TITLE: Help in evaluating $\displaystyle\int\ \cos^2\Big(\arctan\big(\sin\left(\text{arccot}(x)\right)\big)\Big)\ \text{d}x$
QUESTION [6 upvotes]: Is there an easy way to prove this result?

$$\int\ \cos^2\Big(\arctan\big(\sin\left( \text{arccot}(x)\right)\big)\Big)\ \text{d}x = x - \frac{1}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt{2}}\right)$$

I tried some substitutions but I got nothing helpful, like:

$x = \cot (z)$

I also tried the crazy one:

$x = \cot(\arcsin(\tan(\arccos(z))))$

Any hint?
Thank you!

REPLY [4 votes]: HINT:
$$\cos \left(\arctan \left(x\right)\right)=\frac{\sqrt{1+x^2}}{1+x^2}$$
So
$$\begin{align}
\\& \int \:\left(\cos \left(\arctan \left(\sin \left(\operatorname{arccot}\left(x\right)\right)\right)\right)\right)^2dx
\\& = \int \frac{x^2+1}{x^2+2}dx
\\& = \color{red}{\frac{\sqrt{2}x-\arctan \left(\frac{\sqrt{2}x}{2}\right)}{\sqrt{2}}+C}
\end{align}$$