# Compute $\int\frac{dx}{\sqrt{\tan x}}$ [closed]

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.

## closed as off-topic by Did, N. F. Taussig, Shaun, Mark Fantini, Travis WillseJan 18 '15 at 13:32

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• Try $t=\sqrt{\tan x}$ as a first substitution. – mickep Jan 18 '15 at 11:42
• After you make the $t = \sqrt{\tan x}$ substitution. You should find answers in this question useful. – achille hui Jan 18 '15 at 11:56
• I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – Lord_Farin Jan 18 '15 at 12:07

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 $2/(1+t^4)$.

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 $$1/2\,\sqrt {2}\arctan \left( \sqrt {\tan \left( x \right) }\sqrt {2}+1 \right) +1/2\,\sqrt {2}\arctan \left( \sqrt {\tan \left( x \right) } \sqrt {2}-1 \right) +1/4\,\sqrt {2}\ln \left( {\frac {\tan \left( x \right) +\sqrt {\tan \left( x \right) }\sqrt {2}+1}{\tan \left( x \right) -\sqrt {\tan \left( x \right) }\sqrt {2}+1}} \right)$$

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 $$\int\frac{1/t^2+1}{1/t^2+t^2}dt$$

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 $\arctan(1\pm\sqrt{2}\,u)$.