The integral int_{1/4}^{3/4} cos left(2 cot^{ | vedclass

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(D) Let $I = \int_{1/4}^{3/4} \cos \left(2 \cot^{-1} \sqrt{\frac{1-x}{1+x}}ight) dx$. Using the identity $\cot^{-1} 	heta = 	an^{-1} (1/	heta)$,w

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$-1/4$

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(D) Let $I = \int_{1/4}^{3/4} \cos \left(2 \cot^{-1} \sqrt{\frac{1-x}{1+x}}ight) dx$. Using the identity $\cot^{-1} 	heta = 	an^{-1} (1/	heta)$,we have $\cot^{-1} \sqrt{\frac{1-x}{1+x}} = 	an^{-1} \sqrt{\frac{1+x}{1-x}}$. Thus,the integrand becomes $\cos \left(2 	an^{-1} \sqrt{\frac{1+x}{1-x}}ight)$. Using the formula $\cos(2 	an^{-1} 	heta) = \frac{1-	heta^2}{1+	heta^2}$,we set $	heta = \sqrt{\frac{1+x}{1-x}}$. Then $	heta^2 = \frac{1+x}{1-x}$. Substituting this into the formula: $\frac{1 - \frac{1+x}{1-x}}{1 + \frac{1+x}{1-x}} = \frac{\frac{1-x-1-x}{1-x}}{\frac{1-x+1+x}{1-x}} = \frac{-2x}{2} = -x$. Now,evaluate the integral: $I = \int_{1/4}^{3/4} (-x) dx = -\left[ \frac{x^2}{2} ight]_{1/4}^{3/4}$. $I = -\frac{1}{2} \left( \left(\frac{3}{4}ight)^2 - \left(\frac{1}{4}ight)^2 ight) = -\frac{1}{2} \left( \frac{9}{16} - \frac{1}{16} ight) = -\frac{1}{2} \left( \frac{8}{16} ight) = -\frac{1}{2} 	imes \frac{1}{2} = -\frac{1}{4}$.

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

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