Integral of $$$- 8 \cos{\left(t \right)} - 1$$$

Find $$$\int \left(- 8 \cos{\left(t \right)} - 1\right)\, dt$$$.

Integrate term by term:

$${\color{red}{\int{\left(- 8 \cos{\left(t \right)} - 1\right)d t}}} = {\color{red}{\left(- \int{1 d t} - \int{8 \cos{\left(t \right)} d t}\right)}}$$

Apply the constant rule $$$\int c\, dt = c t$$$ with $$$c=1$$$:

$$- \int{8 \cos{\left(t \right)} d t} - {\color{red}{\int{1 d t}}} = - \int{8 \cos{\left(t \right)} d t} - {\color{red}{t}}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=8$$$ and $$$f{\left(t \right)} = \cos{\left(t \right)}$$$:

$$- t - {\color{red}{\int{8 \cos{\left(t \right)} d t}}} = - t - {\color{red}{\left(8 \int{\cos{\left(t \right)} d t}\right)}}$$

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

$$- t - 8 {\color{red}{\int{\cos{\left(t \right)} d t}}} = - t - 8 {\color{red}{\sin{\left(t \right)}}}$$

Therefore,

$$\int{\left(- 8 \cos{\left(t \right)} - 1\right)d t} = - t - 8 \sin{\left(t \right)}$$

Add the constant of integration:

$$\int{\left(- 8 \cos{\left(t \right)} - 1\right)d t} = - t - 8 \sin{\left(t \right)}+C$$

$$$\int \left(- 8 \cos{\left(t \right)} - 1\right)\, dt = \left(- t - 8 \sin{\left(t \right)}\right) + C$$$A