Question

Integrate the function $x{\cos }^{-1}x$.

Answer 1

$\begin{array}{c}I=\int x{\cos }^{-1}x\\ letx=\cos \theta \\ dx=-\sin \theta d\theta \\ \Rightarrow \int \cos \theta \cdot \theta =\frac{-1}{2}\int \theta \sin 2\theta d\theta \end{array}$

now using integration by parts

$\begin{array}{c}I=\frac{-1}{2}\left[\theta \int \sin 2\theta d\theta -\int \left(\frac{d\theta }{d\theta }\int \sin 2\theta d\theta \right)d\theta \right]\\ I=\frac{-1}{2}\left[\frac{-\theta }{2}\cos 2\theta +\frac{1}{2}\frac{\sin 2\theta }{2}\right]+C\\ I=\frac{\theta }{4}\cos 2\theta -\frac{1}{8}\sin 2\theta +C\\ I=\frac{\theta }{4}\left(2{\cos }^2\theta -1\right)\frac{-1}{8}2\cos \theta \sin \theta +c\\ I=\frac{\left(2x^2-1\right)}{4}{\cos }^{-1}x-\frac{x}{4}\sqrt{1-x^2}+c\end{array}$