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

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

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

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

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

Thus,

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

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

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

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

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

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

The integral becomes

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

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

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

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

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

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

Solving it, we get that

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

Therefore,

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

Add the constant of integration:

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

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