Integral of $$$- \frac{\sin{\left(2 x \right)}}{4}$$$

Find $$$\int \left(- \frac{\sin{\left(2 x \right)}}{4}\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{1}{4}$$$ and $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$:

$${\color{red}{\int{\left(- \frac{\sin{\left(2 x \right)}}{4}\right)d x}}} = {\color{red}{\left(- \frac{\int{\sin{\left(2 x \right)} d x}}{4}\right)}}$$

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

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

The integral becomes

$$- \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{4} = - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{4}$$

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

$$- \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{4} = - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{4}$$

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

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

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

$$\frac{\cos{\left({\color{red}{u}} \right)}}{8} = \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{8}$$

Therefore,

$$\int{\left(- \frac{\sin{\left(2 x \right)}}{4}\right)d x} = \frac{\cos{\left(2 x \right)}}{8}$$

Add the constant of integration:

$$\int{\left(- \frac{\sin{\left(2 x \right)}}{4}\right)d x} = \frac{\cos{\left(2 x \right)}}{8}+C$$

$$$\int \left(- \frac{\sin{\left(2 x \right)}}{4}\right)\, dx = \frac{\cos{\left(2 x \right)}}{8} + C$$$A