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

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Answer: B. -1/4 \begin{aligned} & \int\limits_{\frac{1}{4}}^{\frac{3}{4}} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}ight) d x \ & x=\cos 2 	heta \ & \Rightarrow d x=(-2 \sin 2 	heta \mathrm{d} 	heta) \end{aligned} Take limit as \alpha and \beta \begin{aligned} & -2 \int\limits_\alpha^\beta \cos 2 	heta \cdot \sin 2 	heta d 	heta \ & =\int\limits_\alpha^\beta \sin 4 	heta d 	heta \ & =\left.\frac{-\cos 4 	heta}{4}ight|_\alpha ^\beta \ & =-\left.\frac{1}{4}\left(2 \cdot\left(x^2ight)-1ight)ight|_{1 / 4} ^{3 / 4} \ & =-\left.\frac{1}{4}\left(2 x^2-1ight)ight|_{1 / 4} ^{3 / 4} \ & =-\frac{1}{4}\left(\frac{18}{16}-1-\frac{2}{16}+1ight) \ & =-\frac{1}{4} \end{aligned}

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

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