Question:

M-1 : Let 1 is algebraic function $$\int \underbrace{1}_{II} \cdot \underbrace{\sin^{-1} x}_{I} \, dx = \sin^{-1} x (x) - \int \left( \frac{1}{\sqrt{1-x^2}} \right) (x) \, dx$$ Put $1 - x^2 = t \Rightarrow - 2x \, dx = dt$ $$\therefore \int \sin^{-1} x \, dx = x \sin^{-1}(x) + \frac{1}{2} \int \frac{dt}{\sqrt{t}}$$ $$= x \sin^{-1} x + \frac{1}{2} \frac{\sqrt{t}}{\left(\frac{1}{2}\right)} + C = x \sin^{-1} x + \sqrt{1-x^2} + C$$