# Evaluate the closed form of $\int_{0}^{\pi/2}\arctan\left(\sqrt{\sin(2x)\over \sin^2(x)}\right)\mathrm dx$

Given that:

$$I=\int_{0}^{\pi/2}\arctan\left(\sqrt{\sin(2x)\over \sin^2(x)}\right)\mathrm dx$$

and

$$J=\int_{0}^{\pi/2}\arctan\left(\sqrt{\sin(2x)\over \cos^2(x)}\right)\mathrm dx$$

Q1: How do we evaluate the closed for $I$?

Q2: Show that $I=J$.

Recall $$\arctan(x)+\arctan(y)=\arctan\left({x+y\over 1-xy}\right)$$

$$I+J=\int_{0}^{\pi/2}\arctan\left(\sqrt{2\cot x}+\sqrt{2\tan x}\right)dx$$

I am not sure what to do next...

Later on we notice that this integral has the same closed form given by @Jack D'aurizio

$$\int_{0}^{\pi/2}\arctan(2\tan^2x)\mathrm dx=\pi\arctan\left({1\over 2}\right)$$

• Use $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ – lab bhattacharjee May 30 '17 at 11:29
• It seems it can be computed only numerically. – Jon May 30 '17 at 12:21
• The substitution $u = \pi/2-x$ shows that $I=J. \qquad$ – Michael Hardy May 30 '17 at 13:03
• @Jon: that's not true, $\pi\arctan\frac{1}{2}$ is a nice closed form. – Jack D'Aurizio May 30 '17 at 13:38
• @JackD'Aurizio I was cheated by cas tools. Of course, I upvoted your fine solution. – Jon May 30 '17 at 14:21

$$\begin{eqnarray*}I=\int_{0}^{\pi/2}\arctan\sqrt{2\cot x}\,dx &\stackrel{x\mapsto\frac{\pi}{2}-x}{=}&\int_{0}^{\pi/2}\arctan\sqrt{2\tan x}\,dx\tag{1}\\&\stackrel{x\mapsto\arctan u}{=}&\int_{0}^{+\infty}\frac{\arctan\sqrt{2u}}{1+u^2}\,du\\&\stackrel{u\mapsto v^2/2}{=}&\int_{0}^{+\infty}\frac{v\arctan v}{1+\frac{1}{4}v^4}\,dv\end{eqnarray*}$$ and this can be easily solved through differentiation under the integral sign, by computing $$\int_{0}^{+\infty}\frac{v^2\,dv}{(1+a^2 v^2)\left(1+\frac{1}{4}v^4\right)} =\frac{\left(2^{1/4}+a \left(-2+2^{3/4} a\right)\right) \pi }{2+4 a^4}\tag{2}$$ then applying $\int_{0}^{1}\ldots\,da$ to such expression. The final outcome is: $$I = \color{red}{\frac{\pi}{2}\arctan\frac{4}{3}}=\pi\arctan\frac{1}{2}.\tag{3}$$ $(1)$ also proves $I=J$.