The integral ∫_1 / 4^3 / 4 cos (2 cot ^-1 √ 1-x 1+x ) d x is equal to

$\displaystyle \begin{aligned} & I=\int_{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right) d x\right) \\ & \int_{1 / 4}^{3 / 4} \cos \left(2\left(\tan ^{-1} \sqrt{\frac{1+x}{1+x}}\right)\right) d x \\ & \left.\int_{1 / 4}^{3 / 4} \frac{1-\tan ^2\left(\tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)}{1+\tan ^2\left(\tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right.}\right) d x\end{aligned}$​I=∫1/43/4​cos(2cot−1(1+x1−x​​)dx)∫1/43/4​cos(2(tan−11+x1+x​​))dx∫1/43/4​1+tan2(tan−11−x1+x​​1−tan2(tan−11−x1+x​​)​​dx​ $\displaystyle \begin{aligned} & =\int_{1 / 4}^{3 / 4} \frac{1-\left(\frac{1+x}{1-x}\right)}{1+\left(\frac{1+x}{1-x}\right)} d x=\int_{1 / 4}^{3 / 4} \frac{-2 x}{2} d x \\ & =\int_{1 / 4}^{3 / 4}(-x) d x=-\left(\frac{x^2}{2}\right)_{1 / 4}^{3 / 4} \\ & =-\frac{1}{2}\left[\frac{9}{16}-\frac{1}{16}\right] \\ & =-\frac{1}{4}\end{aligned}$​=∫1/43/4​1+(1−x1+x​)1−(1−x1+x​)​dx=∫1/43/4​2−2x​dx=∫1/43/4​(−x)dx=−(2x2​)1/43/4​=−21​[169​−161​]=−41​​