Integral of $$$e^{5 x} \sin{\left(3 x \right)}$$$
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Find $$$\int e^{5 x} \sin{\left(3 x \right)}\, dx$$$.
Solution
For the integral $$$\int{e^{5 x} \sin{\left(3 x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.
Let $$$\operatorname{u}=\sin{\left(3 x \right)}$$$ and $$$\operatorname{dv}=e^{5 x} dx$$$.
Then $$$\operatorname{du}=\left(\sin{\left(3 x \right)}\right)^{\prime }dx=3 \cos{\left(3 x \right)} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{5 x} d x}=\frac{e^{5 x}}{5}$$$ (steps can be seen »).
The integral becomes
$${\color{red}{\int{e^{5 x} \sin{\left(3 x \right)} d x}}}={\color{red}{\left(\sin{\left(3 x \right)} \cdot \frac{e^{5 x}}{5}-\int{\frac{e^{5 x}}{5} \cdot 3 \cos{\left(3 x \right)} d x}\right)}}={\color{red}{\left(\frac{e^{5 x} \sin{\left(3 x \right)}}{5} - \int{\frac{3 e^{5 x} \cos{\left(3 x \right)}}{5} d x}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{3}{5}$$$ and $$$f{\left(x \right)} = e^{5 x} \cos{\left(3 x \right)}$$$:
$$\frac{e^{5 x} \sin{\left(3 x \right)}}{5} - {\color{red}{\int{\frac{3 e^{5 x} \cos{\left(3 x \right)}}{5} d x}}} = \frac{e^{5 x} \sin{\left(3 x \right)}}{5} - {\color{red}{\left(\frac{3 \int{e^{5 x} \cos{\left(3 x \right)} d x}}{5}\right)}}$$
For the integral $$$\int{e^{5 x} \cos{\left(3 x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.
Let $$$\operatorname{u}=\cos{\left(3 x \right)}$$$ and $$$\operatorname{dv}=e^{5 x} dx$$$.
Then $$$\operatorname{du}=\left(\cos{\left(3 x \right)}\right)^{\prime }dx=- 3 \sin{\left(3 x \right)} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{5 x} d x}=\frac{e^{5 x}}{5}$$$ (steps can be seen »).
The integral becomes
$$\frac{e^{5 x} \sin{\left(3 x \right)}}{5} - \frac{3 {\color{red}{\int{e^{5 x} \cos{\left(3 x \right)} d x}}}}{5}=\frac{e^{5 x} \sin{\left(3 x \right)}}{5} - \frac{3 {\color{red}{\left(\cos{\left(3 x \right)} \cdot \frac{e^{5 x}}{5}-\int{\frac{e^{5 x}}{5} \cdot \left(- 3 \sin{\left(3 x \right)}\right) d x}\right)}}}{5}=\frac{e^{5 x} \sin{\left(3 x \right)}}{5} - \frac{3 {\color{red}{\left(\frac{e^{5 x} \cos{\left(3 x \right)}}{5} - \int{\left(- \frac{3 e^{5 x} \sin{\left(3 x \right)}}{5}\right)d x}\right)}}}{5}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{3}{5}$$$ and $$$f{\left(x \right)} = e^{5 x} \sin{\left(3 x \right)}$$$:
$$\frac{e^{5 x} \sin{\left(3 x \right)}}{5} - \frac{3 e^{5 x} \cos{\left(3 x \right)}}{25} + \frac{3 {\color{red}{\int{\left(- \frac{3 e^{5 x} \sin{\left(3 x \right)}}{5}\right)d x}}}}{5} = \frac{e^{5 x} \sin{\left(3 x \right)}}{5} - \frac{3 e^{5 x} \cos{\left(3 x \right)}}{25} + \frac{3 {\color{red}{\left(- \frac{3 \int{e^{5 x} \sin{\left(3 x \right)} d x}}{5}\right)}}}{5}$$
We've arrived to an integral that we already saw.
Thus, we've obtained the following simple equation with respect to the integral:
$$\int{e^{5 x} \sin{\left(3 x \right)} d x} = \frac{e^{5 x} \sin{\left(3 x \right)}}{5} - \frac{3 e^{5 x} \cos{\left(3 x \right)}}{25} - \frac{9 \int{e^{5 x} \sin{\left(3 x \right)} d x}}{25}$$
Solving it, we get that
$$\int{e^{5 x} \sin{\left(3 x \right)} d x} = \frac{\left(5 \sin{\left(3 x \right)} - 3 \cos{\left(3 x \right)}\right) e^{5 x}}{34}$$
Therefore,
$$\int{e^{5 x} \sin{\left(3 x \right)} d x} = \frac{\left(5 \sin{\left(3 x \right)} - 3 \cos{\left(3 x \right)}\right) e^{5 x}}{34}$$
Add the constant of integration:
$$\int{e^{5 x} \sin{\left(3 x \right)} d x} = \frac{\left(5 \sin{\left(3 x \right)} - 3 \cos{\left(3 x \right)}\right) e^{5 x}}{34}+C$$
Answer
$$$\int e^{5 x} \sin{\left(3 x \right)}\, dx = \frac{\left(5 \sin{\left(3 x \right)} - 3 \cos{\left(3 x \right)}\right) e^{5 x}}{34} + C$$$A