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Math Help - Calculating integral

  1. #1
    MHF Contributor Siron's Avatar
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    Calculating integral

    Hi, I've to calculate:

    \int \sqrt{\tan(x)}dx

    My solution:

    \int \sqrt{\tan(x)}dx=\int \frac{\sqrt{\sin(x)}}{\sqrt{\cos(x)}}dx

    Maybe now the substitution:
    Let \sqrt{\cos(x)}=t then  \frac{-\sin(x)dx}{2\sqrt{\cos(x)}}=dt \Leftrightarrow -\sin(x)dx=2\sqrt{\cos(x)}dt
    and so -\sqrt{\sin(x)}\cdot \sqrt{\sin(x)}dx=2tdt \Leftrightarrow \sqrt{\sin(x)}dx=\frac{2tdt}{-\sqrt{\sin(x)}}

    I get now:
    \int ... = -\int \frac{2dt}{\sqrt{\sin(x)}}

    Now the problem is that my integral is in function of t and \sqrt{\sin(x)} is not.

    So:
    \sqrt{\cos(x)}=t \Leftrightarrow \cos(x)=t^2 \Leftrightarrow x=\arccos(t^2) and x>\frac{\pi}{2}

    Substituted in the integral:
    \int ... = -\int \frac{2dt}{\sqrt{\sin(x)}} = - \int \frac{2dt}{\sqrt{\sin(\arccos(t^2)}}= - \int \frac{2dt}{\sqrt{\sqrt{1-t^4}}}= ...?

    Now, I'm stuck. Does anyone has a beter solution or? ...

    Thanks in advance!
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  2. #2
    MHF Contributor
    Prove It's Avatar
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    Re: Calculating integral

    See here, click Show Steps.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Calculating integral

    Quote Originally Posted by Prove It View Post
    See here, click Show Steps.
    Thanks! I think I was trying a to difficult integral for me .
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