I'm having problems calculating this integral.


I don't even know where to begin...


  • $\begingroup$ Now the problem has become quite different. Is the tan in the radical or not? $\endgroup$ – imranfat Dec 6 '16 at 16:25
  • $\begingroup$ From Wolfram: $$\int\frac{dx}{\sqrt{\tan x}}=\frac{1}{2 \sqrt{2}}\left(-2 \tan ^{-1}\left(1-\sqrt{2} \sqrt{\tan(x)}\right)+2 \tan^{-1}\left(\sqrt{2} \sqrt{\tan (x)}+1\right)-\log \left(\tan (x)-\sqrt{2}\sqrt{\tan (x)}+1\right)+\log \left(\tan (x)+\sqrt{2} \sqrt{\tan(x)}+1\right)\right)$$ $\endgroup$ – Ian Miller Dec 6 '16 at 16:26
  • $\begingroup$ The OP edited the question such that the denominator was $\sqrt{\tan(x)}$. I think this most recent edit has obfuscated the question's intentions. $\endgroup$ – teadawg1337 Dec 6 '16 at 16:26
  • $\begingroup$ yes, sorry. It is supposed to be $\sqrt{tan(x)}$ $\endgroup$ – Mc-Ac Dec 6 '16 at 16:28
  • 1
    $\begingroup$ @Mc-Ac Apart from Lab's answer, if you u-sub $\sqrt{tanx}=t$ the integral becomes of the form $\frac{2}{t^4+1}$ and that integral has been worked out elsewhere on this website $\endgroup$ – imranfat Dec 6 '16 at 16:30


$$2\sqrt{\cot x}=\sqrt{\cot x}+\sqrt{\tan x}+\sqrt{\cot x}-\sqrt{\tan x}=\dfrac{\cos x+\sin x}{\sqrt{\cos x\sin x}}+\dfrac{\cos x-\sin x}{\sqrt{\cos x\sin x}}$$

As $\int(\cos x+\sin x)dx=\sin x-\cos x,$ choose $\sin x-\cos x=u$ for the first integral.

May I leave for you the second integral?

  • $\begingroup$ This is effectively a duplicate of THIS ANSWER with the tangent function replaced with the cotangent function. $\endgroup$ – Mark Viola Dec 6 '16 at 16:51

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