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Thread: trig sub

  1. #1
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    trig sub

    Find the following integral:

    $\displaystyle \int e^x \sqrt{1-e^{2x}}$
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by qzno View Post
    Find the following integral:

    $\displaystyle \int e^x \sqrt{1-e^{2x}}\,dx$
    Let $\displaystyle e^x=\sin\theta$

    Thus, $\displaystyle e^x\,dx=\cos\theta\,d\theta$

    The integral becomes $\displaystyle \int\cos\theta\sqrt{1-\sin^2\theta}\,d\theta=\int\cos^2\theta\,d\theta$

    Now recall that $\displaystyle \cos^2\theta=\frac{1+\cos\!\left(2\theta\right)}{2 }$

    Can you continue the problem?
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  3. #3
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    dont you mean that:


    $\displaystyle

    $
    $\displaystyle \int\sin\theta\sqrt{1-\sin^2\theta}\,d\theta
    $

    because

    $\displaystyle
    e^x = sin\theta$
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  4. #4
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    ooh my bad, you did some cancelling there i see haha
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  5. #5
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    I got the answer to be:

    $\displaystyle
    sin^{-1} (e^x) - \frac{e^x \sqrt{1-e^{2x}}}{2} + C$

    Is this correct?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by qzno View Post
    I got the answer to be:

    $\displaystyle
    sin^{-1} (e^x) - \frac{e^x \sqrt{1-e^{2x}}}{2} + C$

    Is this correct?
    Close...

    When you integrate $\displaystyle \tfrac{1}{2}\int 1+\cos\left(2\theta\right)\,d\theta$, you get $\displaystyle \tfrac{1}{2}\theta+\tfrac{1}{4}\sin\left(2\theta\r ight)+C=\tfrac{1}{2}\theta+\tfrac{1}{2}\sin\theta\ cos\theta+C$, where $\displaystyle \sin\left(2\theta\right)=2\sin\theta\cos\theta$

    Now substituting back in, we have $\displaystyle \theta=\sin^{-1}\left(e^x\right)$, $\displaystyle \sin\theta=e^x$, and $\displaystyle \cos\theta=\sqrt{1-e^{2x}}$

    Therefore, $\displaystyle \tfrac{1}{2}\theta+\tfrac{1}{2}\sin\theta\cos\thet a+C=\boxed{\tfrac{1}{2}\left(\sin^{-1}\left(e^x\right)+e^x\sqrt{1-e^{2x}}\right)+C}$
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  7. #7
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    excellent thank you : )
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