Задача №1411

\[ \int{x^2\cos^2{x}}dx =\int{x^2\cdot\frac{1+\cos{2x}}{2}}dx =\frac{1}{2}\int{x^2}dx+\frac{1}{2}\int{x^2\cos{2x}}dx=\\ =\frac{x^3}{6}+\frac{1}{2}\int{x^2\cos{2x}}dx =\left[\begin{aligned}& u=x^2;\,du=2xdx.\\& dv=\cos{2x}dx;\;v=\frac{\sin{2x}}{2}.\end{aligned}\right] =\frac{x^3}{6}+\frac{1}{2}\cdot\left(\frac{x^2\sin{2x}}{2}-\int{x\sin{2x}}dx\right)=\\ =\frac{x^3}{6}+\frac{x^2\sin{2x}}{4}-\frac{1}{2}\int{x\sin{2x}}dx =\left[\begin{aligned}& u=x;\,du=dx.\\& dv=\sin{2x}dx;\;v=-\frac{\cos{2x}}{2}.\end{aligned}\right]=\\ =\frac{x^3}{6}+\frac{x^2\sin{2x}}{4}-\frac{1}{2}\cdot\left(-\frac{x\cos{2x}}{2}+\frac{1}{2}\int{\cos{2x}dx}\right) =\frac{x^3}{6}+\frac{x^2\sin{2x}}{4}+\frac{x\cos{2x}}{4}-\frac{\sin{2x}}{8}+C \]