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Mathematics Question on integral

Prove: ∫01sin−1x dx=π2−1∫^1_0sin^{-1}x\,dx=\frac{π}{2}-1

Answer

Let I=∫01sin−1x dxI=∫^1_0sin^{-1}x\,dx
  ⟹  I=∫01sin−1x 1.dx\implies I=∫^1_0sin^{-1}x\,1.dx
Integrating by parts,we obtain
I=[sin−1x.x]01−∫0111−x2.xdxI=[sin^{-1}x.x]^1_0-∫^1_0\frac{1}{\sqrt{1-x^2}}.xdx
=[xsin−1x]01+12∫01(−2x)1−x2dx=[xsin^{-1}x]^1_0+\frac{1}{2}∫^1_0\frac{(-2x)}{\sqrt{1-x^2}}dx
Let 1−x2=t−2xdx=dt1-x^2=t-2xdx=dt
When x=0,t=1x=0,t=1 and when x=1,t=0x=1,t=0
I=[xsin−1x]01+12∫01dttI=[xsin^{-1}x]^1_0+\frac{1}{2}∫_0^1\frac{dt}{\sqrt{t}}
=[xsin−1x]01+12[2t]01=[xsin^{-1}x]^1_0+\frac{1}{2}[2\sqrt{t}]_0^1
=sin−1(1)+[−1]=sin^{-1}(1)+[-\sqrt{1}]
=π2−1=\frac{π}{2}-1
Hence,the given result is proved.