?
0
SOLUTION: While it is tempting to try to multiply this out and then integrate,
this is very time-intensive. Instead, we use integration by parts with
u = 3t , dv = (2t − 1)6 dt.
Differentiating u and integating dv gives
du = 3dt , v =
5
1
(2t − 1)7 .
14
Thus,
Z
0
1
1
Z 1
3
3t(2t − 1)7 −
3t(2t − 1) dt =
(2t − 1)7 dt
14
14 0
0
1
7 1
3
3t(2t − 1) 8
−
(2t − 1) =
14
14 · 16
0
0
3
= .
14
6
6