Evaluate the following integrals:

Let $I=\int \cos ^{-1}\left(4 x^{3}-3 x\right) d x$

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

$I=-\int \cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right) \sin \theta d \theta$

We know that, $\left(4 \cos ^{3} \theta-3 \cos \theta\right)=\cos 3 \theta$

$=-\int \cos ^{-1}(\cos 3 \theta) \sin \theta d \theta$

$=-\int 3 \theta \sin \theta d \theta$

Using integration by parts,

$=-3\left[\theta \int \sin \theta d \theta-\int \frac{d}{d \theta} \theta \int \sin \theta d \theta\right]$

$=3\left[-\theta \cos \theta+\int \cos \theta d \theta\right.$

$=3 \theta \cos \theta-3 \sin \theta+c$

$I=3 x \cos ^{-1} x-3 \sqrt{1-x^{2}}+c$