Evaluating the integral, correct?

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In summary: You should have done \int e^{-x}\cos{2x}dxinstead of\int e^{-x}\cos{2x}dxIn summary, the integrand in the homework equation is x^2/m sinmx. Evaluating this integral requires multiple stages of integration by parts. The 'm' is a constant, so that will not be involved in the integration. Making the integral (1/m) · integral[ x sin(mx) ] dx . Now u = x and dv =
  • #1
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Homework Statement



Evaluate the integral

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

Homework Equations

Evaluating the integral, correct?

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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  • #3
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...
 
  • #4
i heard that i should do parts with xsinmx / m, should i do that?
 
  • #5
Zack88 said:
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.
 
  • #6
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?
 
  • #7
Zack88 said:
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...
 
  • #8
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 I am lost
 
  • #9
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!
 
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  • #10
Ok, I'm officially done typing! Sorry about that, had too many typographical errors.
 
  • #11
[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
 
  • #12
Zack88 said:
[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]
 
  • #13
ok i see now, thank you, sad part is I am not done with this question and i have one more just like it :(
 
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  • #14
Zack88 said:
ok i see now, thank you, sad part is I am not done with this question and i have one more just like it :(
Post it and we'll work on it step by step.
 
  • #15
[integral] e^-x cos 2x dx

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

dv = cos 2x
v = sin 2x / 2
 
  • #17
[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!
 
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  • #18
ok so [tex]\frac{1}{2}\int e^{-x}\sin{2x}dx[/tex] turns into -1/2e^-xcos2x + c?
 
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  • #19
You're supposed to have "2" Parts just like any Integration by Parts.

What is the other term?
 
  • #20
so i don't take the integral just yet? i do another integration by parts?
 
  • #21
Zack88 said:
so i don't take the integral just yet? i do another integration by parts?
Evaluate [tex]\frac{1}{2}\int e^{-x}\sin{2x}dx[/tex] how you normally would.
 
  • #22
u = e^-x
du = -e^-x

dv = sin 2x
v= -1/2 cos 2x
 
  • #23
Yes, so now you've solved it?
 
  • #24
not quite, I am not sure how to put it all together I know i'll have 1/2e^-xsin2x but I am not sure how to put in the new u, du, dv, and v
 
  • #25
Zack88 said:
not quite, I am not sure how to put it all together I know i'll have 1/2e^-xsin2x but I am not sure how to put in the new u, du, dv, and v
Look at what I did in post 9.
 
  • #26
ok I am not done yet but have so far...

1/2e^-xsin2x + 1/2 (1/2 e^-x cos2 x

and that is where i stopped i didnt understand how you got + 1/m [integra] cosmx for post 9
 
  • #27
By Parts.

[tex]uV-\int Vdu[/tex]
 
  • #28
lol sorry i saw [tex](\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)[/tex] as two different things
 
  • #29
so now i have 1/2 e^-x sin 2x + 1/2 (1/2 e^-x cos 2x + [integral] 1/2 e^-x cos 2x)
 
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  • #30
so now do I finally take the integral (no more parts) to come out with the final answer?
 
  • #31
Do you notice that you get your original Integral back? Move it to the left side and add/subtract your Integral, then divide by the constant, and you're done.
 
  • #32
1/2 e^-x sin 2x + 1/2 (1/2 e^-x cos 2x + [integral] 1/2 e^-x cos 2x)

so take e^-x cos 2x from within the parenthesis and move it to the outside of the parenthesis leaving 1/2 [integral 1/2 x^-x cos 2x) in the parenthesis
 
  • #33
It's basically ...

a, b, c ... = terms from by doing Parts

[tex]I=a+b+c+\frac{1}{2}(d+I)[/tex]

Now distribute your constant 1/2

[tex]I=a+b+c+\frac{1}{2}d+\frac{1}{2}I[/tex]

Bring your Original Integral to the left side, and divide by the constant.

[tex]\frac{1}{2}I=a+b+c+\frac{1}{2}d[/tex]

[tex]I=2a+2b+2c+d[/tex]

Solved!
 
  • #34
ok i get the concept but now I am confused at what should be a, b, c, I know what d is and of course I. what I see is:

a = 1/2
b = e^-x
c = sin 2x
 
  • #35
a, b, c ... etc are just uV, they're solved and do not need to be integrated.
 
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