Math 222
Homework 2 Solutions
R n
Problem 1.7.1: Evaluate x ln x dx where n =
6 1.
Spring 2013
Solution Integrate by parts, letting u = ln x and dv = xn dx. With these choices, du =
R
n+1
and v = xn+1 . The integral becomes uv − vdu, giving us
Z n+1
xn+1 ln x
x
dx
1
−
n+1
n+1
x
Z
xn+1 ln x
1
−
xn dx
n+1
n+1
1
x
dx
xn+1
xn+1 ln x
−
+C
n+1
(n + 1)2
Problem 1.7.3: Evaluate
R
eax cos bx dx where a, b 6= 0.
Solution Let u = cos bx and dv = eax dx. Then du = −b sin bxdx and v = a1 eax . Integrating
by parts, we have
Z
Z
−b
1 ax
ax
eax sin bx dx.
e cos bx dx = e cos bx −
a
a
We can handle the integral on the right in a similar way. Let u = sin bx and dv = eax dx.
Then du = b cos bxdx and v = a1 eax . This gives us
Z
Z
1 ax
b eax sin bx b
ax
ax
e cos bx dx = e cos bx +
−
e cos bx dx + C
a
a
a
a
Notice that the integral we started with has reappeared on the right-hand side of the
equation. If we distribute the ab and add this term to both sides, we get
Z
1
b
b2
eax cos bx dx = eax cos bx + 2 eax sin bx + C
1+ 2
a
a
a
Z
1 ax
e cos bx + ab2 eax sin bx
∴ eax cos bx dx = a
+C
2
1 + ab 2
Problem 1.7.5: Use §6.3 to evaluate sin2 x dx. Show that the answer is the same as the
answer you get using the half-angle formula.
Solution The reduction formula is
Z
Z
1
n−1
n
n−1
sin x dx = − sin
x cos x +
sinn x dx
n
n
We are dealing with the case n = 2. This gives us
Z
Z
1
2−1
2
sin x dx = − sin x cos x +
sin0 x dx
2
2
1
Since sin0 x = 1, this is simply
Z
sin x cos x x
+ +C
sin2 x dx = −
2
2
The identities sin2 x =
different way.
1−cos(2x)
2
and sin(2x) = 2 sin x cos x let you derive this answer in a
Rπ
Problem 1.7.7: Investigate the numbers An = 0 sinn x dx.
Rπ
Rπ 0
dx = π − 0 = 0. Moreover,
sin
x
dx
=
Solution
For
part
(a),
note
that
A
=
0
0
0
Rπ
π
sin
x
dx
=
(−
cos
x)
|
=
2.
0
0
For part (b), we use the reduction formula. The reduction formula tells us that
1
4
I5 = − sin4 x cos x + I3
5
5
and also that
2
1
I3 = − sin2 x cos x + I1 .
3
3
It follows that
1
4
I5 = − sin4 x cos x +
5
5
1 2
2
− sin x cos x + (− cos x)
3
3
16
Evaluating this at the endpoints 0 and π gives us A5 = 45 23 (2) = 15
. The process for A6
6 16
32
5π
and A7 is similar. You should find that A6 = 16 and A7 = 7 15 = 35 .
In part (c), note that sin x is non-negative over the interval [0, π]. Because sin x ≤ 1, it
follows that sinn x ≤ sinn−1 x. (To see why, think about what happens with ( 12 ), ( 12 )2 , etc.)
Since the height of the function sinn x is less than that of sinn−1 x over this interval, the
overall area should be smaller (here we are using the fact that sinn x is strictly smaller than
sinn−1 x at most points in this interval; if they were equal everywhere, we couldn’t make the
same conclusion about the areas). Hence An < An−1 , as claimed.
32
< 5π
< 16
. Multiplying
For part (d), notice that A7 < A6 < A5 by part (c). Hence 35
16
15
16
through by 15 , we see that
512
256
<π<
175
75
Problem 1.7.9: Prove the formula
m 6= −1.
R
xm (ln x)n dx =
xm+1 (ln x)n
m+1
−
n
m+1
R
xm (ln x)n−1 dx for
Solution We integrate by parts. Let u = (ln x)n and dv = xm dx. Then du =
1
by the chain rule and v = m+1
xm+1 . This gives us
Z
Z
(ln x)n−1
xm+1 (ln x)n
n
m
n
x (ln x) dx =
−
xm+1
dx.
m+1
m+1
x
Notice that
xm+1
x
= xm , verifying the reduction formula.
2
n(ln x)n−1
dx
x
R
Problem 1.7.11: Find (ln x)2 dx.
Solution This is simply the case of the formula from the previous problem when m = 0
and n = 2. Plugging these values in, we have
Z
Z
2
2
(ln x) dx = x(ln x) − 2 ln x dx
R
In class we saw that ln x dx = x ln x − x + C by using integration by parts. Combining
these, we have
Z
(ln x)2 dx = x(ln x)2 − 2(x ln x − x) + C.
Problem 1.7.13: Evaluate
R
x−1 ln x dx by another method.
Solution A simple substitution works. Let u = ln x. Then du = x−1 dx, so we are left with
R
2
2
u du = u2 + C. In terms of x, we get (ln2x) + C.
R dx
Problem 1.7.15: Find (1+x
2 )3 .
Solution This calculation is done on page 18 of the course notes.
R xdx
Problem 1.7.17: Find (1+x
2 )4 .
Solution Instead of using the reduction formula, we perform a substitution. Let u = 1 + x2 .
Then du = 2xdx. The integral in question becomes
Z
1
1
u−4 du = − u−3 + C.
2
6
In terms of x, we have
Z
1
xdx
1
=−
+C
2
4
(1 + x )
6 (1 + x2 )3
R
dx
Problem 1.7.18: Find (49+x
2 )3 .
Solution We’d like to transform the integral to be able to apply the relevant reduction
formula. To do so, we must remove the pesky 49 by dividing numerator and denominator
by 493 :
Z
dx
493
49+x2 3
49
Next, we set u = x7 , giving du =
into
dx
.
7
1
75
1
= 3
49
Z
dx
1+
3
x 2
7
Noticing that 493 = 76 , this transforms the integral
Z
du
.
(1 + u2 )3
We evaluated this integral in a previous problem (as well as on page 18 of the course
notes). All that remains is to substitute x7 for u.
3
Problem 1.7.20: Given that the reduction formula
In+1 =
x
1
2n − 1
In
+
2
n
2n (1 + x )
2n
holds for all positive values of n, find a relationship between I1/2 =
R√
1 + x2 dx.
R
√ dx
1+x2
and I−1/2 =
Solution Following the hint, we set n = − 12 in the reduction formula, yielding
√
I1/2 = −x 1 + x2 + 2I−1/2
Z
∴
√
dx
√
= −x 1 + x2 + 2
1 + x2
Z √
1 + x2 dx
R
R dx
Problem 1.7.21: Using integration by parts, we arrive at dx
=
1
+
. Subtracting the
x
x
integral from both sides seems to result in the impossible equality 0 = 1. What gives?
Solution Throughout the course of the argument, the integrals are indefinite, and so are
determined only up to an arbitrary constant +C. Certainly the equality 0 = 1 + C holds for
some constant C, thereby resolving the issue.
Problem 1.9.1: Express each of the following rational functions as the sum of a polynomial
with a proper rational function.
Solution Using polynomial long division, we find that
1
x + xx−1
2 −1 = x + x+1
x3
x3 −4
= 1+
4
x3 −4
and that
x3 −1
x2 −1
=
Problem 1.9.2: Compute the following integrals by completing the square.
1
Solution (a) Note that x2 +6x+8
= (x+3)1 2 −1 . The denominator is the difference of two squares,
1
1
and so it equals (x+3−1)(x+3+1)
= (x+2)(x+4)
. Using partial fractions, we set this equal to
B
A
+ x+4 for some constants A, B. Multiplying through by the denominator (x + 2)(x + 4)
x+2
gives 1 = A(x + 4) + B(x + 2). Equating coefficients, we have A + B = 0 and 4A + 2B = 1.
The solutions to this system of equations are A = 21 and B = − 12 . Finally, after integrating
we are left with
(c) Note that
this substitution,
1
5x2 +20x+25
1
1
ln |x + 2| − ln |x + 4| + C
2
2
1
1
1
= 5 x2 +4x+5 = 5 (x+2)1 2 +1 . Let u = x + 2. Then du = dx. Under
Z
Z
dx
1
du
1
=
= arctan(x + 2) + C
2
2
5x + 20x + 25
5
1+u
5
R
x2 +3
Problem 1.9.3: Evaluate x(x+1)(x−1)
dx.
4
2
x +3
B
C
Solution First, we find constants A, B, and C satisfying x(x+1)(x−1)
= Ax + x+1
+ x−1
.
Clearing denominators gives us Ax2 + Bx2 + Cx2 − Bx + Cx − A = x2 + 3. Equating
coefficients, we see that A + B + C = 1, −B + C = 0, and −A = 3. Solving these equations
gives us A = −3, B = 2, and C = 2. Integrating this is easy:
−3 ln |x| + 2 ln |x + 1| + 2 ln |x − 1| + C
R 2 +3
.
Problem 1.9.5: Find the integral x2x(x−1)
2
+3
C
Solution As before, we first find constants A, B, and C satisfying x2x(x−1)
= Ax + xB2 + x−1
.
2
2
2
Clearing denominators gives us Ax − Ax + Bx − B + Cx = x + 3. Equating coefficients,
we see that A + C = 1, −A + B = 0, and −B = 3. Solving these equations gives us A = −3,
B = −3, and C = 4. Again, once we know the constants the integral is straightforward:
−3 ln |x| +
3
+ 4 ln |x − 1| + C
x
Problem 1.9.6: Simplicio tried so hard and got so far // but in the end, it didn’t even
matter.
Solution Notice that Simplicio tried to apply partial fractions to a rational function that
wasn’t proper: the degree of the numerator wasn’t strictly smaller than that of the denomi4x2
A
B
nator. In fact, suppose we could find A and B satisfying (x−3)(x+1)
= x−3
+ (x+1)
. Clearing
2
the denominators, we’d find that 4x = Ax + A + Bx − 3B. The right-hand side has no
squared terms, while the left-hand side does. This phenomenon explains why we can only
apply the method of partial fractions to proper rational functions.
R −2 4
Problem 1.9.7: Find −5 xx2 −1
dx.
+1
R −2
Solution Using polynomial long division, we find that the integrand equals x2 − 1. −5 x2 −
3
1 dx = ( x3 − x) |−2
−5 = 36.
R 5 dx
.
Problem 1.9.9: Find xx2 −1
Solution Using polynomial long division, we find that the integrand equals x3 + x + x2x−1 .
We can integrate each of these terms individually, using the substitution u = x2 − 1 in the
4
2
third term. Doing so yields x4 + x2 + 21 ln |x2 − 1| + C.
R 3x dx
Problem 1.9.14: Find ee4x −1
.
Solution We begin with the substitution u = ex . Then du = ex dx, so dx = du
. This
u
R u2 du
2
transforms the integral into u4 −1 . We can factor the denominator as (u + 1)(u + 1)(u − 1),
A
B
applying the difference of two squares twice. We’ll try to write the integrand as u+1
+ u+1
+
Cu+D
2
2
2
. Clearing the denominator, we get u = A(u − 1)(u + 1) + B(u + 1)(u + 1) + (Cu +
u2 +1
2
D)(u − 1). Expanding and equating coefficients gives the four equations 0 = A + B + C,
1 = −A + B + D, 0 = A + B − C, and 0 = −A + B − D. This system has the single solution
A = − 14 , B = 41 , C = 0, and D = 12 . After integrating and substituting u = ex , we arrive at
the answer − 41 ln |ex + 1| + 14 ln |ex − 1| + 21 arctan(ex + 1) + C.
5
Problem 1.9.15: Find
R
x
√e dx .
1+e2x
Solution This problem seems exceptionally difficult for this section of the book. Let’s see
x
x
what
R duhappens if we try the substitution u = e . Then du = e dx, so the integral becomes
√
. We’ll learn how to do this integral later on in this chapter.
1+u2
R
Problem 1.9.18: Find x(xdx
2 +1) .
Solution We rewrite x(x21+1) as Ax + Bx+C
for some constants A, B, and C. Clearing dex2 +1
2
nominators, we find that 1 = A(x + 1) + (Bx + C)x. Equating coefficients gives us A = 1,
A + B = 0, and C = 0. From this system of equations, we conclude that A = 1, B = −1,
and C = 0. Integrating yields ln |x| − 12 ln |x2 + 1| + C, where we have used the substitution
u = 1 + x2 to find the second integral.
R
1
dx.
Problem 1.9.21: Find (x−1)(x−2)(x−3)
1
A
B
C
Solution We first rewrite (x−1)(x−2)(x−3)
as x−1
+ x−2
+ x−3
for some constants A, B, and
C. Clearing denominators, we get A(x − 2)(x − 3) + B(x − 1)(x − 3) + C(x − 1)(x − 2) = 1.
Setting x = 1, we find that 2A = 1, so A = 21 . Setting x = 2, we find that B(−1) = 1, so
B = −1. Finally, setting x = 3 gives us 2C = 1, so C = 12 . (You could also multiply these
terms out and equate coefficients, but the Heaviside trick is faster in this case.) Integrating
results in the answer 21 ln |x − 1| − ln |x − 2| + 21 ln |x − 3| + C.
R
x2 +1
dx
Problem 1.9.22: (x−1)(x−2)(x−3)
Solution The procedure is exactly as in the problem above. We get A(x − 2)(x − 3) + B(x −
1)(x − 3) + C(x − 1)(x − 2) = x2 + 1. Setting x = 1, we get 2A = 2, so A = 1. Setting x = 2,
we get −B = 5, so B = −5. Finally, setting x = 3 gives us 2C = 10, so C = 5. Integating
results in the answer ln |x − 1| − 5 ln |x − 2| + 5 ln |x − 3| + C.
R
x3 +1
Problem 1.9.23: (x−1)(x−2)(x−3)
dx
Solution Since the integrand is not a proper rational function, we have to perform polynomial long division before attempting to use partial fractions. Note that (x−1)(x−2)(x−3) =
3 +1
2 −11x+7
2 −11x+7
x3 − 6x2 + 11x − 6 and that x3 −6xx2 +11x−6
= 1 + x36x
. We set x36x
equal to
−6x2 +11x−6
−6x2 +11x−6
A
B
C
+ x−2 + x−3 . Clearing denominators, we arrive at the expression A(x − 2)(x − 3) + B(x −
x−1
1)(x − 3) + C(x − 1)(x − 2) = 6x2 − 11x + 7. Setting x = 1, we get 2A = 2, so A = 1. Setting
x = 2, we get −B = 9, so B = −9. Setting x = 3, we get 2C = 28, so C = 14. Integrating
results in the answer ln |x − 1| − 9 ln |x − 2| + 14 ln |x − 3| + C.
6