Evaluate `int(tanx)/(secx+tanx)dx`

Explanation

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

### Step 1: Rewrite the integral in terms of sine and cosine
We know that:
- \( \tan x = \frac{\sin x}{\cos x} \)
- \( \sec x = \frac{1}{\cos x} \)

Thus, we can rewrite the integral as:
\[ \int \frac{\tan x}{\sec x + \tan x} \, dx = \int \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \, dx \]

### Step 2: Simplify the expression
This simplifies to:
\[ \int \frac{\sin x}{1 + \sin x} \, dx \]

### Step 3: Rationalize the denominator
To simplify further, we can multiply the numerator and the denominator by \( 1 - \sin x \):
\[ \int \frac{\sin x (1 - \sin x)}{(1 + \sin x)(1 - \sin x)} \, dx = \int \frac{\sin x - \sin^2 x}{1 - \sin^2 x} \, dx \]
Since \( 1 - \sin^2 x = \cos^2 x \), we have:
\[ \int \frac{\sin x - \sin^2 x}{\cos^2 x} \, dx \]

### Step 4: Split the integral
Now, we can split the integral into two parts:
\[ \int \frac{\sin x}{\cos^2 x} \, dx - \int \frac{\sin^2 x}{\cos^2 x} \, dx \]
This can be rewritten as:
\[ \int \tan x \sec x \, dx - \int \tan^2 x \, dx \]

### Step 5: Evaluate the first integral
The integral \( \int \tan x \sec x \, dx \) is known to be:
\[ \sec x + C_1 \]

### Step 6: Evaluate the second integral
The integral \( \int \tan^2 x \, dx \) can be rewritten using the identity \( \tan^2 x = \sec^2 x - 1 \):
\[ \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx \]
This gives us:
\[ \tan x - x + C_2 \]

### Step 7: Combine the results
Now, combining both parts, we have:
\[ \sec x - (\tan x - x) + C = \sec x - \tan x + x + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{\tan x}{\sec x + \tan x} \, dx = \sec x - \tan x + x + C \]

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