Evaluate the following integrals:
$\int \sin ^{-1}\left(3 x-4 x^{3}\right) d x$
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,
$=3\left(\theta \int \cos \theta d \theta-\int \frac{d}{d \theta} \theta \int \cos \theta d \theta\right)$
$=3\left(\theta \times \sin \theta-\int \sin \theta d \theta\right)$
$=3(\theta \sin \theta+\cos \theta)+c$
$I=3\left[x \sin ^{-1} x+\sqrt{1-x^{2}}\right]+c$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.