Задача №1411
Условие
Найти интеграл \(\int{x^2\cos^2{x}}dx\).
Решение
\[
\int{x^2\cos^2{x}}dx
=\int{x^2\cdot\frac{1+\cos{2x}}{2}}dx
=\frac{1}{2}\int{x^2}dx+\frac{1}{2}\int{x^2\cos{2x}}dx=\\
=\frac{x^3}{6}+\frac{1}{2}\int{x^2\cos{2x}}dx
=\left[\begin{aligned}& u=x^2;\,du=2xdx.\\& dv=\cos{2x}dx;\;v=\frac{\sin{2x}}{2}.\end{aligned}\right]
=\frac{x^3}{6}+\frac{1}{2}\cdot\left(\frac{x^2\sin{2x}}{2}-\int{x\sin{2x}}dx\right)=\\
=\frac{x^3}{6}+\frac{x^2\sin{2x}}{4}-\frac{1}{2}\int{x\sin{2x}}dx
=\left[\begin{aligned}& u=x;\,du=dx.\\& dv=\sin{2x}dx;\;v=-\frac{\cos{2x}}{2}.\end{aligned}\right]=\\
=\frac{x^3}{6}+\frac{x^2\sin{2x}}{4}-\frac{1}{2}\cdot\left(-\frac{x\cos{2x}}{2}+\frac{1}{2}\int{\cos{2x}dx}\right)
=\frac{x^3}{6}+\frac{x^2\sin{2x}}{4}+\frac{x\cos{2x}}{4}-\frac{\sin{2x}}{8}+C
\]
Ответ:
\(\frac{x^3}{6}+\frac{x^2\sin{2x}}{4}+\frac{x\cos{2x}}{4}-\frac{\sin{2x}}{8}+C\)