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Math Help - evaluating integrals - quick question help

  1. #1
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    evaluating integrals - quick question help

    evaluate integrals:

    { (pie/2,0) ( (1+cos2t)/2 )dt


    cos2t = 2 (cost)^2 - 1

    so,

    { (pie/2,0) (cost)^2 dt

    but how do you approach this problem after this step?

    i forgot the anti derivative of (cost)^2 ...someone help ~ do it for me so i can see

    thx
    Last edited by pghaffari; February 19th 2009 at 12:37 PM.
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  2. #2
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    \cos2t=\cos^2t-\sin^2t=2\cos^2t-1.

    As for your last question, put f(x)=\int_{0}^{\sqrt{x}}{\cos t\,dt}=\sin \sqrt{x} thus f'(x)=\frac{\cos \left( \sqrt{x} \right)}{2\sqrt{x}}. Hence, by using the FTC we have f'(x)=\cos \left( \sqrt{x} \right)\cdot \left( \sqrt{x} \right)', which leads the same result.
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  3. #3
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    bump *new question
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  4. #4
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    You don't need to integrate that, if you can prove that \int_{0}^{\frac{\pi }{2}}{\cos ^{2}t\,dt}=\int_{0}^{\frac{\pi }{2}}{\sin ^{2}t\,dt} then the result follows easily.
    Last edited by Krizalid; February 19th 2009 at 12:52 PM.
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  5. #5
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    what do you mean? i don't get it...

    it says evaluate the integral

    how do i do it !?

    someone please do this one before i have to go !

    ty
    Last edited by Krizalid; February 19th 2009 at 04:20 PM.
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  6. #6
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    Quote Originally Posted by pghaffari View Post
    what do you mean? i don't get it...

    it says evaluate the integral

    how do i do it !?

    someone please do this one before i have to go !

    ty
    By editing your original question multiple times the replies that you get make no sense to someone tying to read and follow the thread. This is annoying and insulting to the members of MHF.

    Do not edit your question once you get replies. If you have new questions, ask them in a new thread. If you have follow-up questions, ask them in the same thread - do NOT edit your original question to ask them.

    This thread is closed.
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