Prove the following Exercise: ^ 1 _ 0 ^ -1 x dx= 2 -1 - Vidyadip

Question

Prove the following Exercise:
$\int^{1}_{0}\sin^{-1}\text{x dx}=\frac{\pi}{2}-1$

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Answer

$\text{Let I}=\int^{1}\limits_{0}\sin^{-1}\text{x dx}$
$\text{I}=\int^{1}\limits_{0}\sin^{-1}\text{x.}\ 1.\text{ dx}$
Integrating by Parts, we obtain
$\text{I}=\Big[\sin^{-1}\text{x}.\text{x}\Big]^{1}_{0}-\int^{1}\limits_{0}\frac{1}{\sqrt{1-\text{x}^{2}}}.\text{x dx}$
$=\Big[\text{x}\sin^{-1}\text{x}\Big]^{1}_{0}+\frac{1}{2}\int^{1}\limits_{0}\frac{(-2\text{x)}}{\sqrt{1-\text{x}^{2}}}\text{dx}$
$\text{Let} 1-\text{x}^{2}=\text{t}\Rightarrow-2\text{x dx}=\text{dt}$
when $\text{x}=0,\text{t}=1$ and when $\text{x}=1,\text{t}=0$
$\text{I}=\Big[\text{x}\sin^{-1}\text{x}\Big]^{1}_{0}+\frac{1}{2}\int^{0}\limits_{1}\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\Big[\text{x}\sin^{-1}\text{x}\Big]^{1}_{0}+\frac{1}{2}\Big[2\sqrt{\text{t}}\Big]^{0}_{1}$
$=\sin^{-1}(1)+\Big[-\sqrt{1}\Big]$
$=\frac{\pi}{2}-1$
Hence, the given result is Proved.

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