Integral of sin(5x+3) dx

1. Let $u = 5 x + 3$.

   Then let $du = 5 dx$ and substitute $\frac{du}{5}$:

   $$\int \frac{\sin{\left(u \right)}}{25}\, du$$

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

      $$\int \frac{\sin{\left(u \right)}}{5}\, du = \frac{\int \sin{\left(u \right)}\, du}{5}$$

      1. The integral of sine is negative cosine:

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

      So, the result is: $- \frac{\cos{\left(u \right)}}{5}$

   Now substitute $u$ back in:

   $$- \frac{\cos{\left(5 x + 3 \right)}}{5}$$

2. Now simplify:

   $$- \frac{\cos{\left(5 x + 3 \right)}}{5}$$

3. Add the constant of integration:

   $$- \frac{\cos{\left(5 x + 3 \right)}}{5}+ \mathrm{constant}$$

The answer is:

$$- \frac{\cos{\left(5 x + 3 \right)}}{5}+ \mathrm{constant}$$