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Evaluating the integral, correct?

by Zack88
Tags: correct, evaluating, integral
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Zack88
#1
Jan18-08, 09:36 PM
P: 62
1. The problem statement, all variables and given/known data

Evaluate the integral

[tex]\int x^2 \cos mx dx[/tex]


2. Relevant equations


Evaluating the integral, correct?

3. The attempt at a solution

[tex]u = x^2[/tex]
[tex]du = 2x[/tex]
[tex]dv = \cos mx[/tex]
[tex]v= \frac {\sin mx }{m}[/tex]

(x^2)(sin mx / m) - [integral] (sin mx / m)(2x)

(x^2)(sin mx / m) - 2 [integral] (sin mx / m) (x)

[My final answer:] (x^2)(sin mx / m) + 2 (cos mx / m) + c
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rocomath
#2
Jan18-08, 09:45 PM
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P: 1,754
Hurts the eyes. If you plan on coming here for help on a regular basis, hopefully you will take the time to learn how to type in LaTeX :-]

http://physicsforums.com/showthread.php?t=8997
dynamicsolo
#3
Jan18-08, 10:00 PM
HW Helper
P: 1,662
I believe you lost a term in evaluating the second term integral in this:

(x^2)(sin mx / m) - 2 [integral] (sin mx / m) (x) .

If you differentiate your final result,

(x^2)(sin mx / m) + 2 (cos mx / m) + c ,

you don't cancel out the additional terms beyond the original integrand...

Zack88
#4
Jan18-08, 10:08 PM
P: 62
Evaluating the integral, correct?

i heard that i should do parts with xsinmx / m, should i do that?
dynamicsolo
#5
Jan18-08, 10:13 PM
HW Helper
P: 1,662
Quote Quote by Zack88 View Post
i heard that i should do parts with xsinmx / m, should i do that?
Yes, you'll need to do that. In general, integration of functions defined by polynomials times sin kx, cos kx, or e^kx need multiple stages of integration by parts.
Zack88
#6
Jan18-08, 10:15 PM
P: 62
ok with doing parts again confuses me since i have xsinmx /m. The u should it be m? but then what is the derivative of m?
dynamicsolo
#7
Jan18-08, 10:23 PM
HW Helper
P: 1,662
Quote Quote by Zack88 View Post
ok with doing parts again confuses me since i have xsinmx /m. The u should it be m? but then what is the derivative of m?
The 'm' is a constant, so that will not be involved in the integration. Make the integral
(1/m) integral[ x sin(mx) ] dx . Now u = x and dv = sin(mx) dx . You can append the multiplicative constant (1/m) afterwards...
Zack88
#8
Jan18-08, 10:37 PM
P: 62
ok so now I have

x^2/m sinmx + 2/m [integral] xsinmx

u = x
du = dx

dv = sin mx
v = 1/m cos mx

and now im lost
rocomath
#9
Jan18-08, 10:45 PM
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P: 1,754
Ok let's start from scratch.

[tex]I=\int x^2\cos{mx}dx[/tex]

[tex]u=x^2[/tex]
[tex]du=2xdx[/tex]

[tex]dV=\cos{mx}dx[/tex]
[tex]V=\frac{1}{m}\sin{mx}[/tex]

[tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\int x\sin{mx}dx[/tex]

Now we have to do Parts again.

[tex]u=x[/tex]
[tex]du=dx[/tex]

[tex]dV=\sin{mx}dx[/tex]
[tex]V=\frac{-1}{m}\cos{mx}[/tex]

[tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)[/tex]

Now you can easily evaluate this Integral!!!
rocomath
#10
Jan18-08, 11:01 PM
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P: 1,754
Ok, I'm officially done typing! Sorry about that, had too many typographical errors.
Zack88
#11
Jan18-08, 11:02 PM
P: 62
[tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\int\cos{mx}dx\right)[/tex]
ok where did the last [tex]\int\cos{mx}dx\right)[/tex] come from
rocomath
#12
Jan18-08, 11:05 PM
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P: 1,754
Quote Quote by Zack88 View Post
[tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\int\cos{mx}dx\right)[/tex]
ok where did the last [tex]\int\cos{mx}dx\right)[/tex] come from
By doing Parts again to evaluate ...

[tex]\int x\sin{mx}dx[/tex]
Zack88
#13
Jan18-08, 11:07 PM
P: 62
ok i see now, thank you, sad part is im not done with this question and i have one more just like it :(
rocomath
#14
Jan18-08, 11:15 PM
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P: 1,754
Quote Quote by Zack88 View Post
ok i see now, thank you, sad part is im not done with this question and i have one more just like it :(
Post it and we'll work on it step by step.
Zack88
#15
Jan18-08, 11:19 PM
P: 62
[integral] e^-x cos 2x dx

u = e^-x
du = -e^-x

dv = cos 2x
v = sin 2x / 2
rocomath
#16
Jan18-08, 11:22 PM
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P: 1,754
Can you check post #9 again http://physicsforums.com/showpost.ph...42&postcount=9 I forgot to type in the constant.

Anyways, your new problem is ...

[tex]\int e^{-x}\cos{2x}dx[/tex]

Right?
rocomath
#17
Jan18-08, 11:28 PM
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P: 1,754
[tex]I=\int e^{-x}\cos{2x}dx[/tex]

[tex]u=e^{-x}[/tex]
[tex]du=-e^{-x}dx[/tex]

[tex]dV=\cos{2x}dx[/tex]
[tex]V=\frac{1}{2}\sin{2x}[/tex]

[tex]I=\frac{1}{2}e^{-x}\sin{2x}+\frac{1}{2}\int e^{-x}\sin{2x}dx[/tex]

Ok done typing. Evaluate your current Integral, notice that your Original Integral reappears? Just move it to the left then divide by the constant and you're done!
Zack88
#18
Jan18-08, 11:34 PM
P: 62
ok so [tex]\frac{1}{2}\int e^{-x}\sin{2x}dx[/tex] turns into -1/2e^-xcos2x + c?


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