Integral of 5(tan^2(x)-4cot(x)) dx

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

   $$\int 5 \left(\tan^{2}{\left(x \right)} - 4 \cot{\left(x \right)}\right)\, dx = 5 \int \left(\tan^{2}{\left(x \right)} - 4 \cot{\left(x \right)}\right)\, dx$$

   1. Integrate term-by-term:

      1. Rewrite the integrand:

         $$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$

      2. Integrate term-by-term:

         1. $$\int \sec^{2}{\left(x \right)}\, dx = \tan{\left(x \right)}$$

         1. The integral of a constant is the constant times the variable of integration:

            $$\int \left(-1\right)\, dx = - x$$

         The result is: $- x + \tan{\left(x \right)}$

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

         $$\int \left(- 4 \cot{\left(x \right)}\right)\, dx = - 4 \int \cot{\left(x \right)}\, dx$$

         1. Rewrite the integrand:

            $$\cot{\left(x \right)} = \frac{\cos{\left(x \right)}}{\sin{\left(x \right)}}$$

         2. Let $u = \sin{\left(x \right)}$.

            Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

            $$\int \frac{1}{u}\, du$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            Now substitute $u$ back in:

            $$\log{\left(\sin{\left(x \right)} \right)}$$

         So, the result is: $- 4 \log{\left(\sin{\left(x \right)} \right)}$

      The result is: $- x - 4 \log{\left(\sin{\left(x \right)} \right)} + \tan{\left(x \right)}$

   So, the result is: $- 5 x - 20 \log{\left(\sin{\left(x \right)} \right)} + 5 \tan{\left(x \right)}$

2. Add the constant of integration:

   $$- 5 x - 20 \log{\left(\sin{\left(x \right)} \right)} + 5 \tan{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$- 5 x - 20 \log{\left(\sin{\left(x \right)} \right)} + 5 \tan{\left(x \right)}+ \mathrm{constant}$$