Integral of sin(5*x)-cos(6*x) dx

1. Integrate term-by-term:

   1. Let $u = 5 x$.

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

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

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

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

         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:

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

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

      $$\int \left(- \cos{\left(6 x \right)}\right)\, dx = - \int \cos{\left(6 x \right)}\, dx$$

      1. Let $u = 6 x$.

         Then let $du = 6 dx$ and substitute $\frac{du}{6}$:

         $$\int \frac{\cos{\left(u \right)}}{6}\, du$$

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

            $$\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{6}$$

            1. The integral of cosine is sine:

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

            So, the result is: $\frac{\sin{\left(u \right)}}{6}$

         Now substitute $u$ back in:

         $$\frac{\sin{\left(6 x \right)}}{6}$$

      So, the result is: $- \frac{\sin{\left(6 x \right)}}{6}$

   The result is: $- \frac{\sin{\left(6 x \right)}}{6} - \frac{\cos{\left(5 x \right)}}{5}$

2. Add the constant of integration:

   $$- \frac{\sin{\left(6 x \right)}}{6} - \frac{\cos{\left(5 x \right)}}{5}+ \mathrm{constant}$$

The answer is:

$$- \frac{\sin{\left(6 x \right)}}{6} - \frac{\cos{\left(5 x \right)}}{5}+ \mathrm{constant}$$