Задача №1422

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

\[ \int\sqrt{1-x^2}dx =x\cdot\sqrt{1-x^2}-\int\sqrt{1-x^2}dx +\arcsin{x}+2C;\\ \int\sqrt{1-x^2}dx=\frac{x}{2}\cdot\sqrt{1-x^2}+\frac{1}{2}\cdot\arcsin{x}+C. \]

\[ \int\frac{x^2dx}{\sqrt{1-x^2}} =-\int\frac{1-x^2-1}{\sqrt{1-x^2}} =-\int\sqrt{1-x^2}dx+\int\frac{dx}{\sqrt{1-x^2}}=\\ =-\frac{x}{2}\cdot\sqrt{1-x^2}-\frac{1}{2}\cdot\arcsin{x}+\arcsin{x}+C =-\frac{x}{2}\cdot\sqrt{1-x^2}+\frac{1}{2}\cdot\arcsin{x}+C \]

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

\[ \int\frac{x^2dx}{\sqrt{1-x^2}} =-x\sqrt{1-x^2}+\arcsin{x}-\int\frac{x^2dx}{\sqrt{1-x^2}};\; \int\frac{x^2dx}{\sqrt{1-x^2}}=-\frac{x\sqrt{1-x^2}}{2}+\frac{\arcsin{x}}{2}+C. \]