Integration Worksheet
MA141-008
November 17, 2011
Stephen Adams
North Carolina State University Department of Mathematics
Answer the following problems.
Z
1
x2 (x3 − 1)4 dx
1.
−1
A u-substitution with u = x3 − 1 gives the differential du = 3x2 dx. Then 31 du = x2 dx. Notice
that when x = −1, u = −2 and when x = 1, u = 0. Substituting this into the original integral
gives
Z 1
Z 0 4
u
2 3
4
x (x − 1) dx =
du
−2 3
−1
5 0
u
=
15 −2
32
= 0− −
15
32
=
15
Z
2.
x ln(x) dx
The integrand is comprised of two terms, x and ln x. Neither of these looks like the derivative
of the other. As a result, we probably want to try integration by parts instead of a usubstitution. Since it is easier to antidifferentiate x than it is to antidifferentiate ln x, a good
1
x2
Rguess is to pick
R u = ln x and dv = x dx. Then du = x dx and v = 2 . Using the formula
u dv − uv − v du, we get
Z
x ln(x) dx =
=
=
Z
3.
0
π/4
sec2 (x) dx
Z 2
x2
x 1
ln x −
· dx
2
2 x
Z
x2
x
ln x −
dx
2
2
x2
x2
ln x −
+C
2
4
Integration Worksheet
MA141-008
Noticing that
November 17, 2011
Stephen Adams
d
[tan x] = sec2 (x), we see that
dx
Z π/4
π/4
sec2 (x) dx = [tan x]0
0
= tan
π
− tan 0
4
= 1
Z
4.
2x dx
d x
[2 ] is ln 2 · 2x . With
There are a couple ways we can evaluate this integral. Recall that
dx
Z
2x
this is mind, one can easily see that 2x dx =
. This technique is essentially performing
ln 2
a u-substitution without explicitly writing out our choice of u. In this case, our u is really
u = 2x so that du = ln 2 · 2x dx. As a result, 2x dx = ln12 du. Therefore
Z
Z
1
x
du
2 dx =
ln 2
u
=
+C
ln 2
2x
=
+C
ln 2
Notice that in this case the derivative of the integrand f (x) = 2x looks very similar to the
integrand itself. This is an indication that u-substitution is a better approach than integration
by parts.
Z
5.
0
4
√
1
dx
2x + 1
Let u = 2x + 1. Then du = 2 dx and hence dx = 12 du. When x = 0, u = 1 and when x = 4,
u = 9. Then substituting into the original integral,
Z 4
Z 9
1
1 −1/2
√
u
du
dx =
2x + 1
0
1 2
h
i9
= u1/2
1
= 3−1
= 2
Z
6.
√
x2 1 − x dx
Integration Worksheet
MA141-008
November 17, 2011
Stephen Adams
Notice that if we let u = 1 − x then du = −dx. Also, we can express x2 in terms of u be
noting that x = 1 − u and hence x2 = 1 − 2u + u2 . Therefore
Z
Z
√
−(1 − 2u + u2 ) · u1/2 du
x2 1 − x dx =
Z
=
(−u1/2 + 2u3/2 − u5/2 ) du
2
4
2
= − u3/2 + u5/2 − u7/2 + C
3
5
7
2
4
2
= − (1 − x)3/2 + (1 − x)5/2 − (1 − x)7/2 + C
3
5
7
Z
7.
arctan x dx
Just like in question #4, the integrand is a single term. Unlike in question #4 the derivative
of the integrand does not resemble the integrand itself. As a result u-substitution will be of
little help. If there is any hope of evaluating the integral we will need to use integration by
parts and hope that the resulting integral will be easier to evaluate. Since there is only a
single term in the integrand, we will need to choose u = arctan x and dv = dx (notice that if
we choose dv = arctan x dx we have no way to find v, for if we could find v we wouldn’t need
1
to use integration by parts to begin with). Then du =
dx and v = x. Then using the
1 + x2
formula for integration by parts,
Z
Z
x
dx
arctan x dx = x arctan x −
1 + x2
1 = x arctan x − ln 1 + x2 + C
2
where this second integral is evaluated using a u-substitution with u = 1 + x2 .
Z
8.
1
9
√
1
√
dx
x(1 + x)2
√
1
1
Let u = 1 + x. Then du = √ dx and hence 2 du = √ dx. Notice that when x = 1, u = 2
2 x
x
and that when x = 9, u = 4. Substituting into the original integral we get
Z 9
Z 4
1
2
√
√ 2 dx =
du
2
x(1 + x)
1
2 u
2 4
= −
u 2
1
= − +1
2
1
=
2
MA141-008
Integration Worksheet
November 17, 2011
Stephen Adams
Z
9.
tan(x) dx
sin x
Notice that tan x =
. Then if we use a u-substitution with u = cos x we get du =
cos x
− sin x dx and hence
Z
Z
1
tan(x) dx =
− du
u
= − ln |u| + C
= − ln | cos x| + C
= ln | sec x| + C
Z
10.
csc2 (x)
dx
cot3 (x)
Let u = cot x. Then du = − csc2 x dx.
Z
Z
1
csc2 (x)
dx =
− 3 du
3
u
cot (x)
1
=
+C
2u2
1
+C
=
2 cot2 x
tan2 x
+C
=
2
Alternatively, we can simplify the integrand using the facts that csc x =
1
and cot x =
sin x
cos x
csc2 (x)
sin3 x
sin x
. Then
=
=
= sec2 x tan x. We can then use a u-substitution
3
2
3
sin x
cos3 x
cot (x)
sin x cos x
with u = sec x and du = sec x tan x dx:
Z
Z
csc2 (x)
dx =
sec2 x tan x dx
cot3 (x)
Z
=
u du
=
=
u2
+C
2
sec2 x
+C
2
Notice that we could also evaluate this integral with a u-substitution of u = tan x and
tan2 x
du = sec2 x dx, and that the resulting integration will yield
+ C.
2
Integration Worksheet
MA141-008
November 17, 2011
Stephen Adams
CHALLENGE: When we evaluate the integral using two different u-substitutions we get two
tan2 x
sec2 x
different antiderivatives. Explain why
+ C and
+ C are both correct antideriva2
2
tan2 x
sec2 x
tives despite the fact that, for arbitrary x,
6=
. Any student who e-mails me the
2
2
correct answer by 2:45 PM on Monday, November 21 will receive five extra credit points on
Test 4. Any student who e-mails me an incorrect answer (or just sends an e-mail stating that
they read this question) will receive two extra credit points on Test 4. My e-mail address is
[email protected].
Z
11.
3
x2 ex dx
Let u = x3 . Then du = 3x2 dx and hence 31 du = x2 dx.
Z
Z
1 u
2 x3
e du
x e dx =
3
1 u
e +C
=
3
1 x3
=
e +C
3
12. lim
n→∞
n
X
i=1
i
n
1+
i 2
n
·
1
n
Notice that this is a limit of a Riemann sum. As such, we can evaluate it by converting it
into a definite integral. Recall that if f is integrable on [a, b], then
Z b
n
X
f (x) dx = lim
f (xi )∆x
a
b−a
n
n→∞
i=1
where ∆x =
and xi = a + i∆x. If we let a = 0 and b = 1, then ∆x =
can then rewrite our limit as
n
n
i
X
X
1
xi
n
lim
lim
∆x.
· = n→∞
i 2 n
n→∞
1 + (xi )2
i=1 1 + n
i=1
x
If we choose f (x) =
, then
1 + x2
Z 1
n
i
X
1
x
n
lim
·
=
dx
2
2
i
n→∞
n
0 1+x
i=1 1 + n
1
1 2
=
ln 1 + x
2
0
1
1
=
ln 2 − ln 1
2
2
1
=
ln 2
2
1
n
and xi = ni . We
Integration Worksheet
MA141-008
November 17, 2011
Stephen Adams
where the integration is done using a u-substitution with u = 1 + x2 .
n X
3i 2 1
·
13. lim
1+
n→∞
n
n
i=1
As before, let a = 0 and b = 1 so that ∆x =
lim
n X
n→∞
i=1
3i
1+
n
2
1
n
and xi = ni .
n
X
1
· = lim
(1 + 3xi )2 · ∆x
n n→∞
i=1
If we choose f (x) = (1 + 3x)2 , then
n X
3i 2 1
·
lim
1+
n→∞
n
n
Z
1
=
(1 + 3x)2 dx
0
i=1
1
=
(1 + 3x)3
9
64 1
=
−
9
9
= 7
14. lim
n→∞
π 2
1
cos
+ cos
n
n
n
2π
n
1
0
nπ 1
n
+ · · · + cos
·
n
n
n
We can evaluate this using a definite integral by realizing that we are evaluating the limit of
a Riemann sum. However, we first need to express this sum in sigma notation:
lim
n→∞
π 2
1
cos
+ cos
n
n
n
2π
n
n
nπ 1
X
n
i
iπ
1
+ · · · + cos
· = lim
cos
· .
n
n
n n→∞
n
n
n
If we let a = 0 and b = 1, then ∆x =
i=1
1
n
and xi = ni . We then have
n
n
X
X
i
iπ
1
1
cos
· = lim
xi cos (πxi ) · .
n→∞
n
n
n n→∞
n
lim
i=1
i=1
MA141-008
Integration Worksheet
November 17, 2011
Stephen Adams
Let f (x) = x cos (πx). Then
n
X
i
1
iπ
lim
·
cos
n→∞
n
n
n
Z
1
x cos (πx) dx
=
0
i=1
1 Z 1
1
1
x sin (πx) −
sin (πx) dx
π
0 π
0
1 1
1
1
x sin (πx) − − 2 cos (πx)
π
π
0
0
cos 0
sin π
cos π
−0 − − 2 − − 2
π
π
π
1
1
0− 2 + 2
π
π
2
− 2
π
=
=
=
=
=
where the integration is done using integration by parts with u = x and dv = cos (πx) dx.