Integral of 8cos(4x) dx

1. The integral of a constant times a function is the constant times the integral of the function:

   $$\int 8 \cos{\left(4 x \right)}\, dx = 8 \int \cos{\left(4 x \right)}\, dx$$

   1. Let $u = 4 x$.

      Then let $du = 4 dx$ and substitute $\frac{du}{4}$:

      $$\int \frac{\cos{\left(u \right)}}{16}\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$$

         1. The integral of cosine is sine:

            $$\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$$

         So, the result is: $\frac{\sin{\left(u \right)}}{4}$

      Now substitute $u$ back in:

      $$\frac{\sin{\left(4 x \right)}}{4}$$

   So, the result is: $2 \sin{\left(4 x \right)}$

2. Add the constant of integration:

   $$2 \sin{\left(4 x \right)}+ \mathrm{constant}$$

The answer is:

$$2 \sin{\left(4 x \right)}+ \mathrm{constant}$$