$\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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Use the search function on this site to avoid posting duplicate questions. – heropup Jul 29 '14 at 13:57
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Should it be $\int \frac{2t^2}{1+t^4}dt$? – Quang Hoang Jul 29 '14 at 13:58
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Your substitution should give $\int \frac{2 t^2}{1 + t^4} dt$. – Bridgeburners Jul 29 '14 at 14:07
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A related problem. You can advance by using partial fraction noticing that
$$ \frac{ 2t^2}{1+t^4} = \frac{1}{t^2+i} + \frac{1}{t^2-i},\quad i=\sqrt{-1} .$$
Mhenni Benghorbal
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1@UmbertoP.: I advanced based on his calculations. Thanks for the comment. – Mhenni Benghorbal Jul 29 '14 at 14:09
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@MhenniBenghorbal Substutution is fine, but how do u complete the integration by introducing $i$ ? – SOORAJ SOMAN May 27 '18 at 16:00