Integral of cos(19x-2)-sin(10x) dx

1. Integrate term-by-term:

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

      $$\int \left(- \sin{\left(10 x \right)}\right)\, dx = - \int \sin{\left(10 x \right)}\, dx$$

      1. Let $u = 10 x$.

         Then let $du = 10 dx$ and substitute $\frac{du}{10}$:

         $$\int \frac{\sin{\left(u \right)}}{10}\, du$$

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

            $$\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{10}$$

            1. The integral of sine is negative cosine:

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

            So, the result is: $- \frac{\cos{\left(u \right)}}{10}$

         Now substitute $u$ back in:

         $$- \frac{\cos{\left(10 x \right)}}{10}$$

      So, the result is: $\frac{\cos{\left(10 x \right)}}{10}$

   1. Let $u = 19 x - 2$.

      Then let $du = 19 dx$ and substitute $\frac{du}{19}$:

      $$\int \frac{\cos{\left(u \right)}}{19}\, du$$

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

         $$\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{19}$$

         1. The integral of cosine is sine:

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

         So, the result is: $\frac{\sin{\left(u \right)}}{19}$

      Now substitute $u$ back in:

      $$\frac{\sin{\left(19 x - 2 \right)}}{19}$$

   The result is: $\frac{\sin{\left(19 x - 2 \right)}}{19} + \frac{\cos{\left(10 x \right)}}{10}$

2. Now simplify:

   $$\frac{\sin{\left(19 x - 2 \right)}}{19} + \frac{\cos{\left(10 x \right)}}{10}$$

3. Add the constant of integration:

   $$\frac{\sin{\left(19 x - 2 \right)}}{19} + \frac{\cos{\left(10 x \right)}}{10}+ \mathrm{constant}$$

The answer is: