Integral of $$$\sin{\left(2 t \right)}$$$

Find $$$\int \sin{\left(2 t \right)}\, dt$$$.

Let $$$u=2 t$$$.

Then $$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (steps can be seen »), and we have that $$$dt = \frac{du}{2}$$$.

The integral becomes

$${\color{red}{\int{\sin{\left(2 t \right)} d t}}} = {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}$$

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

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{2}$$

Recall that $$$u=2 t$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{2} = - \frac{\cos{\left({\color{red}{\left(2 t\right)}} \right)}}{2}$$

Therefore,

$$\int{\sin{\left(2 t \right)} d t} = - \frac{\cos{\left(2 t \right)}}{2}$$

Add the constant of integration:

$$\int{\sin{\left(2 t \right)} d t} = - \frac{\cos{\left(2 t \right)}}{2}+C$$

$$$\int \sin{\left(2 t \right)}\, dt = - \frac{\cos{\left(2 t \right)}}{2} + C$$$A