Question

Prove the following question.

${\int }_0^{1}si{n}^{-1}xdx=\frac{\pi }{2}-1.$

Answer 1

$LetI={\int }_0^{1}si{n}^{-1}xdx={\int }_0^{1}si{n}^{-1}x.1dx$

$\text{Applying rule of integration by parts taking}si{n}^{-1}x\text{as the first function and 1 as second function, we get}$ $I={\left[\right(si{n}^{-1}x\left)x\right]}_0^1-{\int }_0^1\frac{x}{\sqrt{1-x^2}}dxPut1-x^2=t\Rightarrow -2xdx=dtWhenx=0\Rightarrow t=1andwhenx=1\Rightarrow t=0\therefore I={\left[xsi{n}^{-1}x\right]}_0^1+\frac{1}{2}{\int }_1^0+\frac{1}{2}{\int }_1^0\frac{dt}{\sqrt{t}}={\left[xsi{n}^{-1}x\right]}_0^1+\frac{1}{2}{\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right]}_1^0=1si{n}^{-1}\left(1\right)+[-\sqrt{1}]=\frac{\pi }{2}-1.Henceproved.$