
Calculating integral
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:
$\displaystyle \int ... = \int \frac{2dt}{\sqrt{\sin(x)}} =  \int \frac{2dt}{\sqrt{\sin(\arccos(t^2)}}=  \int \frac{2dt}{\sqrt{\sqrt{1t^4}}}= ...$?
Now, I'm stuck. Does anyone has a beter solution or? ...
Thanks in advance! :)

Re: Calculating integral
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Re: Calculating integral
Quote:
Originally Posted by
Prove It See
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Thanks! I think I was trying a to difficult integral for me :).