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.

22

NTU 2015 Calculus I - Solutions to Problem Set 11

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.

!

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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



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

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Highlight



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


(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.


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.

!

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problem set 11 solutions.pdf

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