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Math Help - Finding a definite integral using a given substitution.

  1. #1
    dix
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    Finding a definite integral using a given substitution.

    Hi, I have been given this question and can only get so far, I'm having trouble with the last part. Any help is good thanks

    Last edited by mr fantastic; May 29th 2011 at 12:04 PM. Reason: Re-titled.
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  2. #2
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    First of all, when you let \displaystyle u = \sin{x} \implies du = \cos{x}\,dx, you also need to change the terminals.

    \displaystyle u(0) = 0 and \displaystyle u\left(\frac{\pi}{2}\right) = 1.

    So \displaystyle \int_0^{\frac{\pi}{2}}{2\cos{x}\,e^{\sin{x}}\,dx} = \int_0^1{2e^u\,du} = \left[2e^u\right]_0^1

    Go from here.
    Last edited by Prove It; May 29th 2011 at 07:07 AM.
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  3. #3
    dix
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    Thanks Prove It.

    I don't understand how \displaystyle u(0) = 0 and \displaystyle u\left(\frac{\pi}{2}\right) = 1.

    What does \displaystyle u\left(\frac{\pi}{2}\right) = 1 mean? Do you substitute (\frac{\pi}{2}\right) somewhere?
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    Senior Member bugatti79's Avatar
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    Quote Originally Posted by dix View Post
    Thanks Prove It.

    I don't understand how \displaystyle u(0) = 0 and \displaystyle u\left(\frac{\pi}{2}\right) = 1.

    What does \displaystyle u\left(\frac{\pi}{2}\right) = 1 mean? Do you substitute (\frac{\pi}{2}\right) somewhere?
    Yes, note that \displaystyle u(0) = 0 is the same as writing when x =0 (ie the zero in the bracket) then u = 0 because we know that u = sin x. Same idea for the upper limit.
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  5. #5
    dix
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    Oh right, thanks. I had my calculator in degrees instead of radians so it came out different.

    I'm supposed to end up with 2(e-1), could you tell me where i've went wrong here?

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    You are evaluating \displaystyle \left[2e^u\right]_0^1, not \displaystyle \left[2e^{\sin{x}}\right]_0^1. The whole point in changing your terminals when you make the substitution is so that you DON'T have to convert the integral back to a function of \displaystyle x.
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  7. #7
    dix
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    Oh right. Thanks again.
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