Задача №1418
Условие
Найти интеграл \(\int{e^{3x}(\sin{2x}-\cos{2x})}dx\).
Решение
\[
\int{e^{3x}(\sin{2x}-\cos{2x})}dx
=\left[\begin{aligned}& u=e^{3x};\,du=3e^{3x}dx.\\& dv=(\sin{2x}-\cos{2x})dx;\;v=-\frac{\cos{2x}}{2}-\frac{\sin{2x}}{2}=-\frac{\cos{2x}+\sin{2x}}{2}.\end{aligned}\right]=\\
=-\frac{e^{3x}(\cos{2x}+\sin{2x})}{2}+\frac{3}{2}\int{e^{3x}(\cos{2x}+\sin{2x})}dx
=\left[\begin{aligned}& u=e^{3x};\,du=3e^{3x}dx.\\& dv=(\cos{2x}+\sin{2x})dx;\;v=\frac{\sin{2x}-\cos{2x}}{2}.\end{aligned}\right]=\\
=-\frac{e^{3x}(\cos{2x}+\sin{2x})}{2}+\frac{3}{2}\cdot\left(\frac{e^{3x}(\sin{2x}-\cos{2x})}{2}-\frac{3}{2}\int{e^{3x}(\sin{2x}-\cos{2x})}dx\right)=\\
=-\frac{e^{3x}(\cos{2x}+\sin{2x})}{2}+\frac{3e^{3x}(\sin{2x}-\cos{2x})}{4}-\frac{9}{4}\int{e^{3x}(\sin{2x}-\cos{2x})}dx.
\]
\[
\int{e^{3x}(\sin{2x}-\cos{2x})}dx
=-\frac{e^{3x}(\cos{2x}+\sin{2x})}{2}+\frac{3e^{3x}(\sin{2x}-\cos{2x})}{4}-\frac{9}{4}\int{e^{3x}(\sin{2x}-\cos{2x})}dx;\\
\frac{13}{4}\int{e^{3x}(\sin{2x}-\cos{2x})}dx
=\frac{e^{3x}(\sin{2x}-5\cos{2x})}{4}+\frac{13}{4}C;\\
\int{e^{3x}(\sin{2x}-\cos{2x})}dx
=\frac{e^{3x}(\sin{2x}-5\cos{2x})}{13}+C.
\]
Ответ:
\(\frac{e^{3x}(\sin{2x}-5\cos{2x})}{13}+C\)