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What did I do Wrong with this integration by u-sub

  1. Jul 10, 2014 #1
    1. The problem statement, all variables and given/known data

    ∫x^2(a^2-x^2)^.5 limits from 0 to a
    2. Relevant equations

    ∫x^2(a^2-x^2)^.5 limits from 0 to a

    3. The attempt at a solution

    It's way to much for me to type in the short amount of time I have so I included a picture of my work. It's neat and easy to read.

    the answer should be (PI/16)*a^4
     

    Attached Files:

  2. jcsd
  3. Jul 10, 2014 #2

    LCKurtz

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    To tell you the truth, I didn't read the whole thing. But I suggest where you have$$
    a^4\int_0^a \sin^2\theta \cos^2\theta~d\theta$$you write that as$$
    a^4\int_0^a \left(\frac {\sin(2\theta}{2}\right )^2~d\theta$$and immediately use the double angle formula for the ##\sin^2(2\theta)##. It will be much easier.
     
  4. Jul 10, 2014 #3
    Which Identity did you use to get that?
     
  5. Jul 10, 2014 #4

    LCKurtz

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    ##\sin(2\theta)=2\sin\theta\cos\theta##
     
  6. Jul 10, 2014 #5
    You did not change the limits when you did the substitution.

    $$a^4\int_0^{\pi/2}\sin^2\theta \cos^2\theta\,d\theta=\frac{a^4}{4}\int_0^{\pi/2} \sin^2(2\theta)\,d\theta$$
     
  7. Jul 10, 2014 #6
    Why does my a become PI/2?
     
  8. Jul 10, 2014 #7
    ##x=\sin\theta##, at ##x=a##, ##a=a\sin\theta \Rightarrow \theta=\pi/2##.
     
  9. Jul 10, 2014 #8
    I understand that. The problem is when I apply it, it just gives me my original integral.
     
  10. Jul 10, 2014 #9
    I don't see how that matters. a/a = 1 and the arcsin(1) = PI/2

    It's the same.
     
  11. Jul 10, 2014 #10
    I know it's a lot to read but can someone please look over my entire work? I am very close to the answer except for the numerator.
     
  12. Jul 10, 2014 #11

    LCKurtz

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    Apparently you don't understand it. Use$$
    \sin^2(2\theta) = \frac {1-\cos(4\theta)}2$$And, as another poster has noticed and I missed, the upper limit for ##\theta## should be ##\frac \pi 2##.
     
  13. Jul 11, 2014 #12

    LCKurtz

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    So, are you going to try it? It's only one simple step left to integrate to the correct answer.
     
  14. Jul 11, 2014 #13
    I did get the answer. I went to sleep last night, where I'm located it was past midnight. I solved it this morning.

    I didn't quite do it the way you guys have suggested.
     

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  15. Jul 11, 2014 #14

    LCKurtz

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    You should try it. Much simpler and you might learn something useful.
     
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