Question

Evaluate : ${\int }_0^1{\sin }^{-1}(x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2})dx,0\le x\le 1$

Answer 1

We have

$E={\int }_0^1{\sin }^{-1}\left(x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}\right)dx$

$E={\int }_0^1{\sin }^{-1}x-{\int }_0^1{\sin }^{-1}\left(\sqrt{x}\right)dx$

Now,

$I=\underset{0}{\overset{1}{\int }}{\sin }^{-1}\left(\sqrt{x}\right)dx$

$I=\underset{0}{\overset{1}{\int }}{\sin }^{-1}\left(1-\sqrt{x}\right)dx$

$I=\underset{0}{\overset{1}{\int }}{\cos }^{-1}\left(\sqrt{x}\right)dx$

$2I=\underset{0}{\overset{1}{\int }}\frac{\pi }{2}dx$

$2I=\frac{\pi }{2}$

$I=\frac{\pi }{4}$

$\underset{0}{\overset{1}{\int }}{\sin }^{-1}x=x{\sin }^{-1}x+\frac{1}{2}\int \frac{-2x}{\sqrt{1-x^2}}dx$

$={\left[x{\sin }^{-1}x\right]}_0^1+{\left[\sqrt{1-x^2}\right]}_0^1$

$=\frac{\pi }{2}-1$

Hence,

$E=\frac{\pi }{2}-1\frac{-\pi }{4}=\frac{\pi }{4}-1$