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 »).

The integral becomes

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

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 »).

Therefore,

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

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