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Math Help - Integration by Parts #1

  1. #1
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    Integration by Parts #1

    Can someone help me with this problem? I already started it but got stuck. Thanks.

    1. Integral sin^-1 xdx
    I let u=sin^-1x dv=dx
    du=dx/squareroot 1-x^2 v=x
    so integral sin^-1 xdx = x(sin^-1x)-integral x dx/squareroot 1-x^2.....
    Last edited by mr fantastic; October 5th 2009 at 03:26 AM. Reason: Re-titled post
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  2. #2
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    Hello, jsu03!

    You're doing fine . . .


    1)\;\;\int\sin^{-1}\!x\,dx

    I let: . \begin{array}{ccccccc} u&=&\sin^{-1}\!x &&  dv&=& dx \\<br />
du&=&\frac{dx}{\sqrt{1-x^2}} &&  v&=&x \end{array}

    So: . \int\sin^{-1}\!x\,dx \;\;=\;\; x\sin^{-1}\!x - \int \frac{x\, dx}{\sqrt{1-x^2}} . . . . Good!

    The new integral is: . \int \left(1-x^2\right)^{-\frac{1}{2}}(x\,dx)

    Now let: u \,=\,1-x^2


    Got it?

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  3. #3
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    if I let u = 1-x^2 then du = -2x and what is dv and v ....I'm confuse...would dv =xdx and v= dx?
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