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Math Help - trig sub

  1. #1
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    trig sub

    Find the following integral:

    \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:

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

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

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

    Now recall that \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:


    <br /> <br />
\int\sin\theta\sqrt{1-\sin^2\theta}\,d\theta<br />

    because

    <br />
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:

    <br />
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:

    <br />
sin^{-1} (e^x) - \frac{e^x \sqrt{1-e^{2x}}}{2} + C

    Is this correct?
    Close...

    When you integrate \tfrac{1}{2}\int 1+\cos\left(2\theta\right)\,d\theta, you get \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 \sin\left(2\theta\right)=2\sin\theta\cos\theta

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

    Therefore, \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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