\(\int \sin ^{-1} x d x=\)

(A)
Let \(\begin{aligned} I &=\int \sin ^{-1} x d x=\int\left(\sin ^{-1} x\right) \cdot 1 \cdot d x \\ &=\sin ^{-1} x \int 1 d x-\int\left(\frac{d}{d x} \sin ^{-1} x \cdot \int 1 d x\right) d x \end{aligned}\)
\(=x \cdot \sin ^{-1} x-\int \frac{x}{\sqrt{1-x^{2}}} d x\)...(1)
Consider, \(\int \frac{x}{\sqrt{1-x^{2}}} d x\)
Now, put \(1-x^{2}=t \Rightarrow-2 x d x=d t\)
\(\therefore \int \frac{x}{\sqrt{1-x^{2}}} d x=\int \frac{-1}{2} \times \frac{1}{\sqrt{t}} d t=\frac{-1}{2} \frac{t^{\frac{1}{2}}}{\frac{1}{2}}=-\sqrt{t}\)
Substituting in \((1)\), we get:
\(I=x \cdot \sin ^{-1} x-.-\sqrt{1-x^{2}}-c=x \cdot \sin ^{-1} x+\sqrt{1-x^{2}}+c\)