`int_(0)^(pi)(x tanx)/(secx+cosx)dx` is

Options

A. `(pi^(2))/4`
B. `(pi^(2))/2`
C. `(3pi^(2))/2`
D. `(pi^(2))/3`

Explanation

To solve the integral \( I = \int_{0}^{\pi} \frac{x \tan x}{\sec x + \cos x} \, dx \), we will utilize the property of definite integrals that states:

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

### Step 1: Apply the property of definite integrals
Let \( I = \int_{0}^{\pi} \frac{x \tan x}{\sec x + \cos x} \, dx \).

Now, we will change the variable \( x \) to \( \pi - x \):

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

### Step 2: Simplify the integrand
Using the trigonometric identities:
- \( \tan(\pi - x) = -\tan x \)
- \( \sec(\pi - x) = -\sec x \)
- \( \cos(\pi - x) = -\cos x \)

We can rewrite the integral as:

\[ I = \int_{0}^{\pi} \frac{(\pi - x)(-\tan x)}{-\sec x - \cos x} \, dx \]

This simplifies to:

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

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

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

Adding these two equations:

\[ 2I = \int_{0}^{\pi} \frac{x \tan x}{\sec x + \cos x} \, dx + \int_{0}^{\pi} \frac{(\pi - x) \tan x}{\sec x + \cos x} \, dx \]

This simplifies to:

\[ 2I = \int_{0}^{\pi} \frac{\pi \tan x}{\sec x + \cos x} \, dx \]

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

\[ I = \frac{1}{2} \int_{0}^{\pi} \frac{\pi \tan x}{\sec x + \cos x} \, dx \]

### Step 5: Simplify the integral
We can further simplify the integrand:

\[ \sec x + \cos x = \frac{1}{\cos x} + \cos x = \frac{1 + \cos^2 x}{\cos x} \]

Thus, we have:

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

### Step 6: Change of variable
Let \( u = \cos x \), then \( du = -\sin x \, dx \). The limits change from \( x = 0 \) to \( x = \pi \), which corresponds to \( u = 1 \) to \( u = -1 \):

\[ I = \frac{\pi}{2} \int_{1}^{-1} \frac{-1}{1 + u^2} \, du = \frac{\pi}{2} \int_{-1}^{1} \frac{1}{1 + u^2} \, du \]

### Step 7: Evaluate the integral
The integral \( \int \frac{1}{1 + u^2} \, du \) is known to be \( \tan^{-1}(u) \):

\[ I = \frac{\pi}{2} \left[ \tan^{-1}(u) \right]_{-1}^{1} = \frac{\pi}{2} \left( \tan^{-1}(1) - \tan^{-1}(-1) \right) = \frac{\pi}{2} \left( \frac{\pi}{4} + \frac{\pi}{4} \right) = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4} \]

### Final Answer
Thus, the value of the integral is:

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

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