We need to evaluate the definite integral: $I = \int (\frac{1}{x} + \frac{4}{x^2} - \frac{5}{\sin^2 x}) dx$

1. Problem Description

We need to evaluate the definite integral:

$I = \int (\frac{1}{x} + \frac{4}{x^2} - \frac{5}{\sin^2 x}) dx$

2. Solution Steps

First, we can split the integral into three separate integrals:

$I = \int \frac{1}{x} dx + \int \frac{4}{x^2} dx - \int \frac{5}{\sin^2 x} dx$

We can rewrite the second term as:

$I = \int \frac{1}{x} dx + 4 \int x^{-2} dx - 5 \int \frac{1}{\sin^2 x} dx$

We know that:

$\int \frac{1}{x} dx = \ln|x| + C_1$

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$.

So, $\int x^{-2} dx = \frac{x^{-2+1}}{-2+1} + C_2 = \frac{x^{-1}}{-1} + C_2 = -\frac{1}{x} + C_2$

Also, we know that $\int \frac{1}{\sin^2 x} dx = \int \csc^2 x dx = -\cot x + C_3$

Substituting these back into our integral:

$I = \ln|x| + 4(-\frac{1}{x}) - 5(-\cot x) + C$

$I = \ln|x| - \frac{4}{x} + 5\cot x + C$

3. Final Answer

$I = \ln|x| - \frac{4}{x} + 5\cot x + C$