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

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

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

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

Let $$$u=2 x$$$.

Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

The integral becomes

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

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

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

Let $$$v=e^{\frac{u}{2}}$$$.

Then $$$dv=\left(e^{\frac{u}{2}}\right)^{\prime }du = \frac{e^{\frac{u}{2}}}{2} du$$$ (steps can be seen »), and we have that $$$e^{\frac{u}{2}} du = 2 dv$$$.

The integral becomes

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

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=2$$$ and $$$f{\left(v \right)} = v \cos{\left(v \right)}$$$:

$$- \frac{{\color{red}{\int{2 v \cos{\left(v \right)} d v}}}}{2} = - \frac{{\color{red}{\left(2 \int{v \cos{\left(v \right)} d v}\right)}}}{2}$$

For the integral $$$\int{v \cos{\left(v \right)} d v}$$$, use integration by parts $$$\int \operatorname{c} \operatorname{dr} = \operatorname{c}\operatorname{r} - \int \operatorname{r} \operatorname{dc}$$$.

Let $$$\operatorname{c}=v$$$ and $$$\operatorname{dr}=\cos{\left(v \right)} dv$$$.

Then $$$\operatorname{dc}=\left(v\right)^{\prime }dv=1 dv$$$ (steps can be seen ») and $$$\operatorname{r}=\int{\cos{\left(v \right)} d v}=\sin{\left(v \right)}$$$ (steps can be seen »).

So,

$$- {\color{red}{\int{v \cos{\left(v \right)} d v}}}=- {\color{red}{\left(v \cdot \sin{\left(v \right)}-\int{\sin{\left(v \right)} \cdot 1 d v}\right)}}=- {\color{red}{\left(v \sin{\left(v \right)} - \int{\sin{\left(v \right)} d v}\right)}}$$

The integral of the sine is $$$\int{\sin{\left(v \right)} d v} = - \cos{\left(v \right)}$$$:

$$- v \sin{\left(v \right)} + {\color{red}{\int{\sin{\left(v \right)} d v}}} = - v \sin{\left(v \right)} + {\color{red}{\left(- \cos{\left(v \right)}\right)}}$$

Recall that $$$v=e^{\frac{u}{2}}$$$:

$$- \cos{\left({\color{red}{v}} \right)} - {\color{red}{v}} \sin{\left({\color{red}{v}} \right)} = - \cos{\left({\color{red}{e^{\frac{u}{2}}}} \right)} - {\color{red}{e^{\frac{u}{2}}}} \sin{\left({\color{red}{e^{\frac{u}{2}}}} \right)}$$

Recall that $$$u=2 x$$$:

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

Therefore,

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

Add the constant of integration:

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

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