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

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

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 $$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$:

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

Let $$$u=4 x$$$.

Then $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{4}$$$.

So,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

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

$$- \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{16} = - \frac{{\color{red}{\sin{\left(u \right)}}}}{16}$$

Recall that $$$u=4 x$$$:

$$- \frac{\sin{\left({\color{red}{u}} \right)}}{16} = - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{16}$$

Therefore,

$$\int{\left(- \frac{\cos{\left(4 x \right)}}{4}\right)d x} = - \frac{\sin{\left(4 x \right)}}{16}$$

Add the constant of integration:

$$\int{\left(- \frac{\cos{\left(4 x \right)}}{4}\right)d x} = - \frac{\sin{\left(4 x \right)}}{16}+C$$

$$$\int \left(- \frac{\cos{\left(4 x \right)}}{4}\right)\, dx = - \frac{\sin{\left(4 x \right)}}{16} + C$$$A