Evaluate $\int \frac{x \sin ^{-1} x}{\left(1-x^{2}\right)^{3 / 2}} d x$
$\int \frac{x \sin ^{-1} x}{\left(1-x^{2}\right) \sqrt{1-x^{2}}} d x$
we can put $\sin ^{-1} x=t ; d x /\left(1-x^{2}\right)^{1 / 2}=d t ;\left(1-x^{2}\right)=\cos ^{2} t$ and $x=\sin t$.
$\int \frac{t \sin t}{\cos ^{2} t} d t=\int t$ tant sect $d t$
By by parts,
$\int t$ tant $\sec t d t=t \sec t-\int \operatorname{sect} d t \ldots \ldots$
$\because \int \operatorname{sect} \tan t d t=\int \frac{\sin t}{\cos ^{2} t} d t$
$=\mathrm{t} \sec \mathrm{t}-\log (\tan \mathrm{t}+\sec \mathrm{t})+\mathrm{C}^{\prime}$
Put cost $=u$
$-\sin t d t=d u$
$=\sin ^{-1} x \sec \left(\sin ^{-1} x\right)-\log \left(\tan \left(\sin ^{-1} x\right)+\sec \left(\sin ^{-1} x\right)\right)+c^{\prime} \int-u^{-2} d u$
$=-\left(-\mathrm{u}^{-1}\right)+\mathrm{C}$
$=\mathrm{sec} \mathrm{t}+\mathrm{C}$
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