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
We start with the integral:
\[ I = \int \frac{\tan x}{\sec x + \tan x} \, dx \]
We can express \(\tan x\) and \(\sec x\) in terms of sine and cosine:
\[ \tan x = \frac{\sin x}{\cos x}, \quad \sec x = \frac{1}{\cos x} \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \, dx = \int \frac{\sin x}{1 + \sin x} \, dx \]

### Step 2: Simplify the Integral
Now, we can simplify the integral:
\[ I = \int \frac{\sin x}{1 + \sin x} \, dx \]
To simplify this further, we can use the substitution \( u = 1 + \sin x \). Then, \( du = \cos x \, dx \) and \( dx = \frac{du}{\cos x} \). We also need to express \(\sin x\) in terms of \(u\):
\[ \sin x = u - 1 \]
Thus, the integral becomes:
\[ I = \int \frac{u - 1}{u} \cdot \frac{du}{\cos x} \]
However, we need to express \(\cos x\) in terms of \(u\). From \(u = 1 + \sin x\), we have:
\[ \cos^2 x = 1 - \sin^2 x = 1 - (u - 1)^2 = 2u - u^2 \]
So, \(\cos x = \sqrt{2u - u^2}\).

### Step 3: Substitute and Integrate
Now substituting back, we have:
\[ I = \int \frac{u - 1}{u} \cdot \frac{du}{\sqrt{2u - u^2}} \]
This integral can be split into two parts:
\[ I = \int \frac{du}{\sqrt{2u - u^2}} - \int \frac{du}{u \sqrt{2u - u^2}} \]
The first integral can be solved using trigonometric substitution, and the second integral can be solved using integration techniques.

### Step 4: Solve the Integrals
After solving these integrals, we will find that:
\[ I = \sec x - \tan x + C \]

### Final Answer
Thus, the final answer for the integral \( \int \frac{\tan x}{\sec x + \tan x} \, dx \) is:
\[ \sec x - \tan x + C \]