Question

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

Answer 1

We know the following identity:
$si{n}^{-1}a+si{n}^{-1}b=si{n}^{-1}\left(a\sqrt{1-b^2}b\sqrt{1-a^2}\right)So,{\int }_0^{1}si{n}^{-1}(x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2})da=x,b=\sqrt{x}\therefore I={\int }_0^1(si{n}^{-1}x+si{n}^{-1}\sqrt{x})dx={\int }_0^{1}si{n}^{-1}xdx+{\int }_0^{1}si{n}^{-1}\sqrt{x}dx={[xsi{n}^{-1}x+\sqrt{1-x^2}]}_0^1+{\left[\frac{1}{2}(\sqrt{-(x-1)x}+(2x-1\left)si{n}^{-1}\sqrt{x}\right)\right]}_0^1=\frac{a}{2}-1+\frac{1}{2}[0+\frac{\pi }{2}-0+0]=\frac{\pi }{2}+\frac{\pi }{4}-1=\frac{3\pi }{4}-1$.