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

Find $$$\int 2 \sin{\left(2 x \right)}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$:

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

Let $$$u=2 x$$$.

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

Thus,

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

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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

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

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

Recall that $$$u=2 x$$$:

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

Therefore,

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

Add the constant of integration:

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

$$$\int 2 \sin{\left(2 x \right)}\, dx = - \cos{\left(2 x \right)} + C$$$A