Evaluate the following integrals:

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

$x=\sin \theta$

$\mathrm{dx}=\cos \theta \mathrm{d} \theta$

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

We know that $3 \sin \theta-4 \sin ^{3} \theta=\sin 3 \theta$

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

We know that, $\int \sin ^{-1}(\sin 3 \theta)=3 \theta$

$=\int 3 \theta \cos \theta d \theta$

$=3 \int \theta \cos \theta d \theta$

Using integration by parts,

$=3\left(\theta \int \cos \theta d \theta-\int \frac{d}{d \theta} \theta \int \cos \theta d \theta\right)$

$=3\left(\theta \times \sin \theta-\int \sin \theta d \theta\right)$

$=3(\theta \sin \theta+\cos \theta)+c$

$I=3\left[x \sin ^{-1} x+\sqrt{1-x^{2}}\right]+c$