Evaluate : `int(I+tan^2)/(I+cot^2x)dx`

Options

A. `tan x - 1 + c`
B. `tan x - x + c `
C. `cot x - x +c`
D. `tan x+c`

Answer

To evaluate the integral \( \int \frac{1 + \tan^2 x}{1 + \cot^2 x} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We start with the expression:
\[ \frac{1 + \tan^2 x}{1 + \cot^2 x} \]
Recall that \( \cot^2 x = \frac{1}{\tan^2 x} \). Thus, we can rewrite \( 1 + \cot^2 x \) as:
\[ 1 + \cot^2 x = 1 + \frac{1}{\tan^2 x} = \frac{\tan^2 x + 1}{\tan^2 x} \]
Now substituting this back into our integral, we have:
\[ \int \frac{1 + \tan^2 x}{\frac{\tan^2 x + 1}{\tan^2 x}} \, dx = \int \frac{(1 + \tan^2 x) \tan^2 x}{\tan^2 x + 1} \, dx \]
Since \( 1 + \tan^2 x = \sec^2 x \), we can simplify further:
\[ \int \tan^2 x \, dx \]

### Step 2: Use the identity for \( \tan^2 x \)
We know that:
\[ \tan^2 x = \sec^2 x - 1 \]
Substituting this into our integral gives:
\[ \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx \]

### Step 3: Integrate the expression
Now we can integrate:
\[ \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx \]
The integral of \( \sec^2 x \) is \( \tan x \), and the integral of \( 1 \) is \( x \):
\[ = \tan x - x + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{1 + \tan^2 x}{1 + \cot^2 x} \, dx = \tan x - x + C \]

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