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Math Help - ∫(sin⁻x)

  1. #1
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    ∫(sin⁻x)

    ∫(sin⁻x)

    the only problem is i have to do it via method of substitution, i dont have to to use integration by parts
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Let use integration by parts.

    Let u = [ arcsin(x) ]^2; dv = dx;<br />
du = 2arcsin(x) (\frac{1}{\sqrt(1 - x^2)}) dx; v = x.

    Which gives us:

    x [arcsin(x)]^2 - \int ( 2x arcsin(x) (\frac{1}{\sqrt(1 - x^2)}) )<br />
    Using integration by parts again.

    Let u = arcsin(x); dv = \frac{2x}{\sqrt(1 - x^2)} dx.<br />
du = \frac{1}{\sqrt(1 - x^2)} dx; v = (-1)\sqrt(1 - x^2)<br />
    x [arcsin(x)]^2 - [ (-1) arcsin(x) \sqrt(1 - x^2) - \int ( [\frac{1}{\sqrt(1 - x^2)}] [(-1)\sqrt{(1 - x^2)^(1/2)}] du )

    x [arcsin(x)]^2 + arcsin(x) \sqrt{(1 - x^2)} + \int ( [\frac{1}{\sqrt{(1 - x^2)}}] [(-1)\sqrt{(1 - x^2)^(1/2)}] du )
    <br />
x [arcsin(x)]^2 + arcsin(x) \sqrt{(1 - x^2)} - \int ( [\frac{1}{\sqrt{(1 - x^2)}}] [\sqrt{(1 - x^2)}] du )

    x [arcsin(x)]^2 + arcsin(x) \sqrt{(1 - x^2)} - \int dx

    So,

    x [arcsin(x)]^2 + arcsin(x) \sqrt{(1 - x^2)} - x + C
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    By substitution:

    let t=arcsinx

    so, sint=x

    dt=\frac{dx}{\sqrt{1-x^2}

    dx=\sqrt{1-sin^2t}dt=costdt

    \int (arcsinx)^2dx=\int t^2costdt

    I let you finishing this (by parts)

    I guess there is no escape from integration by parts...
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    ...

    By substitution:

    let t=arcsinx

    so, sint=x

    dt=\frac{dx}{\sqrt{1-x^2}}

    dx=\sqrt{1-sin^2t}dt=costdt

    \int (arcsinx)^2dx=\int t^2costdt

    I let you finishing this (by parts)

    I guess there is no escape from integration by parts...
    Last edited by Also sprach Zarathustra; June 25th 2010 at 11:22 AM.
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  5. #5
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    i guess the same, i hope the question is misprinted in the substitution exercise,

    thanks everyone for their time.
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