Evaluate the following integrals:


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


Put $x=\sin t ; d x=\cos t d t$

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

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

$=3 \int t \cos t d t$

By by parts,

$=3\left[t \sin t-\int \sin t d t\right]+c$

$=3[t \sin t+\cos t]+c$

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

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