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Math Help - Inverse Sine Integral

  1. #1
    Newbie alysha27's Avatar
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    Inverse Sine Integral

    Hi, I tried this problem and got an answer but I am not sure if it is the correct solution. If someone could check my solution, that'd be great.

    Q: the integral of the inverse sine of 2x dx (ie: of sin^-1(2x) )

    I used integration by parts and also substituiton and this is the answer I ended up with:

    xsin^-1(2x) + 1/4*ln(|1-4x^2|) + C
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  2. #2
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    Quote Originally Posted by alysha27
    Hi, I tried this problem and got an answer but I am not sure if it is the correct solution. If someone could check my solution, that'd be great.

    Q: the integral of the inverse sine of 2x dx (ie: of sin^-1(2x) )

    I used integration by parts and also substituiton and this is the answer I ended up with:

    xsin^-1(2x) + 1/4*ln(|1-4x^2|) + C
    You are almost right you misused the natural log,
    Given,
    \int\sin^{-1}(2x)dx use the substitution u=2x then by the substitution rule we have, \frac{1}{2}\int\sin^{-1}udu
    Now consider this integral as,
    \int(1)\sin^{-1}udu. Use integration by parts to get,
    u\sin^{-1}u-\int\frac{u}{\sqrt{1-u^2}}du.
    Notice this new integral is solvable by substitution, call v=1-u^2, thus,
    u\sin^{-1}u+\frac{1}{2}\int v^{-1/2}dv.
    Thus, by the exponent rule,
    u\sin^{-1}u+v^{1/2} substitute back,
    u\sin^{-1}u+(1-u^2)}^{1/2} substiute back,
    2x\sin^{-1}(2x)+\sqrt{1-4x^2}+C
    Remember to mutiply by 1/2 as in the beginning of problem,
    x\sin^{-1}2x+\frac{1}{2}\sqrt{1-4x^2}+C
    Q.E.D.
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  3. #3
    Newbie alysha27's Avatar
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    Thanks, I hate little mistakes like that..
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