$$\displaystyle \int \frac{\arccos \frac{x}{2}}{4+x^2}\text{d}x$$
My attempt:
$x=2\cos \theta\;\; dx=-2\sin \theta d\theta$
$\displaystyle -\frac{1}{2}\frac{\theta \sin \theta}{1+\cos^2\theta}d\theta$
Integration by parts $\displaystyle \frac{1}{2}\theta\arctan(\cos\theta)-\frac{1}{2}\int \arctan(\cos \theta)d\theta$ How do I solve it from here?