Question

Solve $\int \frac{{\sin }^{-1}x}{x^2}dx$

Answer 1

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

$=-\frac{{\sin }^{-1}x}{x}+\int \frac{1}{x\sqrt{1-x^2}}dx$

= let $x=\sin y\Rightarrow dx=\cos ydy$

$I=-\frac{{\sin }^{-1}x}{x}+\int \frac{\cos y}{\sin y\sqrt{1-{\sin }^{2}y}}dy$

$I=-\frac{{\sin }^{-1}x}{x}+\int cosecydy$

$I=-\frac{{\sin }^{-1}x}{x}-ln|cosecy+\cot y|$

$I=-\frac{{\sin }^{-1}x}{x}-ln|1+\sqrt{1+\sqrt{1-x^2}}|+lnx$