Evaluate `int _(0)^(pi)(x)/(1+ cos^(2)x)dx`.

Answer

To evaluate the integral

\[ I = \int_0^{\pi} \frac{x}{1 + \cos^2 x} \, dx, \]

we will use a symmetry property of definite integrals.

### Step 1: Apply the symmetry property
Using the property of definite integrals, we know that:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

In our case, \( a = 0 \) and \( b = \pi \), so we can rewrite the integral as:

\[ I = \int_0^{\pi} \frac{\pi - x}{1 + \cos^2(\pi - x)} \, dx. \]

### Step 2: Simplify the integral
Next, we simplify the expression \( \cos(\pi - x) \):

\[ \cos(\pi - x) = -\cos x \quad \text{so} \quad \cos^2(\pi - x) = \cos^2 x. \]

Thus, we can rewrite the integral:

\[ I = \int_0^{\pi} \frac{\pi - x}{1 + \cos^2 x} \, dx. \]

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

1. \( I = \int_0^{\pi} \frac{x}{1 + \cos^2 x} \, dx \)
2. \( I = \int_0^{\pi} \frac{\pi - x}{1 + \cos^2 x} \, dx \)

Adding these two equations gives:

\[ 2I = \int_0^{\pi} \frac{x + (\pi - x)}{1 + \cos^2 x} \, dx = \int_0^{\pi} \frac{\pi}{1 + \cos^2 x} \, dx. \]

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

\[ 2I = \pi \int_0^{\pi} \frac{1}{1 + \cos^2 x} \, dx. \]

Thus,

\[ I = \frac{\pi}{2} \int_0^{\pi} \frac{1}{1 + \cos^2 x} \, dx. \]

### Step 5: Evaluate the integral \( \int_0^{\pi} \frac{1}{1 + \cos^2 x} \, dx \)
To evaluate this integral, we can use the substitution \( t = \tan x \):

\[ dx = \frac{dt}{1 + t^2}, \quad \text{and the limits change from } 0 \text{ to } \infty. \]

Thus, we have:

\[ \int_0^{\pi} \frac{1}{1 + \cos^2 x} \, dx = \int_0^{\infty} \frac{1}{1 + \frac{1}{1 + t^2}} \cdot \frac{dt}{1 + t^2} = \int_0^{\infty} \frac{1 + t^2}{2 + t^2} \cdot \frac{dt}{1 + t^2}. \]

This simplifies to:

\[ \int_0^{\infty} \frac{dt}{2 + t^2}. \]

This integral can be evaluated using the formula:

\[ \int \frac{dt}{a^2 + t^2} = \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) + C. \]

In our case, \( a = \sqrt{2} \):

\[ \int_0^{\infty} \frac{dt}{2 + t^2} = \frac{1}{\sqrt{2}} \left[ \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) \right]_0^{\infty} = \frac{1}{\sqrt{2}} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{2\sqrt{2}}. \]

### Step 6: Substitute back to find \( I \)
Substituting back into our expression for \( I \):

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

Thus, the final answer is:

\[ \boxed{\frac{\pi^2}{4\sqrt{2}}}. \]