$ \text { The limits for y are 0 and 1 and for x are 0 and x = } \sqrt{1-y^2}) \Longrightarrow x^2 = 1-y^2 \Longrightarrow x^2 + y^2 = 1 \\ $

$ \text{ Hence, the region of integration is the first quadrant of the circle} x^2 + y^2 =1. \text{If we change the order of integration,y varies from 0 to } \sqrt{1-x^2} \text { and x varies from 0 to 1. Hence, } \\ $

$ I = \int_{x=0}^{1} \int_{y=0}^{\sqrt{1- x^2}} \frac {cos^{-1}x} {\sqrt{1-x^2}\sqrt{1- x^2 - y^2} } dy dx \\ $

$ = \int_{x=0}^{1} \frac {cos^{-1}x} {\sqrt{1-x^2}} \left[ \{sin^{-1}\frac{y} {\sqrt{1-x^2} } \right]_0^{\sqrt{1-x^2}} dx \\ $

$ = \int_{x=0}^{1} \frac {cos^{-1}x} {\sqrt{1-x^2}} \left[\frac{\pi}{2}\right] dx \\ $

$ = \frac{\pi}{2} \int_{x=0}^{1} \frac {cos^{-1}x} {\sqrt{1-x^2}} dx \\ $

$ \text {Put } cos^{-1}x = t \Longrightarrow \frac{dx}{\sqrt{1-x^2}} = - dt \\ $

$ I = \frac{\pi}{2} \int_{\frac{\pi}{2}}^{0} -t dt \\ $

$ I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} t dt \\ $

$ I = \frac{\pi}{2} \left[\frac{t^2}{2} \right ]_0^\frac{\pi}{2} \\ $

$ = \frac{\pi}{4} * \frac{\pi^2}{4} \\ $

$ = \frac{\pi^3}{16} \\ $