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

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

For the integral $$$\int{e^{3 t} \sin{\left(3 t \right)} d t}$$$, 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 $$$\operatorname{dv}=e^{3 t} dt$$$.

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

The integral becomes

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

For the integral $$$\int{e^{3 t} \cos{\left(3 t \right)} d t}$$$, 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 $$$\operatorname{dv}=e^{3 t} dt$$$.

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

The integral becomes

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

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=-1$$$ and $$$f{\left(t \right)} = e^{3 t} \sin{\left(3 t \right)}$$$:

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

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

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

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

Solving it, we get that

$$\int{e^{3 t} \sin{\left(3 t \right)} d t} = \frac{\left(\sin{\left(3 t \right)} - \cos{\left(3 t \right)}\right) e^{3 t}}{6}$$

Therefore,

$$\int{e^{3 t} \sin{\left(3 t \right)} d t} = \frac{\left(\sin{\left(3 t \right)} - \cos{\left(3 t \right)}\right) e^{3 t}}{6}$$

Simplify:

$$\int{e^{3 t} \sin{\left(3 t \right)} d t} = - \frac{\sqrt{2} e^{3 t} \cos{\left(3 t + \frac{\pi}{4} \right)}}{6}$$

Add the constant of integration:

$$\int{e^{3 t} \sin{\left(3 t \right)} d t} = - \frac{\sqrt{2} e^{3 t} \cos{\left(3 t + \frac{\pi}{4} \right)}}{6}+C$$

$$$\int e^{3 t} \sin{\left(3 t \right)}\, dt = - \frac{\sqrt{2} e^{3 t} \cos{\left(3 t + \frac{\pi}{4} \right)}}{6} + C$$$A