Integral of cos(7x+2) dx

1. Let $u = 7 x + 2$.

   Then let $du = 7 dx$ and substitute $\frac{du}{7}$:

   $$\int \frac{\cos{\left(u \right)}}{49}\, du$$

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

      $$\int \frac{\cos{\left(u \right)}}{7}\, du = \frac{\int \cos{\left(u \right)}\, du}{7}$$

      1. The integral of cosine is sine:

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

      So, the result is: $\frac{\sin{\left(u \right)}}{7}$

   Now substitute $u$ back in:

   $$\frac{\sin{\left(7 x + 2 \right)}}{7}$$

2. Now simplify:

   $$\frac{\sin{\left(7 x + 2 \right)}}{7}$$

3. Add the constant of integration:

   $$\frac{\sin{\left(7 x + 2 \right)}}{7}+ \mathrm{constant}$$

The answer is:

$$\frac{\sin{\left(7 x + 2 \right)}}{7}+ \mathrm{constant}$$