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Math Help - arcsin int by parts problem

  1. #1
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    arcsin int by parts problem

    \int xarcsinx^2dx on int [0,1]

    Should I set u=arcsinx^2 and dv=x ??
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  2. #2
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    That is correct. You then need to use substitution on the second integral. Just ask if you get stuck.
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  3. #3
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    start with ...

    t = x^2

    dt = 2x \, dx

    \frac{1}{2} \int 2x \arcsin(x^2) \, dx

    \frac{1}{2} \int \arcsin(t) \, dt

    now go with parts ... let u = \arcsin(t) and dv = dt
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  4. #4
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    u=arcsin(t) -> du= \frac{1}{\sqrt{1-x^2}}
    dv=dt -> v=t

    => (t)arcsin(t) - \frac{1}{2}\int \frac{t}{\sqrt{1-x^2}}du
    = x^2arcsin(x^2) - \frac {x^2}{2}arcsin(u) + C ???

    I'm getting lost here....
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  5. #5
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    you are back-substituting too soon ... get the integral done completely in terms of t, then back substitute when finished.

    u = \arcsin(t)

    du = \frac{1}{\sqrt{1 - t^2}}

    dv = dt

    v = t

    \frac{1}{2} \int \arcsin(t) \, dt = \frac{t\arcsin(t)}{2} - \frac{1}{2}\int \frac{t}{\sqrt{1-t^2}} \, dt

    \frac{1}{2} \int \arcsin(t) \, dt = \frac{t\arcsin(t)}{2} + \frac{1}{4}\int \frac{-2t}{\sqrt{1-t^2}} \, dt

    \frac{1}{2} \int \arcsin(t) \, dt = \frac{t\arcsin(t)}{2} + \frac{\sqrt{1-t^2}}{2} + C

    finish by back substituting x^2 for t ...

    \frac{x^2\arcsin(x^2) + \sqrt{1 - x^4}}{2} + C
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  6. #6
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    where do you get the + 1/4 from???
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  7. #7
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    \frac{1}{2} \int \arcsin(t) \, dt = \frac{t\arcsin(t)}{2} \textcolor{red}{-\frac{1}{2}}\int \frac{t}{\sqrt{1-t^2}} \, dt

    \frac{1}{2} \int \arcsin(t) \, dt = \frac{t\arcsin(t)}{2} + \textcolor{red}{\frac{1}{4}}\int \frac{\textcolor{red}{-2}t}{\sqrt{1-t^2}} \, dt

    the -2 is needed for the substitution required to find the antiderivative of the last integral expression.
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