Evaluate the following integrals:

Solution:

Let $I=\int \frac{\sin ^{-1} x}{x^{2}} d x$

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

Using integration by parts,

$I=\left[\sin ^{-1} x \times \int \frac{1}{x^{2}}-\int\left(\frac{1}{\sqrt{1-x^{2}}}\right) \int \frac{1}{x^{2}} d x\right] d x$

$=\sin ^{-1} x\left(\frac{-1}{x}\right)-\int \frac{1}{\sqrt{1-x^{2}}}\left(\frac{-1}{x}\right) d x$

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

$\mathrm{I}=\frac{-1}{\mathrm{x}} \sin ^{-1} \mathrm{x}+\mathrm{I}_{1}$  ........(1)

Where,

$I_{1}=\int \frac{1}{x \sqrt{1-x^{2}}}$

$1-x^{2}=t^{2}$

$-2 x d x=2 t d t$

$I_{1}=\int \frac{t d t}{\left(1-t^{2}\right) \sqrt{t}}$

$=\frac{1}{2} \log \left|\frac{\mathrm{t}-1}{\mathrm{t}+1}\right|$

$=\frac{1}{2} \log \left|\frac{\sqrt{1-\mathrm{x}^{2}}-1}{\sqrt{1-\mathrm{x}^{2}}+1}\right|+\mathrm{c}_{1}$

$I=\frac{-1}{x} \sin ^{-1} x+\frac{1}{2} \log \left|\frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}\right|+c$

$=\frac{-1}{x} \sin ^{-1} x+\frac{1}{2} \log \left(\frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}+1}\right)\left(\frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}-1}\right)+c$

$=\frac{-1}{x} \sin ^{-1} x+\frac{1}{2} \log \left(\frac{\left(\sqrt{1-x^{2}}-1^{2}\right)}{-x^{2}}\right)+c$

$=\frac{-1}{x} \sin ^{-1} x+\log \left|\frac{1-\sqrt{1-x^{2}}}{x}\right|+c$