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

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

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

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

Thus,

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

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

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

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

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

The integral can be rewritten as

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

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

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

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

Solving it, we get that

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int e^{x} \cos{\left(x \right)}\, dx = \frac{\sqrt{2} e^{x} \sin{\left(x + \frac{\pi}{4} \right)}}{2} + C$$$A