Integral of $$$e^{2 x} \sin{\left(x \right)}$$$

Find $$$\int e^{2 x} \sin{\left(x \right)}\, dx$$$.

For the integral $$$\int{e^{2 x} \sin{\left(x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\sin{\left(x \right)}$$$ and $$$\operatorname{dv}=e^{2 x} dx$$$.

Then $$$\operatorname{du}=\left(\sin{\left(x \right)}\right)^{\prime }dx=\cos{\left(x \right)} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{2 x} d x}=\frac{e^{2 x}}{2}$$$ (steps can be seen »).

The integral becomes

$${\color{red}{\int{e^{2 x} \sin{\left(x \right)} d x}}}={\color{red}{\left(\sin{\left(x \right)} \cdot \frac{e^{2 x}}{2}-\int{\frac{e^{2 x}}{2} \cdot \cos{\left(x \right)} d x}\right)}}={\color{red}{\left(\frac{e^{2 x} \sin{\left(x \right)}}{2} - \int{\frac{e^{2 x} \cos{\left(x \right)}}{2} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = e^{2 x} \cos{\left(x \right)}$$$:

$$\frac{e^{2 x} \sin{\left(x \right)}}{2} - {\color{red}{\int{\frac{e^{2 x} \cos{\left(x \right)}}{2} d x}}} = \frac{e^{2 x} \sin{\left(x \right)}}{2} - {\color{red}{\left(\frac{\int{e^{2 x} \cos{\left(x \right)} d x}}{2}\right)}}$$

For the integral $$$\int{e^{2 x} \cos{\left(x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\cos{\left(x \right)}$$$ and $$$\operatorname{dv}=e^{2 x} dx$$$.

Then $$$\operatorname{du}=\left(\cos{\left(x \right)}\right)^{\prime }dx=- \sin{\left(x \right)} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{2 x} d x}=\frac{e^{2 x}}{2}$$$ (steps can be seen »).

Therefore,

$$\frac{e^{2 x} \sin{\left(x \right)}}{2} - \frac{{\color{red}{\int{e^{2 x} \cos{\left(x \right)} d x}}}}{2}=\frac{e^{2 x} \sin{\left(x \right)}}{2} - \frac{{\color{red}{\left(\cos{\left(x \right)} \cdot \frac{e^{2 x}}{2}-\int{\frac{e^{2 x}}{2} \cdot \left(- \sin{\left(x \right)}\right) d x}\right)}}}{2}=\frac{e^{2 x} \sin{\left(x \right)}}{2} - \frac{{\color{red}{\left(\frac{e^{2 x} \cos{\left(x \right)}}{2} - \int{\left(- \frac{e^{2 x} \sin{\left(x \right)}}{2}\right)d x}\right)}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{1}{2}$$$ and $$$f{\left(x \right)} = e^{2 x} \sin{\left(x \right)}$$$:

$$\frac{e^{2 x} \sin{\left(x \right)}}{2} - \frac{e^{2 x} \cos{\left(x \right)}}{4} + \frac{{\color{red}{\int{\left(- \frac{e^{2 x} \sin{\left(x \right)}}{2}\right)d x}}}}{2} = \frac{e^{2 x} \sin{\left(x \right)}}{2} - \frac{e^{2 x} \cos{\left(x \right)}}{4} + \frac{{\color{red}{\left(- \frac{\int{e^{2 x} \sin{\left(x \right)} d x}}{2}\right)}}}{2}$$

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

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

$$\int{e^{2 x} \sin{\left(x \right)} d x} = \frac{e^{2 x} \sin{\left(x \right)}}{2} - \frac{e^{2 x} \cos{\left(x \right)}}{4} - \frac{\int{e^{2 x} \sin{\left(x \right)} d x}}{4}$$

Solving it, we get that

$$\int{e^{2 x} \sin{\left(x \right)} d x} = \frac{\left(2 \sin{\left(x \right)} - \cos{\left(x \right)}\right) e^{2 x}}{5}$$

Therefore,

$$\int{e^{2 x} \sin{\left(x \right)} d x} = \frac{\left(2 \sin{\left(x \right)} - \cos{\left(x \right)}\right) e^{2 x}}{5}$$

Add the constant of integration:

$$\int{e^{2 x} \sin{\left(x \right)} d x} = \frac{\left(2 \sin{\left(x \right)} - \cos{\left(x \right)}\right) e^{2 x}}{5}+C$$

$$$\int e^{2 x} \sin{\left(x \right)}\, dx = \frac{\left(2 \sin{\left(x \right)} - \cos{\left(x \right)}\right) e^{2 x}}{5} + C$$$A