Evaluate the following integrals:


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


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

Put $x=\sin t$

$d x=\cos t d t$

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

$\int t^{3} \cos t d=\left[t^{3} \sin t-3 \int t^{2} \sin t d t\right]=\left[t^{3} \sin t-3\left[-t^{2} \cos t+2 \int t \cos t d t\right]\right]$

$=\left[t^{3} \sin t+3 t^{2} \cos t-6 \int t \cos t d t\right]=\left[t^{3} \sin t+3 t^{2} \cos t-6[t \sin t+\cos t]\right]+c$

$=\left[t^{3} \sin t+3 t^{2} \cos t-6 t \cos t-6 \cos t\right]+c$

$=\left[\left(\sin ^{-1} x\right)^{3} x+3\left(\sin ^{-1} x\right)^{2} \sqrt{1-x^{2}}-6 x \sin ^{-1} x-6 \sqrt{1-x^{2}}\right]+c$

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