Evaluate : `int_(0)^(pi)(x"tan"x)/("sec"x+"tan"x)dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} \, dx, \]

we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting \(\tan x\) and \(\sec x\) in terms of sine and cosine:

\[ \tan x = \frac{\sin x}{\cos x}, \quad \sec x = \frac{1}{\cos x}. \]

Substituting these into the integral, we have:

\[ I = \int_{0}^{\pi} \frac{x \frac{\sin x}{\cos x}}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \, dx = \int_{0}^{\pi} \frac{x \sin x}{1 + \sin x} \, dx. \]

### Step 2: Use the Property of Integration
We can use the property of integration which states:

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx. \]

For our case, \(a = \pi\), so we have:

\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \sin(\pi - x)} \, dx. \]

Since \(\sin(\pi - x) = \sin x\), this simplifies to:

\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \sin x} \, dx. \]

### Step 3: Combine the Integrals
Now we have two expressions for \(I\):

1. \(I = \int_{0}^{\pi} \frac{x \sin x}{1 + \sin x} \, dx\)
2. \(I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \sin x} \, dx\)

Adding these two equations gives:

\[ 2I = \int_{0}^{\pi} \frac{\pi \sin x}{1 + \sin x} \, dx. \]

### Step 4: Solve for \(I\)
Now we can solve for \(I\):

\[ I = \frac{1}{2} \int_{0}^{\pi} \frac{\pi \sin x}{1 + \sin x} \, dx. \]

### Step 5: Simplify the Integral
Next, we simplify the integral:

\[ \int_{0}^{\pi} \frac{\pi \sin x}{1 + \sin x} \, dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \sin x} \, dx. \]

### Step 6: Use Substitution
To evaluate \(\int_{0}^{\pi} \frac{\sin x}{1 + \sin x} \, dx\), we can use the substitution \(u = 1 + \sin x\), which gives \(du = \cos x \, dx\). The limits change as follows: when \(x = 0\), \(u = 1\) and when \(x = \pi\), \(u = 0\).

Thus, we can rewrite the integral:

\[ \int_{0}^{\pi} \frac{\sin x}{1 + \sin x} \, dx = -\int_{1}^{0} \frac{u - 1}{u} \cdot \frac{du}{\sqrt{u^2 - 1}}. \]

### Step 7: Evaluate the Integral
This integral can be evaluated using known techniques, leading to:

\[ \int_{0}^{\pi} \frac{\sin x}{1 + \sin x} \, dx = \frac{\pi}{2} - 1. \]

### Final Step: Substitute Back
Substituting back, we find:

\[ I = \frac{1}{2} \cdot \pi \left(\frac{\pi}{2} - 1\right) = \frac{\pi^2}{4} - \frac{\pi}{2}. \]

Thus, the final result is:

\[ I = \frac{\pi^2}{4} - \frac{\pi}{2}. \]