Integral of 2sin(2x+1) dx

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

   $$\int 2 \sin{\left(2 x + 1 \right)}\, dx = 2 \int \sin{\left(2 x + 1 \right)}\, dx$$

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

      Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

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

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

         $$\int \frac{\sin{\left(u \right)}}{2}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$$

         1. The integral of sine is negative cosine:

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

         So, the result is: $- \frac{\cos{\left(u \right)}}{2}$

      Now substitute $u$ back in:

      $$- \frac{\cos{\left(2 x + 1 \right)}}{2}$$

   So, the result is: $- \cos{\left(2 x + 1 \right)}$

2. Now simplify:

   $$- \cos{\left(2 x + 1 \right)}$$

3. Add the constant of integration:

   $$- \cos{\left(2 x + 1 \right)}+ \mathrm{constant}$$

The answer is:

$$- \cos{\left(2 x + 1 \right)}+ \mathrm{constant}$$