Question

Evaluate the following integrals:

$\int \frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}\mathrm{d}x \\ \mathrm{Let}\mathrm{the}\mathrm{first}\mathrm{function}\mathrm{be}{\cos }^{-1}x\mathrm{and}\mathrm{second}\mathrm{function}\mathrm{be}\frac{x}{\sqrt{1-x^2}}. \\ \mathrm{First}\mathrm{we}\mathrm{find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{function},\mathrm{i}.\mathrm{e}.,\int \frac{x}{\sqrt{1-x^2}}\mathrm{d}x. \\ \mathrm{Put}t=1-x^2.\mathrm{Then}\mathrm{d}t=-2x\mathrm{d}x \\ \mathrm{Therefore}, \\ \int \frac{x}{\sqrt{1-x^2}}\mathrm{d}x=-\frac{1}{2}\int \frac{1}{\sqrt{t}}\mathrm{d}t \\ =-\sqrt{t} \\ =-\sqrt{1-x^2} \\ \mathrm{Hence},\mathrm{using}\mathrm{integration}\mathrm{by}\mathrm{parts},\mathrm{we}\mathrm{get} \\ \int \frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}\mathrm{d}x=\left({\cos }^{-1}x\right)\int \frac{x}{\sqrt{1-x^2}}\mathrm{d}x-\int \left[\left(\frac{d\left({\cos }^{-1}x\right)}{dx}\right)\int \left(\frac{x}{\sqrt{1-x^2}}\mathrm{d}x\right)\right]\mathrm{d}x \\ =\left({\cos }^{-1}x\right)\left(-\sqrt{1-x^2}\right)-\int \left(\frac{-1}{\sqrt{1-x^2}}\right)\left(-\sqrt{1-x^2}\right)dx \\ =-\sqrt{1-x^2}{\cos }^{-1}x-x+c \\ \mathrm{Hence},\int \frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}\mathrm{d}x=-\sqrt{1-x^2}{\cos }^{-1}x-x+c \end{gathered}$$