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

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

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

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

For the integral $$$\int{\theta e^{- \theta} d \theta}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\theta$$$ and $$$\operatorname{dv}=e^{- \theta} d\theta$$$.

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

Thus,

$$\cos{\left(2 \right)} {\color{red}{\int{\theta e^{- \theta} d \theta}}}=\cos{\left(2 \right)} {\color{red}{\left(\theta \cdot \left(- e^{- \theta}\right)-\int{\left(- e^{- \theta}\right) \cdot 1 d \theta}\right)}}=\cos{\left(2 \right)} {\color{red}{\left(- \theta e^{- \theta} - \int{\left(- e^{- \theta}\right)d \theta}\right)}}$$

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

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

Let $$$u=- \theta$$$.

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

The integral becomes

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

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

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

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

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

Recall that $$$u=- \theta$$$:

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

Therefore,

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

Simplify:

$$\int{\theta e^{- \theta} \cos{\left(2 \right)} d \theta} = - \left(\theta + 1\right) e^{- \theta} \cos{\left(2 \right)}$$

Add the constant of integration:

$$\int{\theta e^{- \theta} \cos{\left(2 \right)} d \theta} = - \left(\theta + 1\right) e^{- \theta} \cos{\left(2 \right)}+C$$

$$$\int \theta e^{- \theta} \cos{\left(2 \right)}\, d\theta = - \left(\theta + 1\right) e^{- \theta} \cos{\left(2 \right)} + C$$$A