Question

Integrate$\int x{\cos }^{-1}xdx$

Answer 1

$\int x{\cos }^{-1}xdx$

Let $x=\cos \theta \Rightarrow dx=-\sin \theta d\theta$

$\int x{\cos }^{-1}xdx$

$=\int \cos \theta {\cos }^{-1}\left(\cos \theta \right)(-\sin \theta )d\theta$

$=-\int \sin \theta \left(\theta \right)\cos \theta d\theta$

$=\frac{-1}{2}\int \theta \left(2\sin \theta \cos \theta \right)d\theta$

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

Integrating by parts,

$\int u.dv=uv-\int v.du$

Put $u=\theta \Rightarrow du=d\theta$

$dv=\sin 2\theta \Rightarrow v=\frac{-\cos 2\theta }{2}+c$

$=\frac{-1}{2}\left[\theta \int \sin 2\theta d\theta -\int \left(\frac{d\theta }{d\theta }\int \sin 2\theta d\theta \right)d\theta \right]$

$=\frac{-1}{2}[\theta \frac{-\cos 2\theta }{2}-\int 1.\frac{-\cos 2\theta }{2}d\theta ]$

$=\frac{-1}{2}[\frac{-\theta }{2}\cos 2\theta +\frac{1}{2}\int \cos 2\theta d\theta ]$

$=\frac{-1}{2}[\frac{-\theta }{2}\cos 2\theta +\frac{1}{2}\frac{\sin 2\theta }{2}]+c$

$=\frac{\theta }{4}\cos 2\theta -\frac{1}{8}\sin 2\theta +c$

$=\frac{\theta }{4}(2{\cos }^2\theta -1)-\frac{1}{8}\times 2\sin \theta \cos \theta +c$

$=\frac{\theta }{4}(2{\cos }^2\theta -1)-\frac{1}{4}\times \sin \theta \cos \theta +c$

$=\frac{\theta }{4}(2{\cos }^2\theta -1)-\frac{1}{4}\cos \theta \sqrt{1-{\cos }^2\theta }+c$

Using $x=\cos \theta \Rightarrow \theta ={\cos }^{-1}x$

$=\frac{{\cos }^{-1}x}{4}(2x^2-1)-\frac{x}{4}\sqrt{1-x^2}+c$

$=\frac{(2x^2-1)}{4}{\cos }^{-1}x-\frac{x}{4}\sqrt{1-x^2}+c$