Evaluate the following definite integral: `int_(-pi)^(pi)(2x(1+sinx))/(1+cos^(2)x)dx`

Answer

To evaluate the definite integral

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

we can break it down into manageable steps.

### Step 1: Split the Integral
We can separate the integral into two parts:

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

### Step 2: Analyze the First Integral
Let’s denote the first integral as \( I_1 \):

\[ I_1 = \int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x} \, dx. \]

To determine if \( I_1 \) is zero, we check if the integrand is an odd function.

The function \( f(x) = \frac{2x}{1+\cos^2 x} \) is odd because:

\[ f(-x) = \frac{2(-x)}{1+\cos^2(-x)} = -\frac{2x}{1+\cos^2 x} = -f(x). \]

Since the integral of an odd function over a symmetric interval around zero is zero, we have:

\[ I_1 = 0. \]

### Step 3: Analyze the Second Integral
Now, let’s denote the second integral as \( I_2 \):

\[ I_2 = \int_{-\pi}^{\pi} \frac{2x \sin x}{1+\cos^2 x} \, dx. \]

We check if \( I_2 \) is even or odd. The function \( g(x) = \frac{2x \sin x}{1+\cos^2 x} \) is odd because:

\[ g(-x) = \frac{2(-x) \sin(-x)}{1+\cos^2(-x)} = \frac{-2x(-\sin x)}{1+\cos^2 x} = -\frac{2x \sin x}{1+\cos^2 x} = -g(x). \]

Thus,

\[ I_2 = 0. \]

### Step 4: Combine Results
Since both \( I_1 \) and \( I_2 \) are zero, we conclude:

\[ I = I_1 + I_2 = 0 + 0 = 0. \]

### Final Result
Thus, the value of the definite integral is:

\[ \boxed{0}. \]