Quiz # 2 Solutions
March Madness!!!!
March 13, 2012
Here are the solutions to the second quiz. Enjoy!
2x+1
1. Notice that ln 3x+4
= ln(2x + 1) − ln(3x + 4). So
2x + 1
d
2
3
d
ln
=
[ln(2x + 1) − ln(3x + 4)] =
−
.
dx
3x + 4
dx
2x + 1 3x + 4
2. Let u = ex . Then du = ex dx, and
tan ex + C.
R
ex sec2 ex dx =
R
sec2 udu = tan u + C =
3. We consider the option that this is an arcsin type integral, and factor out the 4 from
the denominator:
Z
Z
1
1
1
√
p
dx =
dx.
2
2
4−x
1 − (x/2)2
We let u = x/2, so that du = 12 dx, 2du = dx, and
Z
Z
Z
1
1
1
1
2du
p
√
√
=
du = arcsin(u)+C = arcsin(x/2)+C.
dx =
2
2
2
1−u
1 − u2
1 − (x/2)2
Of course, you could also do trig substitution, where x = 2 sin θ.
4. We use integration by parts. Let u = e−x , dv = sin xdx. Then du = −e−x dx, v =
− cos x. Then
Z
Z
−x
−x
e sin xdx = −e cos x − e−x cos xdx.
We use integration by parts
−e−x dx, v = sin x. Then
R −x
e sin xdx =
=
=
again.
Let u = e−x , dv = cos xdx, so that du =
R
−e−x cos x − e−x cos xdxR
−e−x cos x − e−x sin x −R −e−x sin xdx
−e−x cos x − e−x sin x − e−x sin xdx
1
Adding
R
e−x sin xdx to both sides and dividing by 2 gives us the result of
Z
1
e−x sin xdx =
−e−x cos x − e−x sin x + C.
2
5. We treat the integral as an arctangent-type integral, since the denominator is a
polynomial of degree 2 that doesn’t factor (since its degree is 2, this is as a result
of it not having any roots). We start by noticing that, by completing the square,
x2 + 2x + 5 = (x + 1)2 + 4, and factor a 4 out of the denominator:
!
2
1
1
=
x2 + 2x + 5
2 1 + x+1 2
2
So if u =
Z
x+1
2 ,
then du = 21 dx, and so
2
dx =
2
x + 2x + 5
Z
1
2
1
1+
!
Z
dx =
x+1 2
2
2
1
du = arctan(u)+C = arctan
1 + u2
x+1
+C.
2