Question

Solve ${\int }_0^1\left(2{\cot }^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx=$

Answer 1

Let P = ${\int }_0^1(2{\cot }^{-1}\sqrt{\frac{1-x}{1+x}})dx$

Subtitute $x=\cos y=>dx=-\sin ydy$

and $forx=0,y=\frac{\pi }{2}$

$forx=1,y=0$

So, P = ${\int }_{\frac{\pi }{2}}^0(2{\cot }^{-1}\sqrt{\frac{1-\cos y}{1+\cos y}}\left)\right(-\sin ydy)$

=> P = ${\int }_0^{\frac{\pi }{2}}(2\sin y{\cot }^{-1}\sqrt{\frac{1-\cos y}{1+\cos y}})dy$

=> P = ${\int }_0^{\frac{\pi }{2}}(2\sin y{\cot }^{-1}\sqrt{\frac{2{\sin }^2\frac{y}{2}}{2{\cos }^2\frac{y}{2}}})dy$

=> P = ${\int }_0^{\frac{\pi }{2}}(2\sin y{\cot }^{-1}\tan \frac{y}{2})dy$ = ${\int }_0^{\frac{\pi }{2}}\left[2\sin y\right\{{\cot }^{-1}\cot (\frac{\pi }{2}-\frac{y}{2})\}]dy$

=> P = ${\int }_0^{\frac{\pi }{2}}2\sin y(\frac{\pi }{2}-\frac{y}{2})dy$ = ${\int }_0^{\frac{\pi }{2}}\sin y(\pi -y)dy$

=> P = $\pi {\int }_0^{\frac{\pi }{2}}\sin ydy-{\int }_0^{\frac{\pi }{2}}y.\sin ydy$

=> P = $\pi {[-\cos y]}_0^{\frac{\pi }{2}}-{\left[y\right(-\cos y)+\sin y]}_0^{\frac{\pi }{2}}$

Hence, P = $\pi -1$