Question

Prove ${\int }_0^1{\sin }^{-1}xdx=\frac{\pi }{2}-1$

Answer 1

Let $I={\int }_0^1{\sin }^{-1}xdx$
$\Rightarrow I={\int }_0^1{\sin }^{-1}x\cdot 1\cdot dx$
Integrating by parts, we obtain
$I=[{\sin }^{-1}x\cdot x{]}_0^1-{\int }_0^1\frac{1}{\sqrt{1-x^2}}\cdot xdx$
$=[x{\sin }^{-1}x{]}_0^1+\frac{1}{2}{\int }_0^1\frac{(-2x)}{\sqrt{1-x^2}}dx$
Let $1-x^2=t\Rightarrow -2xdx=dt$
When $x=0,t=1$ and when $x=1,t=0$
$I=[x{\sin }^{-1}x{]}_0^1+\frac{1}{2}{\int }_1^0\frac{dt}{\sqrt{t}}$
$=[x{\sin }^{-1}x{]}_0^1+\frac{1}{2}[2\sqrt{t}{]}_1^0$
$={\sin }^{-1}\left(1\right)+[-\sqrt{1}]$
$=\frac{\pi }{2}-1$
Hence, the given result is proved.