Integral of 4cos2x-3sinx dx

1. Integrate term-by-term:

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

      $$\int \left(- 3 \sin{\left(x \right)}\right)\, dx = - 3 \int \sin{\left(x \right)}\, dx$$

      1. The integral of sine is negative cosine:

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

      So, the result is: $3 \cos{\left(x \right)}$

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

      $$\int 4 \cos{\left(2 x \right)}\, dx = 4 \int \cos{\left(2 x \right)}\, dx$$

      1. Let $u = 2 x$.

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

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

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

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

            1. The integral of cosine is sine:

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

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

         Now substitute $u$ back in:

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

      So, the result is: $2 \sin{\left(2 x \right)}$

   The result is: $2 \sin{\left(2 x \right)} + 3 \cos{\left(x \right)}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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