Question:
Evaluate $\int\left(\sin ^{-1} x\right)^{3} d x$
Solution:
$\int\left(\sin ^{-1} x\right)^{3} d x$
Put $x=\sin t$
$d x=\cos t d t$
$\int\left(\sin ^{-1} x\right)^{3} d x=\int\left(\sin ^{-1}(\sin t)\right)^{3} \cos t d t$
$\int t^{3} \cos t d=\left[t^{3} \sin t-3 \int t^{2} \sin t d t\right]=\left[t^{3} \sin t-3\left[-t^{2} \cos t+2 \int t \cos t d t\right]\right]$
$=\left[t^{3} \sin t+3 t^{2} \cos t-6 \int t \cos t d t\right]=\left[t^{3} \sin t+3 t^{2} \cos t-6[t \sin t+\cos t]\right]+c$
$=\left[t^{3} \sin t+3 t^{2} \cos t-6 t \cos t-6 \cos t\right]+c$
$=\left[\left(\sin ^{-1} x\right)^{3} x+3\left(\sin ^{-1} x\right)^{2} \sqrt{1-x^{2}}-6 x \sin ^{-1} x-6 \sqrt{1-x^{2}}\right]+c$