1. ## 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? ...

2. ## Re: Calculating integral

See here, click Show Steps.

3. ## Re: Calculating integral

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