Hi, I've to calculate:

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

My solution:

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

Maybe now the substitution:

Let $\displaystyle \sqrt{\cos(x)}=t$ then $\displaystyle \frac{-\sin(x)dx}{2\sqrt{\cos(x)}}=dt \Leftrightarrow -\sin(x)dx=2\sqrt{\cos(x)}dt$

and so $\displaystyle -\sqrt{\sin(x)}\cdot \sqrt{\sin(x)}dx=2tdt \Leftrightarrow \sqrt{\sin(x)}dx=\frac{2tdt}{-\sqrt{\sin(x)}}$

I get now:

$\displaystyle \int ... = -\int \frac{2dt}{\sqrt{\sin(x)}}$

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

So:

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

Substituted in the integral:

$\displaystyle \int ... = -\int \frac{2dt}{\sqrt{\sin(x)}} = - \int \frac{2dt}{\sqrt{\sin(\arccos(t^2)}}= - \int \frac{2dt}{\sqrt{\sqrt{1-t^4}}}= ...$?

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

Thanks in advance!