Question

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

Answer 1

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

$puttingx=\cos \theta$

$dx=i\sin \theta d\theta$

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

$=-\int \sin \theta \cos \theta \times \theta d\theta$

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

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

$=-\frac{1}{2}\left[\theta \times \frac{-\cos 2\theta }{2}-\int 1\times \frac{-\cos 2\theta }{2}d\theta \right]$

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

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

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

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

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