Integral of $$$\frac{\sin{\left(\theta \right)}}{r}$$$ with respect to $$$\theta$$$

Find $$$\int \frac{\sin{\left(\theta \right)}}{r}\, d\theta$$$.

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

$${\color{red}{\int{\frac{\sin{\left(\theta \right)}}{r} d \theta}}} = {\color{red}{\frac{\int{\sin{\left(\theta \right)} d \theta}}{r}}}$$

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

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

Therefore,

$$\int{\frac{\sin{\left(\theta \right)}}{r} d \theta} = - \frac{\cos{\left(\theta \right)}}{r}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(\theta \right)}}{r} d \theta} = - \frac{\cos{\left(\theta \right)}}{r}+C$$

$$$\int \frac{\sin{\left(\theta \right)}}{r}\, d\theta = - \frac{\cos{\left(\theta \right)}}{r} + C$$$A