Integral Exercise mixed Trigonometry and monimial asked in the math group

Exercise:

Integrate , $\int{x{{\sin }^{-1}}x\,dx}$

Solution: Let $u={{\sin }^{-1}}x\,\,\And \,\,dv=x\Leftrightarrow du=\frac{1}{\sqrt{1-{{x}^{2}}}}dx\,\,\And \,\,v=\frac{1}{2}{{x}^{2}}$

So $\int{x{{\sin }^{-1}}x\,dx}=\frac{1}{2}{{x}^{2}}\arcsin x-\frac{1}{2}\int{\frac{{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx}$

Let $x=\sin \theta \Rightarrow dx=\cos \theta \,d\theta $

So $\int{\frac{{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx}=\int{\frac{{{\sin }^{2}}\theta \cos \theta d\theta }{\sqrt{1-{{\sin }^{2}}\theta }}}=\int{\frac{{{\sin }^{2}}\theta \cos \theta }{\cos \theta }d\theta =\int{{{\sin }^{2}}\theta d\theta }}$

But ${{\sin }^{2}}\theta =\frac{1}{2}\left( 1-\cos 2\theta  \right)$

So $\int{{{\sin }^{2}}\theta }d\theta =\frac{1}{2}\int{d\theta -\frac{1}{2}\int{\cos 2\theta \,d\theta }=\frac{\theta }{2}-\frac{1}{4}\sin 2\theta +c}$

But $x=\sin \theta \Leftrightarrow \theta =\arcsin x$ and $\sin 2\theta =2\sin \theta \cos \theta =2x\cos \theta $

But $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-{{x}^{2}}}$

Thus $\int{\frac{{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx}=\frac{1}{2}\arcsin x+\frac{x}{2}\sqrt{1-{{x}^{2}}}+c$

Therefore, $\int{x{{\sin }^{-1}}x\,dx}=\frac{1}{2}{{x}^{2}}\arcsin x-\frac{1}{4}\arcsin x+\frac{x}{4}\sqrt{1-{{x}^{2}}}+c$