The integral $$\int_\limits{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x$$ is equal to

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