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$\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}} d x=$
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The correct answer is:
$\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c$
Let $I=\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}} d x$
Put $\sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^{2}}} d x=d t$
$\begin{aligned} \therefore I &=\int t^{\frac{3}{2}} d t \\ &=\frac{t^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}=\frac{2}{5} t^{\frac{5}{2}}+c=\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c \end{aligned}$
Put $\sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^{2}}} d x=d t$
$\begin{aligned} \therefore I &=\int t^{\frac{3}{2}} d t \\ &=\frac{t^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}=\frac{2}{5} t^{\frac{5}{2}}+c=\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c \end{aligned}$
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