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Thread: indefinite integral

  1. April 19th 2010, 03:01 PM #1
    tbenne3
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    indefinite integral

    Evaluate the indefinite integral: ______________
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  2. April 19th 2010, 04:25 PM #2
    Deadstar
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    Let x = \frac{7}{2}\sin(u) => u = \arcsin(\tfrac{2}{7}x)

    dx = \frac{7}{2}\cos(u).

    Integral becomes...

    \int \frac{(7/2)\cos(u)}{\sqrt{49 - 49\sin^2(u)}}du

    = \int \frac{(7/2)\cos(u)}{7\sqrt{1-\sin^2(u)}}du

    = \int \frac{(7/2)\cos(u)}{7\cos(u)}du

    = \int \frac{1}{2} du

    = \frac{u}{2}.

    Substitute in u = \arcsin(\tfrac{2}{7}x) to get...

    \frac{1}{2}\arcsin(\tfrac{2}{7}x).

    Epic.
    Last edited by Deadstar; April 19th 2010 at 04:41 PM.
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