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Math Help - Trig sub integrals

  1. #1
    Junior Member symstar's Avatar
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    Trig sub integrals

    \int x\sqrt{1-x^{4}}dx

    Here's the work I've done, hopefully there's not too much bad math in it...
    x=\sin{\theta} so dx=\cos{\theta}d{\theta}

    \int \sin{\theta}\sqrt{(1-\sin^2{\theta})^2}\cos{\theta}d{\theta}
    \int \sin{\theta}\sqrt{(\cos^2{\theta})^2}\cos{\theta}d  {\theta}
    \int \sin{\theta}\cos^2{\theta}\cos{\theta}d{\theta}
    u=\sin{\theta} so du=\cos{\theta}d{\theta}
    \int u(1-u^2)du
    \tfrac{1}{2}u^2 - \tfrac{1}{4}u^4 + C

    ...and I stopped right here, seeing that this wouldn't give me the right answer. I'm not ever certain that I can u-sub in this problem like I did. Can someone point me in the right direction please?
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  2. #2
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     \int {x\sqrt {1 - x^4 } } dx \hfill \\

      {\text{Let us substitute, }}x^2  = \sin \theta  \Rightarrow \theta  = \sin ^{ - 1} \left( {x^2 } \right) \hfill \\

      {\text{so that,  }}2xdx = \cos \theta d\theta  \Rightarrow xdx = \frac{1}<br />
{2}\cos \theta d\theta  \hfill \\

      {\text{Now see, }} = \frac{1}<br />
{2}\int {\left( {\sqrt {1 - \sin ^2 \theta } } \right)} \cos \theta ~d\theta  = \frac{1}<br />
{2}\int {\cos \theta \sqrt {\cos ^2 \theta } } ~d\theta  \hfill \\

       = \frac{1}<br />
{2}\int {\cos ^2 \theta ~d\theta }  \hfill \\

       = \frac{1}<br />
{2}\int {\frac{{1 + \cos 2\theta }}<br />
{2}} ~d\theta  \hfill \\

       = \frac{1}<br />
{4}\left( {\theta  + \sin 2\theta } \right) + c \hfill \\

       = \frac{1}<br />
{4}\left( {\theta  + 2\sin \theta \cos \theta } \right) + c \hfill \\

       = \frac{1}<br />
{4}\left( {\theta  + 2\sin \theta \sqrt {1 - \sin ^2 \theta } } \right) + c \hfill \\

       = \frac{1}<br />
{4}\left[ {\sin ^{ - 1} \left( {x^2 } \right) + 2x^2 \sqrt {1 - x^4 } } \right] + c \hfill \\
    Last edited by Shyam; September 9th 2008 at 01:01 PM.
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by symstar View Post
    \int x\sqrt{1-x^{4}}dx

    Here's the work I've done, hopefully there's not too much bad math in it...
    x=\sin{\theta} so dx=\cos{\theta}d{\theta}

    \int \sin{\theta}\sqrt{(1-\sin^2{\theta})^2}\cos{\theta}d{\theta}
    Using x = sin\theta

    gives

    1-sin^4\theta = (1 + sin^2\theta)(1-sin^2\theta)

    not

    (1-sin^2\theta)^2

    Hope that helps
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  4. #4
    Junior Member symstar's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    Using x = sin\theta

    gives

    1-sin^4\theta = (1 + sin^2\theta)(1-sin^2\theta)

    not

    (1-sin^2\theta)^2

    Hope that helps
    It does help and hopefully I won't do bad math like this in the future.

    Also, thank you Shyam for making sense of this problem for me. I did just want to point out that I'm pretty sure that
    = \frac{1}<br />
{2}\int {\frac{{1 + \cos 2\theta }}<br />
{2}} ~d\theta  \hfill<br />
    actually gives
    = \frac{1}<br />
{4}\left( {\theta  + \tfrac{1}{2}\sin 2\theta } \right) + c \hfill<br />

    but it doesn't change the answer much and it was still a lot of help.
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by symstar View Post
    It does help and hopefully I won't do bad math like this in the future.

    Also, thank you Shyam for making sense of this problem for me. I did just want to point out that I'm pretty sure that
    = \frac{1}<br />
{2}\int {\frac{{1 + \cos 2\theta }}<br />
{2}} ~d\theta \hfill<br />
    actually gives
    = \frac{1}<br />
{4}\left( {\theta + \tfrac{1}{2}\sin 2\theta } \right) + c \hfill<br />

    but it doesn't change the answer much and it was still a lot of help.
    Yep you are correct
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