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$\int \sqrt{\tan (x)}dx $ Let $\tan(x)=t^{2}$

then $dx$ will become $\frac{2t}{1+t^{4}}$

Hence $\int \sqrt{\tan (x)}dx =\int\frac{2t}{1+t^4} dt $

But I cannot proceed from this step.

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  • $\begingroup$ Use the search function on this site to avoid posting duplicate questions. $\endgroup$ – heropup Jul 29 '14 at 13:57
  • $\begingroup$ Should it be $\int \frac{2t^2}{1+t^4}dt$? $\endgroup$ – Quang Hoang Jul 29 '14 at 13:58
  • $\begingroup$ Your substitution should give $\int \frac{2 t^2}{1 + t^4} dt$. $\endgroup$ – Bridgeburners Jul 29 '14 at 14:07
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A related problem. You can advance by using partial fraction noticing that

$$ \frac{ 2t^2}{1+t^4} = \frac{1}{t^2+i} + \frac{1}{t^2-i},\quad i=\sqrt{-1} .$$

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    $\begingroup$ The OP has a typo so $u=t^2$ doesn't do the job. $\endgroup$ – Umberto P. Jul 29 '14 at 14:07
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    $\begingroup$ @UmbertoP.: I advanced based on his calculations. Thanks for the comment. $\endgroup$ – Mhenni Benghorbal Jul 29 '14 at 14:09
  • $\begingroup$ @MhenniBenghorbal Substutution is fine, but how do u complete the integration by introducing $i$ ? $\endgroup$ – ss1729 May 27 '18 at 16:00

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