Question

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

Answer 1

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

Let $x=\cos \theta$

$dx=-\sin \theta d\theta$

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

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

$=-\frac{1}{2}[\theta \int \sin 2\theta d\theta -\int (\frac{d\theta }{d\theta }\int sin2\theta d\theta \left)d\theta \right]$

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

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

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

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

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

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

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