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Math Help - help intergrating

  1. #1
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    I have been doing a parametric arc length problem and have my integrand reduced to (2-2cost)^(1/2). How do I go about integrating this. I tried by parts and u-sub and failed. I think I can use a trig sub but how do I make (2cost) into a square. Is this possible, and if it is is there an easier way.

    Would I use (2)^(1/2)*cos^(1/2)[t]
    Last edited by mr fantastic; March 20th 2009 at 02:09 AM. Reason: Merged posts
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  2. #2
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    Quote Originally Posted by manyarrows View Post
    I have been doing a parametric arc length problem and have my integrand reduced to (2-2cost)^(1/2). How do I go about integrating this. I tried by parts and u-sub and failed. I think I can use a trig sub but how do I make (2cost) into a square. Is this possible, and if it is is there an easier way.
    It might be a good idea to post the whole question.

    To solve the integral you have posted, note that:

    1 - \cos t = 1 - \cos \left( 2 \cdot \frac{t}{2} \right) = 1 - \left[\cos^2 \left( \frac{t}{2} \right) - \sin^2 \left( \frac{t}{2}\right) \right] = 2 \sin^2 \left( \frac{t}{2}\right).
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