# integrating square root of tanx [duplicate]

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

• Use the search function on this site to avoid posting duplicate questions. – heropup Jul 29 '14 at 13:57
• Should it be $\int \frac{2t^2}{1+t^4}dt$? – Quang Hoang Jul 29 '14 at 13:58
• Your substitution should give $\int \frac{2 t^2}{1 + t^4} dt$. – Bridgeburners Jul 29 '14 at 14:07

## 1 Answer

A related problem. You can advance by using partial fraction noticing that

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

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