Mathematics - Area Under Curves Question with Solution | TestHub

The curves meet at $x=\frac{1}{\sqrt{2}}$. Required area $={\int }_0^{1/\sqrt{2}}{\sin }^{-1}xdx+{\int }_{1/\sqrt{2}}^1{\cos }^{-1}xdx={\left[x{\sin }^{-1}x+\sqrt{1-x^2}\right]}_0^{1/\sqrt{2}}+$ ${\left[x{\cos }^{-1}x-\sqrt{1-x^2}\right]}_{1/\sqrt{2}}^1=\left(\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1\right)+\left(\frac{1}{\sqrt{2}}-\frac{\pi }{4\sqrt{2}}\right)=\sqrt{2}-1$.