Problem Set 11 Solutions.pdf
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Transcript of Problem Set 11 Solutions.pdf
Problem 15: (Various from Section 4.5 from [Stewart]).
In this problem we were asked to do substitutions to evaluate various inte-grals.
Solution to 1.
The appropriate substitution is to set u = x2. Then du = 2x dx.!
x sinx2 dx =1
2
!sinx2 2xdx
=1
2
!sinu du
= −1
2cosu+ C
= −1
2cosx2 + C.
Solution to 2.
Set u = 1− 2x. Then du = (−2)dx.
!(1− 2x)9 dx = −1
2
!(1− 2x)9 (−2)dx
= −1
2
!u9 du
= −1
2∗ 1
10u10 + C
= − 1
20(1− 2x)10 + C.
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Solution to 3.
Note that ddx [tan θ] = sec2 θ. So we’ll set u = tan θ.
!sec2 θ tan3 θ dθ =
!u3 du
=1
4u4 + C
=1
4(tan θ)4 + C.
Solution to 4.
Set u = 1 + x3/2. Then du = 32
√xdx.
! √x sin
"1 + x3/2
#dx =
!sinu du
= − cosu+ C
= − cos (1 + x3/2) + C.
Solution to 5.
Set u = 1 + z3. Then du = 3z2dz.!
z2
3√1 + z3
dx =1
3
!(1 + z3)−1/33z2 dz
=1
3
!u−1/3 du
=1
3∗ 3
2∗ u2/3 + C
=1
2(1 + z3)2/3 + C.
Solution to 6.
We can see the derivative of 1+tan t sitting out the front of this expression.So we’ll set u = 1 + tan t.
!dt
cos2 t√1 + tan t
dt =
!1√
1 + tan t
d
dt[1 + tan t] dt
=
!u−1/2 du
= 2u1/2 + C
= 2√1 + tan t+ C.
23
Solution to 7.
It looks like we should make the substitution u = 2x + 5. Then du = 2dx.The issue here is rewriting what is left as a function of u.
!x(2x+ 5)8 dx =
1
2
! $1
2(2x+ 5)− 5
2
%(2x+ 5)8 2dx
=1
2
! $1
2u9 − 5
2u8%
du
=1
2
$1
2∗ 1
10∗ u10 − 5
2∗ 1
9∗ u9
%+ C
=1
40(2x+ 5)10 − 5
36(2x+ 5)9 + C.
Solution to 8.
Set u = lnx. Then du = 1xdx.
!(lnx)2
xdx =
!u2 du
=1
3u3 + C
=1
3(lnu)3 + C.
Solution to 9.
Set u = 1 + ex. Then du = exdx.!
ex√1 + ex dx =
!u1/2du
=2
3u3/2 + C
=2
3(1 + ex)3/2 + C.
24
Solution to 10.
Set u = tanx. Then du = sec2 xdx.!
etanx sec2 x dx =
!eu du
= eu + C
= etanx + C.
Solution to 11.
Recall that:d
dx
&tan−1 x
'=
1
1 + x2.
So we’ll set u = tan−1 x.!
tan−1 x
1 + x2dx =
!u du
=1
2u2 + C
=1
2
(tan−1 x
)2+ C.
Solution to 12.
Set u = x2. Then du = 2x dx.!
x
1 + x4dx =
1
2
!1
1 + (x2)22xdx
=1
2
!1
1 + u2du
=1
2tan−1 u+ C
=1
2tan−1
(x2)+ C.
!
25
Problem 16: (Various from Section 7.1 from [Stewart]).
In this problem we were asked to do an appropriate integration-by-parts toevaluate various indefinite integrals. We’ll put in full detail for a few of thefollowing solutions, but will be more brief in the later solutions.
Solution to 1.
This integral looks like it will be simplified if the factor t was replaced byddt [t] = 1. This observation guides an integration-by-parts calculation:
!te−3t dt = −1
3
!td
dt
&e−3t
'dt
= −1
3
! $d
dt
&te−3t
'− e−3t dt
%
= −1
3
$te−3t −
!e−3t dt
%
= −1
3
$te−3t +
1
3
!e−3t (−3)dt
%
= −1
3
$te−3t +
1
3
!eu du
%(Setting u = −3t.)
= −1
3
$te−3t +
1
3eu + C
%
= −1
3te−3t − 1
9e−3t + C ′.
26
Solution to 2.
To simplify this integral we would like to differentiate the first term, x2+2x.To replace it by a constant we would need to differnetiate twice, so we expectto have to do two integration-by-parts steps.
!(x2 + 2x) cosx dx
=
!(x2 + 2x)
d
dx[sinx] dx
=
!d
dx
&(x2 + 2x) sinx
'dx−
!d
dx
&x2 + 2x
'sinx dx
= (x2 + 2x) sinx−!
(2x+ 2) sinx dx
= (x2 + 2x) sinx− 2
!x sinx dx− 2
!sinx dx.
To solve the second term in the sum, we’ll need to do another integration-by-parts. Continuing:
= (x2 + 2x) sinx− 2
!xd
dx[− cosx] dx+ 2 cosx+ C
= (x2 + 2x) sinx− 2
$−x cosx−
!d
dx[x](− cosx) dx
%+ 2 cosx+ C
= (x2 + 2x) sinx− 2
$−x cosx+
!cosx dx
%+ 2 cosx+ C
= (x2 + 2x) sinx+ 2x cosx− 2 sinx+ 2 cosx+ C.
27
Solution to 3.
This integral is a bit perplexing at first glance, because we would expect theintegrand of an integration-by-parts calculation to consist of two factors.At any rate the integral will be simplified if we can replace the term ln 3
√x
by its derivative. We can do that by interpreting the second factor as ‘1’.Observe:
!ln 3
√x dx =
!1 ∗ ln 3
√x dx
=
!d
dx[x] ∗ ln 3
√x dx
= x ln 3√x−
!x ∗ d
dx
&ln 3
√x'dx
= x ln 3√x−
!x ∗ 1
3√x∗ 1
3x−2/3 dx
= x ln 3√x− 1
3
!x1−1/3−2/3 dx
= x ln 3√x− 1
3
!1 dx
= x ln 3√x− 1
3x+ C.
Solution to 4.
This integral will be similar to the previous one.!
sin−1(x) dx =
!d
dx[x] sin−1 x dx
= x sin−1 x−!
x ∗ d
dx
&sin−1 x
'dx
= x sin−1 x−!
x√1− x2
dx
= x sin−1 x+1
2
!1√
1− x2(−2x)dx
= x sin−1 x+1
2
!u−1/2 du (Setting u = 1− x2.)
= x sin−1 x+ u1/2 + C
= x sin−1 x+*
1− x2 + C.
28
Solution to 5.
To simplify this we aim to differentiate ln p.!
p5 ln p dp =
!d
dp
+1
6p6,ln p dp
=1
6p6 ln p− 1
6
!p6
d
dp[ln p] dp
=1
6p6 ln p− 1
6
!p5 dp
=1
6p6 ln p− 1
36p6 + C.
Solution to 6.
This integral can be simplified by differnetiating the s factor. To do thatwe need to find an antiderivative for 2s.!
s2s ds =1
ln 2
!s ln 2 2s ds
=1
ln 2
!sd
ds[2s] ds
=1
ln 2
$s2s −
!2s ds
%
=1
ln 2s2s − 1
(ln 2)22s + C.
Solution to 7.
!(lnx)2 dx =
!d
dx[x] ∗ (lnx)2 dx
= x(lnx)2 −!
x ∗ 2 ∗ lnx ∗ 1
xdx
= x(lnx)2 − 2
!lnx dx
= x(lnx)2 − 2
!d
dx[x] lnx dx
= x(lnx)2 − 2
$x lnx−
!1 dx
%
= x(lnx)2 − 2 (x lnx− x) + C
= x(lnx)2 − 2x lnx+ 2x+ C.
29
Solution to 8.
This problem took me a fair bit of trial and error before I found an organi-zation of this integral which solved the problem.
!xe2x
(1 + 2x)2dx
=
!1
(1 + 2x)2∗ (xe2x) dx
=
!d
dx
+−1
2∗ (1 + 2x)−1
,∗ xe2x dx
=
$−1
2∗ (1 + 2x)−1xe2x
%−$−1
2
%!(1 + 2x)−1 d
dx[xe2x] dx
=
$−1
2∗ (1 + 2x)−1xe2x
%+
1
2
!1
1 + 2x∗ (1 + 2x)e2x dx
= − 1
2(1 + 2x)xe2x +
1
4
!2e2x dx
= − 1
2(1 + 2x)xe2x +
1
4e2x + C.
!
30
Problem 17: (#7.1.37 from [Stewart]).
Find some combination of techniques to evaluate the indefinite integral:!
cos√x dx.
Solution
Looking at the integral we judge that the only way forward will be to replace√x by a variable u =
√x. To do the substitution we’ll use du = 1
2√xdx. So
our first step will be to introduce this missing term 12√xso we can change
variables.!
cos√x dx
= 2
! √x cos
√x ∗ 1
2√xdx
= 2
!u cosu du
= 2
!ud
du[sinu] du
= 2
$u sinu−
!sinu du
%
= 2u sinu+ 2 cosu+ C
= 2√x sin
√x+ 2 cos
√x+ C.
!
31
Problem 18: (#7.1.41 from [Stewart]).
Find some combination of techniques to evaluate the indefinite integral:!
x ln (1 + x) dx.
Solution
There seem multiple approaches to this. One way is to set u = ln (1 + x).Then du = 1
1+x dx. And
x = (1 + x)− 1 = eln (1+x) − 1 = eu − 1.
So:!
x ln (1 + x) dx
=
!x(1 + x) ln (1 + x)
1
1 + xdx
=
!(eu − 1)euu du
=
!(e2u − eu)u du
=
!d
dx
+1
2e2u − eu
,u du
=
$1
2e2u − eu
%u−
! $1
2e2u − eu
%du
=
$1
2e2u − eu
%u− 1
4e2u + eu + C
=
$1
2e2 ln (1+x) − eln (1+x)
%ln (1 + x)− 1
4e2 ln (1+x) + eln (1+x) + C
=
$1
2eln (1+x)2 − eln (1+x)
%ln (1 + x)− 1
4eln (1+x)2 + eln (1+x) + C
=
$1
2(1 + x)2 − (1 + x)
%ln (1 + x)− 1
4(1 + x)2 + (1 + x) + C
=1
2
(x2 − 1
)ln (1 + x)− x2
4+
x
2+ C ′.
!
32
Problem 19: (#7.1.42 from [Stewart]).
Find some combination of techniques to evaluate the indefinite integral:!
sin (lnx) dx.
Solution
It seems we will need to make the substitution u = lnx. For that we need:du = 1
x dx. So:
!sin (lnx) dx =
!x sin (lnx)
1
xdx
=
!elnx sin (lnx)
1
xdx
=
!eu sinu du.
To determine this integral we need to use the trick we met in lectures:!
eu sinu du = eu sinu−!
eu cosu du
= eu sinu−$eu cosu+
!eu sinu du
%
= eu (sinu− cosu)−!
eu sinu du.
Solving this for-eu sinu du, and substituting back into the above calcula-
tion, we get that!
sin (lnx) dx =1
2eu (sinu− cosu) + C =
1
2x (sin lnx− cos lnx) + C.
!
33
Problem 20: (#7.1.48 from [Stewart]).
(a) Prove the following “reduction formula”. For every n ∈ N, with n atleast 2,
!cosn x dx =
1
ncosn−1 x sinx+
n− 1
n
!cosn−2 x dx.
(b) Use Part (a) to evaluate-cos2 x dx.
(c) Use Part (a) and (b) to evaluate-cos4 x dx.
Solutions
Part (a).
Let n be a positive integer at least 2. Then:!
cosn x dx
=
!cosn−1 x cosx dx
=
!cosn−1 x
d
dx[sinx] dx
= sinx cosn−1 x−!
sinxd
dx
&cosn−1 x
'dx
= sinx cosn−1 x− (n− 1)
!sinx ∗ (− sinx) cosn−2 x dx
= sinx cosn−1 x+ (n− 1)
!(1− cos2 x) cosn−2 x dx
= sinx cosn−1 x+ (n− 1)
!cosn−2 x dx− (n− 1)
!cosn x dx.
Solving for-cosn x dx we deduce:
!cosn x dx =
1
nsinx cosn−1 x+
n− 1
n
!cosn−2 x dx .
Part (b).
Specializing this formula to n = 2 we obtain:!
cos2 x dx =1
2sinx cosx+
1
2
!cos0 x dx =
1
2sinx cosx+ C.
34
Part (c).
!cos4 x dx =
1
4sinx cos3 x+
3
4
!cos2 x dx
=1
4sinx cos3 x+
3
4∗ 1
2sinx cosx+ C
=1
4sinx cos3 x+
3
8sinx cosx+ C.
Problem 21: (#7.1.52 from [Stewart]).
Prove the following reduction formula:!
xnex dx = xnex − n
!xn−1ex dx.
Solution
Let n be an integer.!
xnex dx =
!xn
d
dx[ex] dx
= xnex −!
d
dx[xn] ex dx
= xnex − n
!xn−1ex dx.
!
35