Integral of 6sin(3x+12) dx

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

   $$\int 6 \sin{\left(3 x + 12 \right)}\, dx = 6 \int \sin{\left(3 x + 12 \right)}\, dx$$

   1. Let $u = 3 x + 12$.

      Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

      $$\int \frac{\sin{\left(u \right)}}{3}\, du$$

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

         $$\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$$

         1. The integral of sine is negative cosine:

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

         So, the result is: $- \frac{\cos{\left(u \right)}}{3}$

      Now substitute $u$ back in:

      $$- \frac{\cos{\left(3 x + 12 \right)}}{3}$$

   So, the result is: $- 2 \cos{\left(3 x + 12 \right)}$

2. Now simplify:

   $$- 2 \cos{\left(3 x + 12 \right)}$$

3. Add the constant of integration:

   $$- 2 \cos{\left(3 x + 12 \right)}+ \mathrm{constant}$$

The answer is:

$$- 2 \cos{\left(3 x + 12 \right)}+ \mathrm{constant}$$