$\begin{align}  & \text{Let }x=\cos 2\theta ,\text{ }\frac{dx}{d\theta }=-2\sin 2\theta ,\text{ }dx=-2\sin 2\theta d\theta  \\ & \text{Changing the limit also to }\theta  \\ & \text{When }x=1,\text{ }\theta =0 \\ & \text{when }x=-1,\text{ }\theta =\frac{\pi }{2} \\ & \int_{-1}^{1}{\sqrt{\frac{1+x}{1-x}}dx}=\int_{\tfrac{\pi }{2}}^{0}{\sqrt{\frac{1+\cos 2\theta }{1-\cos 2\theta }}}(-2\sin 2\theta d\theta ) \\ & \text{Note} \\ & \cos 2\theta =1-2{{\sin }^{2}}\theta  \\ & \cos 2\theta =2{{\cos }^{2}}\theta -1 \\ & \sin 2\theta =2\sin \theta \cos \theta  \\ & \int_{\tfrac{\pi }{2}}^{0}{\sqrt{\frac{1+\cos 2\theta }{1-\cos 2\theta }}}(-2\sin 2\theta d\theta )=\int_{\tfrac{\pi }{2}}^{0}{\sqrt{\frac{1+2{{\cos }^{2}}\theta -1}{1+2{{\sin }^{2}}\theta -1}}}(-4\sin \theta \cos \theta d\theta ) \\ & \int_{\tfrac{\pi }{2}}^{0}{\sqrt{\frac{1+x}{1-x}}dx}=\int_{\tfrac{\pi }{2}}^{0}{\sqrt{\frac{2{{\cos }^{2}}\theta }{2{{\sin }^{2}}\theta }}(-4\sin \theta \cos \theta d\theta )} \\ & \int_{\tfrac{\pi }{2}}^{0}{\sqrt{\frac{1+x}{1-x}}dx}=\int_{\tfrac{\pi }{2}}^{0}{\frac{\cos \theta }{\sin \theta }(-4\sin \theta \cos \theta d\theta )} \\ & \int_{\tfrac{\pi }{2}}^{0}{\sqrt{\frac{1+x}{1-x}}dx}=-4\int_{\tfrac{\pi }{2}}^{0}{{{\cos }^{2}}\theta d\theta }=-4\int_{\tfrac{\pi }{2}}^{0}{\tfrac{1}{2}(\cos 2\theta +1)d\theta } \\ & \int_{\tfrac{\pi }{2}}^{0}{\sqrt{\frac{1+x}{1-x}}dx}=-2\int_{\tfrac{\pi }{2}}^{0}{(\cos 2\theta +1)d\theta }=-2\left[ \frac{\sin 2\theta }{2}+\theta  \right]_{\tfrac{\pi }{2}}^{0} \\ & \int_{\tfrac{\pi }{2}}^{0}{\sqrt{\frac{1+x}{1-x}}dx}=-2\left[ \left( \frac{\sin 2(0)}{2}+0 \right)-\left( \frac{\sin 2(\tfrac{\pi }{2})}{2}+\frac{\pi }{2} \right) \right] \\ & \int_{\tfrac{\pi }{2}}^{0}{\sqrt{\frac{1+x}{1-x}}dx}=-2[0+0-0-\tfrac{\pi }{2}]=\pi  \\\end{align}$