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Math Help - More Inverse Trig.

  1. #1
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    Exclamation More Inverse Trig.

    Use the substitution u = \arcsin (x/a) to find the integral of \sqrt(a^2-x^2) between x = 0 and x = b.

    I've never confronted an integral as hard as this. Could someone please please show me how to do this???
    Last edited by mr fantastic; November 15th 2009 at 11:13 AM. Reason: Deleted useless bits of post title
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Use the substitution u = \arcsin (x/a) to find the integral of \sqrt(a^2-x^2) between x = 0 and x = b.

    I've never confronted an integral as hard as this. Could someone please please show me how to do this???
    \sin{u} = \frac{x}{a}

    x = a\sin{u}

    dx = a\cos{u} \, du


    \int \sqrt{a^2 - x^2} \, dx

    \int \sqrt{a^2 - a^2\sin^2{u}} \cdot a\cos{u} \, du

    \int \sqrt{a^2(1 - \sin^2{u})} \cdot a\cos{u} \, du

    a^2 \int \cos^2{u} \, du

    use the double angle identity \cos^2{u} = \frac{1+\cos(2u)}{2} to help in finding the antiderivative.
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