Evaluate `int(4 cot x - 5 tan x)^2 dx`

Answer

To evaluate the integral \( \int (4 \cot x - 5 \tan x)^2 \, dx \), we will follow these steps:

### Step 1: Expand the integrand
We start by expanding the square:
\[ (4 \cot x - 5 \tan x)^2 = (4 \cot x)^2 - 2(4 \cot x)(5 \tan x) + (5 \tan x)^2 \]
Calculating each term:
- \( (4 \cot x)^2 = 16 \cot^2 x \)
- \( (5 \tan x)^2 = 25 \tan^2 x \)
- \( -2(4 \cot x)(5 \tan x) = -40 \)

Thus, we have:
\[ (4 \cot x - 5 \tan x)^2 = 16 \cot^2 x + 25 \tan^2 x - 40 \]

### Step 2: Rewrite in terms of sine and cosine
Using the identities \( \cot x = \frac{\cos x}{\sin x} \) and \( \tan x = \frac{\sin x}{\cos x} \), we can express \( \cot^2 x \) and \( \tan^2 x \):
\[ \cot^2 x = \frac{\cos^2 x}{\sin^2 x}, \quad \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \]

Thus, we rewrite the integral:
\[ \int (16 \cot^2 x + 25 \tan^2 x - 40) \, dx = \int \left( 16 \frac{\cos^2 x}{\sin^2 x} + 25 \frac{\sin^2 x}{\cos^2 x} - 40 \right) \, dx \]

### Step 3: Simplify the integral
We can use the identity \( \tan^2 x + 1 = \sec^2 x \) and \( \cot^2 x + 1 = \csc^2 x \):
\[ \cot^2 x = \csc^2 x - 1, \quad \tan^2 x = \sec^2 x - 1 \]
Substituting these into the integral:
\[ \int (16 (\csc^2 x - 1) + 25 (\sec^2 x - 1) - 40) \, dx \]
This simplifies to:
\[ \int (16 \csc^2 x + 25 \sec^2 x - 16 - 25 - 40) \, dx = \int (16 \csc^2 x + 25 \sec^2 x - 81) \, dx \]

### Step 4: Integrate term by term
Now we can integrate each term separately:
1. \( \int \csc^2 x \, dx = -\cot x \)
2. \( \int \sec^2 x \, dx = \tan x \)
3. The integral of a constant \( -81 \) is \( -81x \)

Thus, we have:
\[ \int (16 \csc^2 x + 25 \sec^2 x - 81) \, dx = 16(-\cot x) + 25\tan x - 81x + C \]

### Final Answer
Combining everything, we get:
\[ -16 \cot x + 25 \tan x - 81x + C \]