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

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

Let $$$u=e^{x}$$$.

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

Thus,

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

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

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

Recall that $$$u=e^{x}$$$:

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

Therefore,

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

Add the constant of integration:

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

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