Задача №1415

\[ \int(\arcsin{x})^2dx =\left[\begin{aligned}& u=(\arcsin{x})^2;\,du=\frac{2\arcsin{x}dx}{\sqrt{1-x^2}}.\\& dv=dx;\;v=x.\end{aligned}\right] =x\cdot(\arcsin{x})^2-\int\frac{2x\arcsin{x}dx}{\sqrt{1-x^2}}=\\ =\left[\begin{aligned}& u=\arcsin{x};\,du=\frac{dx}{\sqrt{1-x^2}}.\\& dv=\frac{2xdx}{\sqrt{1-x^2}}=-\left(1-x^2\right)^{-\frac{1}{2}}d\left(1-x^2\right);\;v=-2\sqrt{1-x^2}.\end{aligned}\right]=\\ =x\cdot(\arcsin{x})^2-\left(-2\arcsin{x}\sqrt{1-x^2}+\int{2dx}\right) =x\cdot(\arcsin{x})^2+2\arcsin{x}\sqrt{1-x^2}-2x+C \]