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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.
- (7.1 #2) โซ ln x x dx
Let u = ln x, then du = (^1) x dx, and we rewrite the integral as: โซ (^) 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: โซ cos(x)esin(x)^ dx =
eu^ du = eu^ + C
= esin^ x^ + C
4 x^3 โ 4 x x^4 โ 2 x^2 + 3 dx
Let u = x^4 โ 2 x^2 + 3, then du = 4x^3 โ 4 x dx, and we can write the integral as: โซ 4 x^3 โ 4 x x^4 โ 2 x^2 + 3 dx =
du u = ln u + C
= ln(x^4 โ 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 โ 1) (4x^2 โ 4 x)^2 dx
First weโll factor the denominator as (4x^2 โ 4 x)^2 = [4(x^2 โ x)]^2 = 16(x^2 โ x)^2 giving us 1 16
โซ (^) (2x โ 1) (x^2 โ x)^2 dx
Now we let u = x^2 โ x, then du = 2x โ 1, and we can write the integral as 1 16
โซ (^) (2x โ 1) (x^2 โ x)^2 dx =
โซ (^) du u^2 = โ
uโ^1 + C
= โ 16(x^12 โx) + C
x
x + 1 dx
Let u = x + 1, then du = dx (and note that x = u โ 1). Then we can write the integral as โซ x
x + 1 dx =
(u โ 1)u^1 /^2 du
=
u^3 /^2 โ u^1 /^2 du
= 2 5 u^5 /^2 โ 2 3 u^3 /^2 + C
= 25 (x + 1)^5 /^2 โ 23 (x + 1)^3 /^2 + C
x sin x dx
Let u = x and dv = sin x dx. Then du = dx and v = โ cos x. Integration by parts gives us: โซ x sin x dx =
u dv
= uv โ
v du
= โx cos x +
cos x dx
= โx 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 = โ
e^2 x^ cos(3x) +
e^2 x^ sin(3x) 13 9
e^2 x^ sin(3x) dx = โ
e^2 x^ cos(3x) +
e^2 x^ sin(3x) โซ
e^2 x^ sin(3x) dx = โ 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: โซ (^) ln(sin x) tan x dx =
u du
=
u^2 + C
=
(ln(sin x))^2 + C
โซ (^) ฯ
0
x (sin x + cos x) dx
Let u = x and dv = sin x + cos x dx. Then du = dx and v = โ cos x + sin x. We can then rewrite the integral as: โซ (^) ฯ
0
x (sin x + cos x) dx =
โซ (^) ฯ
0
u dv
= uv |ฯ 0 โ
โซ (^) ฯ
0
v du
= (โx cos x + x sin x) |ฯ 0 โ
โซ (^) ฯ
0
sin x โ cos x dx = (โx cos x + x sin x) |ฯ 0 + (cos x + sin x)|ฯ 0 = (โฯ(โ1) + 0) โ (0) + (โ1 + 0) โ (1 โ 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: โซ 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 = โ cos w. So by integration by parts, we have: โซ e^2 x^ sin(ex) dx =
w sin w dw
= โw cos w +
cos w dw = โw cos w + sin w + C = โex^ 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
