Integral of $$$e^{t} \cos{\left(3 t \right)}$$$

Find $$$\int e^{t} \cos{\left(3 t \right)}\, dt$$$.

For the integral $$$\int{e^{t} \cos{\left(3 t \right)} d t}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\cos{\left(3 t \right)}$$$ and $$$\operatorname{dv}=e^{t} dt$$$.

Then $$$\operatorname{du}=\left(\cos{\left(3 t \right)}\right)^{\prime }dt=- 3 \sin{\left(3 t \right)} dt$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{t} d t}=e^{t}$$$ (steps can be seen »).

Therefore,

$${\color{red}{\int{e^{t} \cos{\left(3 t \right)} d t}}}={\color{red}{\left(\cos{\left(3 t \right)} \cdot e^{t}-\int{e^{t} \cdot \left(- 3 \sin{\left(3 t \right)}\right) d t}\right)}}={\color{red}{\left(e^{t} \cos{\left(3 t \right)} - \int{\left(- 3 e^{t} \sin{\left(3 t \right)}\right)d t}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=-3$$$ and $$$f{\left(t \right)} = e^{t} \sin{\left(3 t \right)}$$$:

$$e^{t} \cos{\left(3 t \right)} - {\color{red}{\int{\left(- 3 e^{t} \sin{\left(3 t \right)}\right)d t}}} = e^{t} \cos{\left(3 t \right)} - {\color{red}{\left(- 3 \int{e^{t} \sin{\left(3 t \right)} d t}\right)}}$$

For the integral $$$\int{e^{t} \sin{\left(3 t \right)} d t}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\sin{\left(3 t \right)}$$$ and $$$\operatorname{dv}=e^{t} dt$$$.

Then $$$\operatorname{du}=\left(\sin{\left(3 t \right)}\right)^{\prime }dt=3 \cos{\left(3 t \right)} dt$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{t} d t}=e^{t}$$$ (steps can be seen »).

Therefore,

$$e^{t} \cos{\left(3 t \right)} + 3 {\color{red}{\int{e^{t} \sin{\left(3 t \right)} d t}}}=e^{t} \cos{\left(3 t \right)} + 3 {\color{red}{\left(\sin{\left(3 t \right)} \cdot e^{t}-\int{e^{t} \cdot 3 \cos{\left(3 t \right)} d t}\right)}}=e^{t} \cos{\left(3 t \right)} + 3 {\color{red}{\left(e^{t} \sin{\left(3 t \right)} - \int{3 e^{t} \cos{\left(3 t \right)} d t}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=3$$$ and $$$f{\left(t \right)} = e^{t} \cos{\left(3 t \right)}$$$:

$$3 e^{t} \sin{\left(3 t \right)} + e^{t} \cos{\left(3 t \right)} - 3 {\color{red}{\int{3 e^{t} \cos{\left(3 t \right)} d t}}} = 3 e^{t} \sin{\left(3 t \right)} + e^{t} \cos{\left(3 t \right)} - 3 {\color{red}{\left(3 \int{e^{t} \cos{\left(3 t \right)} d t}\right)}}$$

We've arrived to an integral that we already saw.

Thus, we've obtained the following simple equation with respect to the integral:

$$\int{e^{t} \cos{\left(3 t \right)} d t} = 3 e^{t} \sin{\left(3 t \right)} + e^{t} \cos{\left(3 t \right)} - 9 \int{e^{t} \cos{\left(3 t \right)} d t}$$

Solving it, we get that

$$\int{e^{t} \cos{\left(3 t \right)} d t} = \frac{\left(3 \sin{\left(3 t \right)} + \cos{\left(3 t \right)}\right) e^{t}}{10}$$

Therefore,

$$\int{e^{t} \cos{\left(3 t \right)} d t} = \frac{\left(3 \sin{\left(3 t \right)} + \cos{\left(3 t \right)}\right) e^{t}}{10}$$

Add the constant of integration:

$$\int{e^{t} \cos{\left(3 t \right)} d t} = \frac{\left(3 \sin{\left(3 t \right)} + \cos{\left(3 t \right)}\right) e^{t}}{10}+C$$

$$$\int e^{t} \cos{\left(3 t \right)}\, dt = \frac{\left(3 \sin{\left(3 t \right)} + \cos{\left(3 t \right)}\right) e^{t}}{10} + C$$$A