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Math Help - \int{\frac{x^2}{\sqrt{4x-x^2}}}dx

  1. #1
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    \int{\frac{x^2}{\sqrt{4x-x^2}}}dx

    Having trouble with this one.

    \int{\frac{x^2}{\sqrt{4x-x^2}}}dx
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  2. #2
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    Quote Originally Posted by th%$&873 View Post
    Having trouble with this one.

    \int{\frac{x^2}{\sqrt{4x-x^2}}}dx
    Note that 4x - x^2 = 4 - (x + 2)^2.

    Make the substitution x + 2 = 2 \sin t.

    You might want to check the details (as I rushed this) - I get something like 8 \int (\sin t - 1)^2 \, dt.

    Expand, integrate term-by-term and finally substitute back that x + 2 = 2 \sin t (tedious, I know, but not too hard).

    Ask for more help if you get stuck at any point.

    Edit: There is a small mistake - you should spot it easily. A small detail that doesn't change the general game plan .....
    Last edited by mr fantastic; February 18th 2008 at 11:04 AM.
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by th%$&873 View Post
    Having trouble with this one.

    \int{\frac{x^2}{\sqrt{4x-x^2}}}dx
    \int{\frac{x^2}{\sqrt{4x-x^2}}}\:dx = \int \frac{x^2}{\sqrt{4-(x-2)^2}}\:dx

    Let x-2=2\sin u then,

    \int \frac{x^2}{\sqrt{4-(x-2)^2}}dx=4\int \sin^2 u+2\sin u+1\:du, now you integrate each term

    now you should get, 4\int \sin^2 u+2\sin u+1\:du=6u-2\sin u\cos u-8\cos u

    Re-substitute, each term and your final answer should/will be:

    \int{\frac{x^2}{\sqrt{4x-x^2}}}\;dx=6\arcsin\left(\frac{x-2}{2}\right)-4\sqrt{4x-x^2}-\frac{(x-2)\sqrt{4x-x^2}}{2}+C
    Last edited by polymerase; February 21st 2008 at 07:53 AM.
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  4. #4
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    Quote Originally Posted by polymerase View Post
    \int{\frac{x^2}{\sqrt{4x-x^2}}}\:dx = \int \frac{x^2}{\sqrt{4-(x-2)^2}}\:dx

    Let x-2=2\sin\:u then,

    \int \frac{x^2}{\sqrt{4-(x-2)^2}}dx=4\int \sin^2\:u+2\sin\:u+1\:du, now you integrate each term

    now you should get, 4\int \sin^2\:u+2\sin\:u+1\:du=6u-2\sin u\cos u-8\cos\:u

    Re-substitute, each term and your final answer should/will be:

    \int{\frac{x^2}{\sqrt{4x-x^2}}}\;dx=6arcsin\left(\frac{x-2}{2}\right)-4\sqrt{4x-x^2}-\frac{(x-2)(\sqrt{4x-x^2})}{2}+C
    Thanks for the catch.
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  5. #5
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    I don't really understand either of your methods (you combined two steps?), but I used u-substitution then trig-substitution to get 6\theta-8\cos\:\theta-\sin 2\theta.
    It appears that polymerase used the double-angle formula to get -2\sin u\cos u instead of -\sin 2 u? I just wasn't sure what to do with that last term. I also made a small mistake with the coefficient of the first term, but I caught that the second time around.
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  6. #6
    Senior Member polymerase's Avatar
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    Quote Originally Posted by th%$&873 View Post
    I don't really understand either of your methods (you combined two steps?), but I used u-substitution then trig-substitution to get 6\theta-8\cos\:\theta-\sin 2\theta.
    It appears that polymerase used the double-angle formula to get -2\sin u\cos u instead of -\sin 2 u? I just wasn't sure what to do with that last term. I also made a small mistake with the coefficient of the first term, but I caught that the second time around.
    You're right, I did use the double angle formula, the reason is because I needed it when I'm re-substituting and expressing everything in terms of x. Thats all.
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