Question

Найти неопределенный интеграл ∫ x^2 sin(4x)dx

Answer 1

Простой интеграл, берется два раза по частям.
$\int {x^2sin4x} \, dx = \left\begin{Vmatrix}u = x^2&du=2xdx\\dv=sin4xdx&v=-\frac{1}{4}cos4x\end{Vmatrix}\right = -\frac{1}{4}x^2cos4x +$$\frac{1}{2} \int {xcos4x} \, dx = \left\begin{Vmatrix}u = x&du=dx\\dv=cos4xdx&v=\frac{1}{4}sin4x\end{Vmatrix}\right = -\frac{1}{4}x^2cos4x +$$\frac{1}{2}(\frac{1}{4}xsin4x - \frac{1}{4}\int {sin4x} \, dx ) = -\frac{1}{4}x^2cos4x + \frac{1}{2}(\frac{1}{4}xsin4x + \frac{1}{16}cos4x) +$$C = -\frac{1}{4}x^2cos4x + \frac{1}{8}xsin4x + \frac{1}{32}cos4x + C$.