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Math Help - Integration by Parts

  1. #1
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    Integration by Parts

    Hello,

    \int_{ 0}^{1/2} {cos^{-1}(x)dx}

    let
    u = cos^{-1}x and dv=dx

    du = \frac{-1}{\sqrt{1-x^2}}dx and v=x

    Then:

    = [xarccos(x)]_{0}^{1/2} - \int_{0}^{1/2} \frac{-x}{\sqrt{1-x^2}}dx

    However at this point I would use u-sub to find the value of the second integral, but the answer has changed the limits of integration? I'm confused. Any help would be appreciated.
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  2. #2
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    Re: Integration by Parts

    then use the u-sub to find the antiderivative, then back sub ...

    x\arccos{x} - \frac{1}{2} \int \frac{-2x}{\sqrt{1-x^2}} \, dx

    \left[x\arccos{x} - \sqrt{1-x^2}\right]_0^{\frac{1}{2}}

    \left(\frac{\pi}{6} - \frac{\sqrt{3}}{2}\right) - (-1)
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  3. #3
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    Re: Integration by Parts

    Quote Originally Posted by skeeter View Post
    then use the u-sub to find the antiderivative, then back sub ...

    x\arccos{x} - \frac{1}{2} \int \frac{-2x}{\sqrt{1-x^2}} \, dx

    \left[x\arccos{x} - \sqrt{1-x^2}\right]_0^{\frac{1}{2}}

    \left(\frac{\pi}{6} - \frac{\sqrt{3}}{2}\right) - (-1)
    The answer in the solution manual is:

    = [xarccos(x)]_{0}^{1/2} + \int_{1}^{3/4} t^{\frac{-1}{2}}[\frac{-1}{2}dt]

    Is it wrong? -- notice when it did the u-sub it changed the limits of integration...
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  4. #4
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    Re: Integration by Parts

    Quote Originally Posted by sjmiller View Post
    The answer in the solution manual is:

    = [xarccos(x)]_{0}^{1/2} + \int_{1}^{3/4} t^{\frac{-1}{2}}[\frac{-1}{2}dt]

    Is it wrong? -- notice when it did the u-sub it changed the limits of integration...
    no, it's not wrong.

    t = 1-x^2

    lower limit, x = 0

    reset lower limit, t = 1

    upper limit, x = \frac{1}{2}

    reset upper limit, t = \frac{3}{4}
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