Integral of sin(3+4x) dx

1. Let $u = 4 x + 3$.

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

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

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

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

      1. The integral of sine is negative cosine:

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

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

   Now substitute $u$ back in:

   $$- \frac{\cos{\left(4 x + 3 \right)}}{4}$$

2. Add the constant of integration:

   $$- \frac{\cos{\left(4 x + 3 \right)}}{4}+ \mathrm{constant}$$

The answer is:

$$- \frac{\cos{\left(4 x + 3 \right)}}{4}+ \mathrm{constant}$$