Integral of $$$2 i \pi d n t \theta$$$ with respect to $$$t$$$

Find $$$\int 2 i \pi d n t \theta\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=2 i \pi d n \theta$$$ and $$$f{\left(t \right)} = t$$$:

$${\color{red}{\int{2 i \pi d n t \theta d t}}} = {\color{red}{\left(2 i \pi d n \theta \int{t d t}\right)}}$$

Apply the power rule $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$$2 i \pi d n \theta {\color{red}{\int{t d t}}}=2 i \pi d n \theta {\color{red}{\frac{t^{1 + 1}}{1 + 1}}}=2 i \pi d n \theta {\color{red}{\left(\frac{t^{2}}{2}\right)}}$$

Therefore,

$$\int{2 i \pi d n t \theta d t} = i \pi d n t^{2} \theta$$

Add the constant of integration:

$$\int{2 i \pi d n t \theta d t} = i \pi d n t^{2} \theta+C$$

$$$\int 2 i \pi d n t \theta\, dt = i \pi d n t^{2} \theta + C$$$A