Integral of (6sinx+6cosx)dx dx

1. Integrate term-by-term:

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

      $$\int 6 \sin{\left(x \right)}\, dx = 6 \int \sin{\left(x \right)}\, dx$$

      1. The integral of sine is negative cosine:

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

      So, the result is: $- 6 \cos{\left(x \right)}$

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

      $$\int 6 \cos{\left(x \right)}\, dx = 6 \int \cos{\left(x \right)}\, dx$$

      1. The integral of cosine is sine:

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

      So, the result is: $6 \sin{\left(x \right)}$

   The result is: $6 \sin{\left(x \right)} - 6 \cos{\left(x \right)}$

2. Now simplify:

   $$- 6 \sqrt{2} \cos{\left(x + \frac{\pi}{4} \right)}$$

3. Add the constant of integration:

   $$- 6 \sqrt{2} \cos{\left(x + \frac{\pi}{4} \right)}+ \mathrm{constant}$$

The answer is:

$$- 6 \sqrt{2} \cos{\left(x + \frac{\pi}{4} \right)}+ \mathrm{constant}$$