Show that

Let $I=\int\left(\sin ^{-1} x\right)^{2} \cdot 1 d x$

Taking $\left(\sin ^{-1} x\right)^{2}$ as first function and 1 as second function and integrating by parts, we obtain

$I=\left(\sin ^{-1} x\right) \int 1 d x-\int\left\{\frac{d}{d x}\left(\sin ^{-1} x\right)^{2} \cdot \int 1 \cdot d x\right\} d x$

$=\left(\sin ^{-1} x\right)^{2} \cdot x-\int \frac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}} \cdot x d x$

$=x\left(\sin ^{-1} x\right)^{2}+\int \sin ^{-1} x \cdot\left(\frac{-2 x}{\sqrt{1-x^{2}}}\right) d x$

$=x\left(\sin ^{-1} x\right)^{2}+\left[\sin ^{-1} x \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\left\{\left(\frac{d}{d x} \sin ^{-1} x\right) \int \frac{-2 x}{\sqrt{1-x^{2}}} d x\right\} d x\right]$

$=x\left(\sin ^{-1} x\right)^{2}+\left[\sin ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \frac{1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x\right]$

$=x\left(\sin ^{-1} x\right)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-\int 2 d x$

$=x\left(\sin ^{-1} x\right)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-2 x+\mathrm{C}$