Let intlimits_0^1 {{{tan }^{ - 1}}left( {frac | vedclass

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(D) Let $I = \int_{0}^{1} 	an^{-1}\left(\frac{	an x - 2\cot x}{3}ight) dx$. Using the identity $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A

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$\alpha - \frac{\pi }{2} + \frac{1}{2}$

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(D) Let $I = \int_{0}^{1} 	an^{-1}\left(\frac{	an x - 2\cot x}{3}ight) dx$. Using the identity $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,we can rewrite the integrand. Note that $\frac{	an x - 2\cot x}{3} = \frac{	an x - 2/	an x}{1 + (	an x)(2/	an x)} = \frac{	an x - 2/	an x}{1 + 2}$. Thus,$	an^{-1}\left(\frac{	an x - 2\cot x}{3}ight) = 	an^{-1}(	an x) - 	an^{-1}(2\cot x) = x - 	an^{-1}(2\cot x)$. However,a simpler approach is to use $	an^{-1}(\frac{	an x}{2}) - 	an^{-1}(\cot x) = tan^{-1}\left(\frac{\frac{	an x}{2} - \cot x}{1 + \frac{	an x}{2} \cot x}ight) = 	an^{-1}\left(\frac{	an x - 2\cot x}{2(1 + 1/2)}ight) = 	an^{-1}\left(\frac{	an x - 2\cot x}{3}ight)$. Therefore,the integral is $\int_{0}^{1} 	an^{-1}\left(\frac{	an x}{2}ight) dx - \int_{0}^{1} 	an^{-1}(\cot x) dx$. Given $\int_{0}^{1} 	an^{-1}\left(\frac{	an x}{2}ight) dx = \alpha$,we have $I = \alpha - \int_{0}^{1} (\frac{\pi}{2} - x) dx$. $I = \alpha - [\frac{\pi}{2}x - \frac{x^2}{2}]_{0}^{1} = \alpha - (\frac{\pi}{2} - \frac{1}{2}) = \alpha - \frac{\pi}{2} + \frac{1}{2}$.

Let $\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\frac{{\tan x}}{2}} \right)} dx = \alpha $. Then $\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\frac{{\tan x - 2\cot x}}{3}} \right)} dx$ is equal to:

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