Задача №1419
Условие
Найти интеграл \(\int{e^{ax}\cos{nx}}dx\).
Решение
Если \(a=n=0\), то:
\[
\int{e^{ax}\cos{nx}}dx
=\int{dx}=x+C.
\]
Если \(a=0\), \(n\neq{0}\), то:
\[
\int{e^{ax}\cos{nx}}dx
=\int\cos{nx}dx=\frac{\sin{nx}}{n}+C.
\]
Если \(a\neq{0}\), \(n=0\), то:
\[
\int{e^{ax}\cos{nx}}dx
=\int{e^{ax}}dx
=\frac{e^{ax}}{a}+C
\]
Если \(a\neq{0}\), \(n\neq{0}\), то:
\[
\int{e^{ax}\cos{nx}}dx
=\left[\begin{aligned}& u=e^{ax};\,du=ae^{ax}dx.\\& dv=\cos{nx}dx;\;v=\frac{\sin{nx}}{n}.\end{aligned}\right]
=\frac{e^{ax}\sin{nx}}{n}-\frac{a}{n}\int{e^{ax}\sin{nx}}dx=\\
=\left[\begin{aligned}& u=e^{ax};\,du=ae^{ax}dx.\\& dv=\sin{nx}dx;\;v=-\frac{\cos{nx}}{n}.\end{aligned}\right]
=\frac{e^{ax}\sin{nx}}{n}-\frac{a}{n}\cdot\left(-\frac{e^{ax}\cos{nx}}{n}+\frac{a}{n}\int{e^{ax}\cos{nx}}dx\right)=\\
=\frac{e^{ax}\sin{nx}}{n}+\frac{ae^{ax}\cos{nx}}{n^2}-\frac{a^2}{n^2}\int{e^{ax}\cos{nx}}dx.
\]
\[
\int{e^{ax}\cos{nx}}dx
=\frac{e^{ax}\sin{nx}}{n}+\frac{ae^{ax}\cos{nx}}{n^2}-\frac{a^2}{n^2}\int{e^{ax}\cos{nx}}dx;\\
\int{e^{ax}\cos{nx}}dx=
\frac{e^{ax}(n\sin{nx}+a\cos{nx})}{n^2+a^2}+C.
\]
Ответ:
\(\frac{e^{ax}(n\sin{nx}+a\cos{nx})}{n^2+a^2}+C\)