Question

Evaluate

${\int }_0^1\frac{x{\left({\sin }^{-1}x\right)}^2}{\sqrt{{1-x}^2}}dx$

Answer 1

We have,

${\int }_0^1\frac{x\left({\sin }^{-1}x\right)}{\sqrt{1-x^2}}dx$

Let

${\sin }^{-1}x=t$

$\frac{1}{\sqrt{1-x^2}}dx=dt$

Then, $x=\sin t$

Therefore,

${\int }_0^{\frac{\pi }{2}}t\sin tdt$

Using ILATE and we get,

${\int }_0^{\frac{\pi }{2}}t.\sin tdt$

$\Rightarrow t{\int }_0^{\frac{\pi }{2}}\sin tdt-{\int }_0^{\frac{\pi }{2}}(\frac{dt}{dt}.{\int }_0^{\frac{\pi }{2}}\sin tdt)dt$

$\Rightarrow t{{(-\cos t)}_0}^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}(-\cos t)dt$

$\Rightarrow (\frac{\pi }{2}-0)(-\cos \frac{\pi }{2}+\cos 0)+{\int }_0^{\frac{\pi }{2}}\left(\cos t\right)dt$

$\Rightarrow \frac{\pi }{2}(-0+1)+{{\left(\sin t\right)}_0}^{\frac{\pi }{2}}$

$\Rightarrow \frac{\pi }{2}+(\sin \frac{\pi }{2}-\sin 0)$

$\Rightarrow \frac{\pi }{2}+(1-0)$

$\Rightarrow \frac{\pi }{2}+1$

$\Rightarrow 1+\frac{\pi }{2}$

Hence, this is the answer.