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

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

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

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

The integral becomes

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

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

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

For the integral $$$\int{u \sin{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{q} \operatorname{dv} = \operatorname{q}\operatorname{v} - \int \operatorname{v} \operatorname{dq}$$$.

Let $$$\operatorname{q}=u$$$ and $$$\operatorname{dv}=\sin{\left(u \right)} du$$$.

Then $$$\operatorname{dq}=\left(u\right)^{\prime }du=1 du$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\sin{\left(u \right)} d u}=- \cos{\left(u \right)}$$$ (steps can be seen »).

The integral can be rewritten as

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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