Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{{\sin }^{-1}x}{x^2}dx \\ \mathrm{Putting}x=\sin \theta \\ \Rightarrow \theta ={\sin }^{-1}x \\ \&dx=\cos \theta d\theta \\ \therefore I=\int \frac{\theta .\cos \theta }{{\sin }^2\theta }d\theta \\ =\int \theta .\left(\frac{\cos \theta }{\sin \theta }\right)\times \frac{1}{\sin \theta }d\theta \\ =\int \underset{I}{\theta }.\underset{\mathrm{II}}{\mathrm{cosec}\theta }\cot \theta d\theta \\ =\theta \int \mathrm{cosec}\theta \cot \theta d\theta -\int \left\{\frac{d}{d\theta }\left(\theta \right)\int \mathrm{cosec}\theta \cot \theta d\theta \right\}d\theta \\ =\theta \left(-\mathrm{cosec}\theta \right)-\int 1.\left(-\mathrm{cosec}\theta \right)d\theta \\ =-\theta \mathit{}\mathrm{cosec}\mathit{}\theta +\int \mathrm{cosec}\theta d\theta \\ =-\theta \mathit{}\mathrm{cosec}\theta +\ln \left|\mathrm{cosec}\theta -\cot \theta \right|+C \\ =\frac{-\theta }{\sin \theta }+\ln \left|\frac{1-\cos \theta }{\sin \theta }\right|+C \\ =\frac{-\theta }{\sin \theta }+\ln \left|\frac{1-\sqrt{1-{\sin }^2\theta }}{\sin \theta }\right|+C \\ =\frac{-{\sin }^{-1}x}{x}+\ln \left|\frac{1-\sqrt{1-x^2}}{x}\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\mathit{\because }\mathit{}\theta ={\sin }^{-1}\mathit{}x\right] \end{gathered}$$