Official solution
\[\begin{aligned}
& \int_\limits{\frac{1}{4}}^{\frac{3}{4}} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x \\
& x=\cos 2 \theta \\
& \Rightarrow d x=(-2 \sin 2 \theta \mathrm{d} \theta)
\end{aligned}\]
Take limit as \[\alpha\] and \[\beta\]
\[\begin{aligned}
& -2 \int_\limits\alpha^\beta \cos 2 \theta \cdot \sin 2 \theta d \theta \\
& =\int_\limits\alpha^\beta \sin 4 \theta d \theta \\
& =\left.\frac{-\cos 4 \theta}{4}\right|_\alpha ^\beta \\
& =-\left.\frac{1}{4}\left(2 \cdot\left(x^2\right)-1\right)\right|_{1 / 4} ^{3 / 4} \\
& =-\left.\frac{1}{4}\left(2 x^2-1\right)\right|_{1 / 4} ^{3 / 4} \\
& =-\frac{1}{4}\left(\frac{18}{16}-1-\frac{2}{16}+1\right) \\
& =-\frac{1}{4}
\end{aligned}\]