The value of the integral `int_(0)^(pi) (x sin x)/(1+cos^(2)x)dx`, is

Options

A. `(pi^(2))/(2)`
B. `(pi^(2))/(4)`
C. `(pi^(2))/(8)`
D. `(pi^(2))/(16)`

Explanation

To solve the integral \( I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx \), we can use 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’s denote the integral as \( I \):

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

Now, we will replace \( x \) with \( \pi - x \):

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

### Step 2: Simplify the expression

Using the identities \( \sin(\pi - x) = \sin x \) and \( \cos(\pi - x) = -\cos x \), we can rewrite the integral:

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

### Step 3: Combine the two integrals

Now we have two expressions for \( I \):

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

Adding these two equations together:

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

### Step 4: Factor out the common terms

This simplifies to:

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

### Step 5: Solve for \( I \)

Thus, we can express \( I \) as:

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

### Step 6: Change of variable

Now, we can use the substitution \( t = \cos x \), which gives \( dt = -\sin x \, dx \). The limits change from \( x = 0 \) to \( x = \pi \) corresponding to \( t = 1 \) to \( t = -1 \):

\[ I = \frac{\pi}{2} \int_{1}^{-1} \frac{-dt}{1 + t^2} \]

Reversing the limits gives:

\[ I = \frac{\pi}{2} \int_{-1}^{1} \frac{dt}{1 + t^2} \]

### Step 7: Evaluate the integral

The integral \( \int \frac{dt}{1 + t^2} \) is known to be \( \tan^{-1}(t) \):

\[ I = \frac{\pi}{2} \left[ \tan^{-1}(t) \right]_{-1}^{1} \]

Calculating the limits:

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

### Final Answer

Thus, the value of the integral is:

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