integral question

[j_madison]

\int \sqrt{tan(x)} dx


this problem has me stumped. any suggestions would be greatly appreciated. thanks

[devious_]

I remember doing this a while back. It's pretty long-winded. Anyway, here's a push in the right direction:

Use the substitution u=\sqrt{\tan(x)}, then:
u^2=\tan(x)[/itex]
[tex]2u\,du=\sec^2(x)\,dx
2u\,du=1+\tan^2(x)\,dx
dx = \frac{2u}{1+u^4}\,du