18.014 Problem Set 9 Solutions
Sam Elder
November 18, 2015
Problem 1 (30 points). Compute the following indefinite integrals.
Z
sin x + cos x
dx.
Part 1.1.
2 + sin x − cos x
Solution. Make the substitution u = 2 + sin x − cos x. Then du = (cos x + sin x)dx, so the integral is
Z
du
= log|u| + C = log|2 + sin x − cos x| + C .
u
Z
Part 1.2.
x log x dx, for x > 0.
Solution. We integrate by parts: Let u = log x and v = 21 x2 , which are both differentiable on x > 0. Then
u dv = x log x dx, so we have
Z
Z
x log x dx =
Z
Part 1.3.
Z
u dv = uv −
1
v du = x2 log x −
2
Z
1
1 21
1
x dx = x2 log x − x2 + C .
2 x
2
4
x
q
dx.
p
2
1 + x + (1 + x2 )3
Solution. We make a substitution: y = 1 + x2 , so dy = 2x dx and
Z
Z
Z
dy
1
x
dy
p
q
=
dx
=
√ ·p
√ .
p
3/2
2 y
1
+
y
2
y
+
y
2
2
3
1 + x + (1 + x )
Now we make the substitution z = 1 +
Z
Z
Part 1.4.
√
y, so dz =
dy
√
2 y.
This integral becomes
q
q
p
√
dz
√
√ = 2 z + C = 2 1 + y + C = 2 1 + 1 + x2 + C .
z
tan−1 x dx.
Solution. We integrate by parts, with u = tan−1 x and v = x. Then
Z
Z
Part 1.5.
tan−1 x dx = x tan−1 x −
Z
x
1
dx = x tan−1 x − log(1 + x2 ) + C .
2
1+x
2
ex sin x dx.
1
Solution. We integrate by parts a couple times. First, with u = sin x and v = ex :
Z
Z
x
x
e sin x dx = e sin x − ex cos x dx.
Seeing a similar formula, we integrate by parts again, this time with u = cos x and v = ex :
Z
Z
ex cos x dx = ex cos x + ex sin x dx.
Putting these together,
Z
ex sin x dx = ex sin x − ex cos x −
Z
Part 1.6.
Z
ex sin x dx =⇒
Z
ex sin x =
1 x
e (sin x − cos x) + C .
2
1
dx.
1 + 2 cos x
Solution. This is a problem where a magic substitution works. Let u = tan 21 x. Then x = 2 tan−1 u, so
2
1 − u2
2
2 x
du,
and
cos
x
=
2
cos
−
1
=
−
1
=
. Therefore,
dx =
2
1 + u2
1 + u2
1 + u2
Z
Z
Z
1
2
1
2
dx =
du
=
du.
2
2
1−u
1 + 2 cos x
3 − u2
1 + 2 1+u2 1 + u
Now we need to use partial fractions to split this up.
√
√
2
1/ 3
1/ 3
=√
+√
.
3 − u2
3+u
3−u
Finally, we compute
Z
√
√
√
√
Z
3 + u
3 + tan 1 x 1/ 3
1/ 3
1
1
2 √
√
dx +
dx = √ log √
+ C = √ log √
+C .
3 − u
3 − tan 21 x 3+u
3−u
3
3
Note: This fraction does not simplify further as the example in the textbook does.
Problem 2. Show that
Z
1
x
t2
dt =
1 + t2 + t4 + t6
1/x
Z
1
t2
dt
1 + t2 + t4 + t6
for any x > 0.
Solution. We simply make the substitution of u = 1/t. Then t = u1 , so dt = − u12 du, and
Z
1
x
t2
dt =
1 + t2 + t4 + t6
1
Z
1/x
Z
−u−2
1
du
−2
−4
−6
1 + u + u + u u2
1
=−
1/x
Z
u6−2 /u2
du
u6 + u6−2 + u6−4 + 1
1/x
=
1+
1
as desired.
2
u2
u2
du,
+ u4 + u6
Problem 3. Prove the formula
Z
Z
u
2n − 3
1
1
du
=
+
du
(1 + u2 )n
2(n − 1)(1 + u2 )n−1
2(n − 1)
(1 + u2 )n−1
for any integer n ≥ 2.
Solution. We start with the smaller exponent and integrate by parts: Let v = (1 + u2 )−(n−1) , so
Z
Z
1
u
−2u2 (n − 1)
du
=
−
du
2
n−1
2
n−1
(1 + u )
(1 + u )
(1 + u2 )n
Z
Z
u
1
1 + u2 − 1
du
=
+
2(n
−
1)
du
2
n−1
2
n−1
(1 + u )
(1 + u )
(1 + u2 )n
Z
Z
Z
u
1
1
1
du
=
+
2(n
−
1)
du
−
du
(1 + u2 )n−1
(1 + u2 )n−1
(1 + u2 )n−1
(1 + u2 )n
Z
Z
u
1
1
0=
+ (2n − 3)
du − 2(n − 1)
du
2
n−1
2
n−1
(1 + u )
(1 + u )
(1 + u2 )n
Z
Z
1
u
1
2(n − 1)
du =
+ (2n − 3)
du
(1 + u2 )n
(1 + u2 )n−1
(1 + u2 )n−1
Z
Z
1
u
2n − 3
1
du =
+
du,
(1 + u2 )n
2(n − 1)(1 + u2 )n−1
2(n − 1)
(1 + u2 )n−1
as desired.
Problem 4 (5.11-26). Compute f (0), given that f (π) = 2 and
Rπ
0
[f (x) + f 00 (x)] sin x = 5.
Solution. Seeing an integral of a product involving a derivative, we integrate by parts. First, if u = f (x)
and v = − cos x,
Z π
Z π
Z π
0
f (x) sin x dx = −f (π) cos(π) + f (0) cos(0) +
f (x) cos x dx = f (π) + f (0) +
f 0 (x) cos x dx.
0
0
0
Then we integrate by parts again, with u = f 0 (x) and v = sin x:
Z π
Z π
Z
0
0
0
00
f (x) cos x dx = f (π) sin(π) − f (0) sin(0) −
f (x) sin x dx = −
0
0
Combining these, we have
Z
π
f 00 (x) sin x dx.
0
π
(f (x) + f 00 (x)) sin x dx = f (π) + f (0).
0
Therefore, 5 = 2 + f (0), so f (0) = 3 .
Problem 5 (6.26-28). A function, called the integral logarithm and denoted by Li, is defined as follows:
Z x
dt
if x ≥ 2.
Li(x) =
2 log t
This function occurs in analytic number theory where it is proved that Li(x) is a very good approximation
to the number of primes ≤ x. Derive the following properties of Li(x):
Part 5.1.
x
Li(x) =
+
log x
Z
2
3
x
dt
2
2 − log 2 .
log t
1
and v = t. Then
log t
Z x
Z x
x
−1
dt
2
x
2
Li(x) =
−
−
t
dt
=
+
2
2 − log 2 ,
log x log 2
log
x
t
log
t
log
t
2
2
Solution. We integrate by parts. Let u =
as desired.
Part 5.2.
n−1
Li(x) =
X k!x
x
+
+ n!
log x
logk+1 x
k=1
Z
2
x
dt
+ Cn ,
logn+1 t
where Cn is a constant depending on n. Find this constant.
Solution. We induct on n to claim that Cn = −
n−1
X
k=0
2 · k!
.
logk+1 2
Base Case. n = 1 amounts to the claim of the previous part, since
1
Inductive Step. We again integrate by parts, with u =
Z
n!
2
x
log
n+1
t
2·0!
log0+1 2
=
2
log 2 .
and v = t:
Z x
dt
n!x
n! · 2
−(n + 1)t
−
− n!
n+1 =
n+1
n+1
n+2 dt
log
t
log
x log
2
t
2 t log
Z x
n!x
n! · 2
dt
=
−
+ (n + 1)!
n+1
n+1
n+2
log
x log
2
t
2 log
n−1
Z x
n−1
X
X k! · 2
x
n!x
n! · 2
dt
k+1
Li(x) =
+
k!xlog
x+
−
+ (n + 1)!
−
n+1
n+1
n+2
log x
log
x log
2
t
logk+1 2
2 log
k=1
k=0
Z x
n
n
X
X
x
k!x
dt
k! · 2
Li(x) =
+
+
(n
+
1)!
−
,
n+2
k+1
log x
t k=0 logk+1 2
log
x
2 log
k=1
as desired.
Induction is complete.
Part 5.3. Show that there is a constant b such that
R log x
b
et /t dt = Li(x).
Solution. Make the substitution u = log t, so dt = eu du and
Z
Li(x) =
2
x
dt
=
log t
as desired with the constant b = log 2.
4
Z
log x
log 2
eu
du,
u