Name:
Math 1220–003
Calculus II Exam 2
Place all books and papers under your seat or otherwise out of sight.
No calculators are allowed on this test.
Show all work for parital credit.
Good luck!
Problem
Score
1
/20
2
/20
3
/20
4
/20
5
/20
total
/100
Exam 2
Math 1220–003
1. Evaluate the following using trigonometric substitution:
Za
1
dx
√
2
0
a − x2
We use the substitution
x = a sin θ
dx = a cos θ
x
θ = sin−1
a
Now substitute in:
Z
a
0
1
dx =
√
2
a − x2
Z
sin−1 ( aa )
sin−1 0a
π
2
( )
Z
=
0
π
2
Z
=
0
π
2
Z
=
0
π
2
Z
=
a cos θ
√
a2 − a2 sin2 θ
a cos θ
dx
√
a 1 − sin2 θ
a cos θ
dx
√
a cos2 θ
a cos θ
dx
a cos θ
1 dx
0
π
= [t]02
π
=
2
If they simply jump to
Z
a
0
a
1
x
dx = sin−1
√
2
2
a 0
a −x
then they have not used trig sub and they are wrong.
dx
Exam 2
Math 1220–003
2. Evaluate the following integral:
π
2
Z
π
4
cos2 x sin2 x dx
Use the double angle formulas to perform a reduction, since m and n are both even:
Z
π
2
π
4
cos2 x sin2 x dx =
Z
π
2
π
4
1 + cos 2x
2
1
4
Z
π
2
1
=
4
Z
π
4
π
2
=
π
4
1 − cos 2x
dx
2
(1 + cos 2x) (1 − cos 2x) dx
1 − cos2 2x dx
Now we reduce using the half angle formulas again.
1
4
Z
=
π
2
1
4
Z
=
π
4
π
2
1
=
8
Z
π
4
π
2
=
=
=
=
=
=
=
π
4
1−
1 + cos 4x
dx
2
1−
1 1
− cos 4x dx
2 2
1 − cos 4x dx
π2
1
1
x − sin 4x
π
8
4
4
1 π 1
π
1 π 1
π
− sin 4
−
− sin 4
8 2 4
2
8 4 4
4
1 π 1
1 π 1
− sin (2π) −
− sin (π)
8 2 4
8 4 4
1 π
1 π
−
8 2
8 4
π
π
−
16 32
π
π
−
16 32
π
32
Exam 2
Math 1220–003
3. Evaluate the following integral:
Z
4x2 − 4x + 6
dx
(x − 2)(x2 + 3)
This is clearly a partial fractions problem.
4x2 − 4x + 6
A
Bx + C
=
+
(x − 2)(x2 + 3) x − 2 x2 + 3
4x2 − 4x + 6 = A(x2 + 3) + (Bx + C)(x − 2)
Now plug in 3 x-values. At x = 2:
4(2)2 − 4(2) + 6 = A((2)2 + 3) + (B(2) + C)((2) − 2)
14 = 7A
A=2
x = 0:
4(0)2 − 4(0) + 6 = A((0)2 + 3) + (B(0) + C)((0) − 2)
6 = 2 · 3 + C · −2
6 = 6 − 2C
2C = 0
C=0
x = 1:
4(1)2 − 4(1) + 6 = A((1)2 + 3) + (B(1) + C)((1) − 2)
6 = 2 · 4 + B · −1
6 = 8−B
B=2
Now integrate:
Z
4x2 − 4x + 6
dx =
(x − 2)(x2 + 3)
Z
2x
2
+
dx
x − 2 x2 + 3
= 2 ln |x − 2| + ln |x2 + 3| + C
They can simplify that to:
(x − 2)2 +C
= ln 2
x + 3 Exam 2
Math 1220–003
4. Evaluate the following integral using integration by parts:
Z
ex cos x dx
This problem is hard. It’s easy to set up the first step, though, and that’s where most of the points
will be.
u = ex
dv = cos xdx
x
du = e dx
v = sin x
Now the integration by parts formula is:
Z
Z
u dv = uv −
Z
v du
Z
x
x
e cos x dx = e sin x − ex sin x dx
R
Now we need to integrate ex sin x dx. Let’s use parts again:
u = ex
dv = sin xdx
x
du = e dx
v = − cos x
So again:
Z
Z
u dv = uv −
Z
v du
Z
x
x
e sin x dx = −e cos x + ex cos x dx
Collecting everything together:
Z
x
x
Z
ex sin x dx
"
#
Z
x
x
x
= e sin x − −e cos x + e cos x dx
Z
x
x
= e sin x + e cos x − ex cos x dx
Z
Z
x
x
e cos x dx = e [sin x + cos x] − ex cos x dx
Z
2 ex cos x dx = ex [sin x + cos x]
Z
1
ex cos x dx = ex [sin x + cos x]
2
e cos x dx = e sin x −
But that’s a specific antiderivative; the final answer needs a +C:
Z
ex cos x dx =
1 x
e [sin x + cos x] + C
2
Exam 2
Math 1220–003
5. Evaluate the following integral:
Z
y
dy
p
1−y
There are several techniques that will work. You will get some points for choosing an appropriate technique, even if you don’t successfully evaluate the integral.
The first (and best) option is integration by parts. There’s no other obvious choice, and integration by parts is always our fallback.
1
dy
dv = p
1−y
p
v = −2 1 − y
u=y
du = dy
Plugging into the integration by parts formula:
Z
Z
u dv = uv − v du
Z
Z
p
p
y
dy = −2y 1 − y − −2 1 − ydy
p
1−y
Z
p
p
= −2y 1 − y + 2
1 − ydy
p
3
4
= −2y 1 − y − (1 − y) 2 + C
3
p
2
= −2 1 − y y + (1 − y) + C
3
p
1
2
= −2 1 − y y +
+C
3
3
It can also be done using a clever trig substitution.
y = sin2 θ
dy = 2 sin θ cos θdθ
√
θ = sin−1 y
Now plug in:
Z
y
dy =
p
1−y
Z
sin2 θ
2 sin θ cos θ dθ
√
1 − sin2 θ
sin2 θ
2 sin θ cos θ dθ
√
cos2 θ
Z
sin2 θ
=
2 sin θ cos θ dθ
cos θ
Z
= 2 sin3 θ dθ
Z
=
Exam 2
Math 1220–003
Z
sin θ 1 − cos2 θ dθ
Z
sin θ − sin θ cos2 θ dθ
=2
=2
2
= −2 cos θ + cos3 θ + C
3
√ √ 2
= −2 cos sin−1 y + cos3 sin−1 y + C
3
√
√ p
A bit of trig tells us that cos sin−1 (x) = 1 − x2 , so cos sin−1 y = 1 − y, and:
Z
p
3
y
2
dy = −2 1 − y + (1 − y) 2 + C
p
3
1−y
p
1
= −2 1 − y 1 − (1 − y) + C
3
p
2
1
+C
= −2 1 − y y +
3
3