Question

Prove that

$\int \frac{{\sin }^{-1}x}{{\left(1-x^2\right)}^{\frac{3}{2}}}dx$

Answer 1

$I=\int \frac{{\sin }^{-1}x}{(1-x^2{)}^{\frac{3}{2}}}dx$

Let ${\sin }^{-1}x=u\Rightarrow \frac{1}{\sqrt{1-x^2}}dx=du$

So : $\int \frac{{\sin }^{-1}x}{(1-x^2)(1-x^2{)}^{\frac{1}{2}}}dx=\int \frac{u}{(1-{\sin }^{2}u)}du$

$=\int \frac{u}{{\cos }^{2}u}dx=\int ucose{c}^{2}udu$

$\Rightarrow u\int cose{c}^{2}udu-\int (\int cose{c}^{2}udu)du$

$=-u\cot u+\int \cot udu$

$=-u\cot u+\int \cot udu$

$=-u\cot u+In\sin u+a$

$=-u\cot u+In\sin u+a$

$I=-{\sin }^{-1}x\cot {\sin }^{-1}u+lnx+a$