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Math 1352-11 Solutions: Integration of Various Functions - Prof. Victoria Ellen Howle, Assignments of Mathematics

Solutions to selected integration problems from math 1352-11, including the use of substitution and integration by parts. Topics covered include logarithmic, trigonometric, and exponential integrals.

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Pre 2010

Uploaded on 03/19/2009

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Math 1352-11 โ€” WW04 Solutions
October 8, 2008
Assigned problems: 7.1 โ€“ 2, 4, 8, 12, 28; 7.2 โ€“ 2, 4, 8, 16, 22, 26
Always read through the solution sets even if your answer was correct.
1. (7.1 #2)
Zln x
xdx
Let u= ln x, then du =1
xdx, and we rewrite the integral as:
Zln x
xdx =Zu du
=1
2u2+C
=1
2(ln x)2+C
2. (7.1 # 4)
Zcos(x)esin(x)dx
Let u= sin x, then du = cos x dx, and we can write the integral as:
Zcos(x)esin(x)dx =Zeudu
=eu+C
=esin x+C
3. (7.1 # 8)
Z4x3โˆ’4x
x4โˆ’2x2+ 3 dx
Let u=x4โˆ’2x2+ 3, then du = 4x3โˆ’4x dx, and we can write the integral as:
Z4x3โˆ’4x
x4โˆ’2x2+ 3 dx =Zdu
u
= ln u+C
=ln(x4โˆ’2x2+ 3) + C
Copyright 2008, Victoria Howle and Department of Mathematics & Statistics, Texas Tech University.
All rights reserved. No part of this document may be reproduced, redistributed, or transmitted
in any manner without the permission of the instructor.
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Download Math 1352-11 Solutions: Integration of Various Functions - Prof. Victoria Ellen Howle and more Assignments Mathematics in PDF only on Docsity!

Math 1352-11 โ€” WW04 Solutions

October 8, 2008

Assigned problems: 7.1 โ€“ 2, 4, 8, 12, 28; 7.2 โ€“ 2, 4, 8, 16, 22, 26

Always read through the solution sets even if your answer was correct.

  1. (7.1 #2) โˆซ ln x x dx

Let u = ln x, then du = (^1) x dx, and we rewrite the integral as: โˆซ (^) ln x x dx =

u du

=

u^2 + C

= 12 (ln x)^2 + C

cos(x)esin(x)^ dx

Let u = sin x, then du = cos x dx, and we can write the integral as: โˆซ cos(x)esin(x)^ dx =

eu^ du = eu^ + C

= esin^ x^ + C

4 x^3 โˆ’ 4 x x^4 โˆ’ 2 x^2 + 3 dx

Let u = x^4 โˆ’ 2 x^2 + 3, then du = 4x^3 โˆ’ 4 x dx, and we can write the integral as: โˆซ 4 x^3 โˆ’ 4 x x^4 โˆ’ 2 x^2 + 3 dx =

du u = ln u + C

= ln(x^4 โˆ’ 2 x^2 + 3) + C

Copyright 2008, Victoria Howle and Department of Mathematics & Statistics, Texas Tech University. All rights reserved. No part of this document may be reproduced, redistributed, or transmitted

(2x โˆ’ 1) (4x^2 โˆ’ 4 x)^2 dx

First weโ€™ll factor the denominator as (4x^2 โˆ’ 4 x)^2 = [4(x^2 โˆ’ x)]^2 = 16(x^2 โˆ’ x)^2 giving us 1 16

โˆซ (^) (2x โˆ’ 1) (x^2 โˆ’ x)^2 dx

Now we let u = x^2 โˆ’ x, then du = 2x โˆ’ 1, and we can write the integral as 1 16

โˆซ (^) (2x โˆ’ 1) (x^2 โˆ’ x)^2 dx =

โˆซ (^) du u^2 = โˆ’

uโˆ’^1 + C

= โˆ’ 16(x^12 โˆ’x) + C

x

x + 1 dx

Let u = x + 1, then du = dx (and note that x = u โˆ’ 1). Then we can write the integral as โˆซ x

x + 1 dx =

(u โˆ’ 1)u^1 /^2 du

=

u^3 /^2 โˆ’ u^1 /^2 du

= 2 5 u^5 /^2 โˆ’ 2 3 u^3 /^2 + C

= 25 (x + 1)^5 /^2 โˆ’ 23 (x + 1)^3 /^2 + C

x sin x dx

Let u = x and dv = sin x dx. Then du = dx and v = โˆ’ cos x. Integration by parts gives us: โˆซ x sin x dx =

u dv

= uv โˆ’

v du

= โˆ’x cos x +

cos x dx

= โˆ’x cos x + sin x + C

Copyright 2008, Victoria Howle and Department of Mathematics & Statistics, Texas Tech University. All rights reserved. No part of this document may be reproduced, redistributed, or transmitted

to the left-hand side and solve for the value of the integral: ( 1 +

e^2 x^ sin(3x) dx = โˆ’

e^2 x^ cos(3x) +

e^2 x^ sin(3x) 13 9

e^2 x^ sin(3x) dx = โˆ’

e^2 x^ cos(3x) +

e^2 x^ sin(3x) โˆซ

e^2 x^ sin(3x) dx = โˆ’ 133 e^2 x^ cos(3x) + 132 e^2 x^ sin(3x) + C

ln(sin x) tan x dx

There are multiple ways to do this integral, but I found substitution to be easiest. Let u = ln(sin x), then using the Chain Rule, du = (^) sin^1 x ยท cos x dx = cos sin^ xx dx = (^) tan^1 x dx. So we can rewrite the integral as: โˆซ (^) ln(sin x) tan x dx =

u du

=

u^2 + C

=

(ln(sin x))^2 + C

โˆซ (^) ฯ€

0

x (sin x + cos x) dx

Let u = x and dv = sin x + cos x dx. Then du = dx and v = โˆ’ cos x + sin x. We can then rewrite the integral as: โˆซ (^) ฯ€

0

x (sin x + cos x) dx =

โˆซ (^) ฯ€

0

u dv

= uv |ฯ€ 0 โˆ’

โˆซ (^) ฯ€

0

v du

= (โˆ’x cos x + x sin x) |ฯ€ 0 โˆ’

โˆซ (^) ฯ€

0

sin x โˆ’ cos x dx = (โˆ’x cos x + x sin x) |ฯ€ 0 + (cos x + sin x)|ฯ€ 0 = (โˆ’ฯ€(โˆ’1) + 0) โˆ’ (0) + (โˆ’1 + 0) โˆ’ (1 โˆ’ 0)

e^2 x^ sin(ex) dx

Copyright 2008, Victoria Howle and Department of Mathematics & Statistics, Texas Tech University. All rights reserved. No part of this document may be reproduced, redistributed, or transmitted

First we will do a substitution. Let w = ex, then dw = ex^ dx. Then the integral can be rewritten as: โˆซ e^2 x^ sin(ex) dx =

w sin w dw

This new integral is like the one in problem 6 (7.2 #2). So as in that problem, we will now use integration by parts. Let u = w and dv = sin w dw. Then du = dw and v = โˆ’ cos w. So by integration by parts, we have: โˆซ e^2 x^ sin(ex) dx =

w sin w dw

= โˆ’w cos w +

cos w dw = โˆ’w cos w + sin w + C = โˆ’ex^ cos(ex) + sin(ex) + C

Copyright 2008, Victoria Howle and Department of Mathematics & Statistics, Texas Tech University. All rights reserved. No part of this document may be reproduced, redistributed, or transmitted