Compute $\int\frac{dx}{\sqrt{\tan x}}$

Question

Compute $\displaystyle \int\dfrac{dx}{\sqrt{\tan x}}$.

Can you help me! , I don't have an idea for solve this problems.

I think, set $t=\tan x$ but i can't solve it.

Answer 1

If you set $t=\sqrt{\tan{x}}$ you get $\frac{dt}{dx}=\frac{1+t^4}{2t}$ and it boils down to find a primitive of $2/(1+t^4)$.

Answer 2

with $t=\sqrt{\tan(x)}$ we get $$dt=\frac{1+\tan(x)^2}{2\sqrt{\tan(x)}}dx$$ we geet $$dx=\frac{2\sqrt{\tan(x)}}{1+\tan(x)^2} dt$$ thus we obtain $$dx=\frac{2t}{1+t^4}dt$$ the result should be $$1/2\,\sqrt {2}\arctan \left( \sqrt {\tan \left( x \right) }\sqrt {2}+1 \right) +1/2\,\sqrt {2}\arctan \left( \sqrt {\tan \left( x \right) } \sqrt {2}-1 \right) +1/4\,\sqrt {2}\ln \left( {\frac {\tan \left( x \right) +\sqrt {\tan \left( x \right) }\sqrt {2}+1}{\tan \left( x \right) -\sqrt {\tan \left( x \right) }\sqrt {2}+1}} \right) $$

Answer 3

HINT:

As marwalix has suggested

$$\frac2{1+t^4}=\frac{1-t^2+1+t^2}{1+t^4}=\frac{1/t^2-1}{1/t^2+t^2}+\frac{1/t^2+1}{1/t^2+t^2}$$

For the first integral,

as $\int\left(1/t^2+1\right)dx=t-1/t,$ write the numerator as $\left(t-1/t\right)^2+2$

Can you take up the second integral $$\int\frac{1/t^2+1}{1/t^2+t^2}dt$$

Answer 4

By replacing $x$ with $\arctan t$ we get: $$\int \frac{dt}{\sqrt{t}(t^2+1)}$$ and by replacing $t$ with $u^2$ we get: $$\int\frac{2\,du}{u^4+1}=\frac{1}{\sqrt{2}}\left(\int\frac{\sqrt{2}-u}{1-\sqrt{2}\,u+u^2}\,du+\int\frac{\sqrt{2}+u}{1+\sqrt{2}\,u+u^2}\,du\right).$$ The last two integrals simply depend on $\log(1\pm\sqrt{2}\,u+u^2)$ and $\arctan(1\pm\sqrt{2}\,u)$.