Integral of $$$6 e^{- \frac{x}{2}} \sin{\left(2 x \right)}$$$

Find $$$\int 6 e^{- \frac{x}{2}} \sin{\left(2 x \right)}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=6$$$ and $$$f{\left(x \right)} = e^{- \frac{x}{2}} \sin{\left(2 x \right)}$$$:

$${\color{red}{\int{6 e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x}}} = {\color{red}{\left(6 \int{e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x}\right)}}$$

For the integral $$$\int{e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\sin{\left(2 x \right)}$$$ and $$$\operatorname{dv}=e^{- \frac{x}{2}} dx$$$.

Then $$$\operatorname{du}=\left(\sin{\left(2 x \right)}\right)^{\prime }dx=2 \cos{\left(2 x \right)} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{- \frac{x}{2}} d x}=- 2 e^{- \frac{x}{2}}$$$ (steps can be seen »).

Therefore,

$$6 {\color{red}{\int{e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x}}}=6 {\color{red}{\left(\sin{\left(2 x \right)} \cdot \left(- 2 e^{- \frac{x}{2}}\right)-\int{\left(- 2 e^{- \frac{x}{2}}\right) \cdot 2 \cos{\left(2 x \right)} d x}\right)}}=6 {\color{red}{\left(- \int{\left(- 4 e^{- \frac{x}{2}} \cos{\left(2 x \right)}\right)d x} - 2 e^{- \frac{x}{2}} \sin{\left(2 x \right)}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-4$$$ and $$$f{\left(x \right)} = e^{- \frac{x}{2}} \cos{\left(2 x \right)}$$$:

$$- 6 {\color{red}{\int{\left(- 4 e^{- \frac{x}{2}} \cos{\left(2 x \right)}\right)d x}}} - 12 e^{- \frac{x}{2}} \sin{\left(2 x \right)} = - 6 {\color{red}{\left(- 4 \int{e^{- \frac{x}{2}} \cos{\left(2 x \right)} d x}\right)}} - 12 e^{- \frac{x}{2}} \sin{\left(2 x \right)}$$

For the integral $$$\int{e^{- \frac{x}{2}} \cos{\left(2 x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\cos{\left(2 x \right)}$$$ and $$$\operatorname{dv}=e^{- \frac{x}{2}} dx$$$.

Then $$$\operatorname{du}=\left(\cos{\left(2 x \right)}\right)^{\prime }dx=- 2 \sin{\left(2 x \right)} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{- \frac{x}{2}} d x}=- 2 e^{- \frac{x}{2}}$$$ (steps can be seen »).

The integral becomes

$$24 {\color{red}{\int{e^{- \frac{x}{2}} \cos{\left(2 x \right)} d x}}} - 12 e^{- \frac{x}{2}} \sin{\left(2 x \right)}=24 {\color{red}{\left(\cos{\left(2 x \right)} \cdot \left(- 2 e^{- \frac{x}{2}}\right)-\int{\left(- 2 e^{- \frac{x}{2}}\right) \cdot \left(- 2 \sin{\left(2 x \right)}\right) d x}\right)}} - 12 e^{- \frac{x}{2}} \sin{\left(2 x \right)}=24 {\color{red}{\left(- \int{4 e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x} - 2 e^{- \frac{x}{2}} \cos{\left(2 x \right)}\right)}} - 12 e^{- \frac{x}{2}} \sin{\left(2 x \right)}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = e^{- \frac{x}{2}} \sin{\left(2 x \right)}$$$:

$$- 24 {\color{red}{\int{4 e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x}}} - 12 e^{- \frac{x}{2}} \sin{\left(2 x \right)} - 48 e^{- \frac{x}{2}} \cos{\left(2 x \right)} = - 24 {\color{red}{\left(4 \int{e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x}\right)}} - 12 e^{- \frac{x}{2}} \sin{\left(2 x \right)} - 48 e^{- \frac{x}{2}} \cos{\left(2 x \right)}$$

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

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

$$6 \int{e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x} = - 96 \int{e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x} - 12 e^{- \frac{x}{2}} \sin{\left(2 x \right)} - 48 e^{- \frac{x}{2}} \cos{\left(2 x \right)}$$

Solving it, we get that

$$\int{e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x} = \frac{2 \left(- \sin{\left(2 x \right)} - 4 \cos{\left(2 x \right)}\right) e^{- \frac{x}{2}}}{17}$$

Thus,

$$6 {\color{red}{\int{e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x}}} = 6 {\color{red}{\left(\frac{2 \left(- \sin{\left(2 x \right)} - 4 \cos{\left(2 x \right)}\right) e^{- \frac{x}{2}}}{17}\right)}}$$

Therefore,

$$\int{6 e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x} = \frac{12 \left(- \sin{\left(2 x \right)} - 4 \cos{\left(2 x \right)}\right) e^{- \frac{x}{2}}}{17}$$

Add the constant of integration:

$$\int{6 e^{- \frac{x}{2}} \sin{\left(2 x \right)} d x} = \frac{12 \left(- \sin{\left(2 x \right)} - 4 \cos{\left(2 x \right)}\right) e^{- \frac{x}{2}}}{17}+C$$

$$$\int 6 e^{- \frac{x}{2}} \sin{\left(2 x \right)}\, dx = \frac{12 \left(- \sin{\left(2 x \right)} - 4 \cos{\left(2 x \right)}\right) e^{- \frac{x}{2}}}{17} + C$$$A