# Evaluating $\int_0^{\pi /2}\left(\frac{1}{\sqrt{\tan(x)}}+\frac{1}{\sqrt{\arctan(x)}}\right) dx$ [closed]

I've come across the following integral: $$\int_0^{\pi /2}\left(\frac{1}{\sqrt{\tan(x)}}+\frac{1}{\sqrt{\arctan(x)}}\right) dx$$

I haven't been able to make any of the obvious methods work (or make things easier, for that matter). Any hint would be appreciated!

## closed as off-topic by Carl Mummert, Tom-Tom, TravisJ, user223391, wythagorasAug 27 '15 at 15:29

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• But tan(0) = 0 and arctan(0) = 0 ... doesn't seem to be well defined – user261263 Aug 26 '15 at 21:12
• @EugenCovaci the same is true for $\dfrac{1}{\sqrt x}$, yet integrating over $(0,c)$ with $0<c<\infty$ still works out fine. – user170231 Aug 26 '15 at 21:14
• The tangent part is easy to calculate; you can use the beta function. The arctangent part is... well, I will be surprised if it has either an antiderivative or a closed form. – Sangchul Lee Aug 26 '15 at 21:18
• Nothing to say more than Sangchul Lee. Maybe just: the integral of the tangent part equals $\frac{\pi}{\sqrt{2}}$. – Jack D'Aurizio Aug 26 '15 at 21:26
• Where did you come across the integral? Please edit the question to include additional background and motivation. Also, please include a description of what you have already tried. These things help make the question more compelling, help others find it, and help others write more useful answers. – Carl Mummert Aug 27 '15 at 11:05