`int_(0)^(pi) (x sin x)/(1+ cos^(2) x)dx`=

Options

A. `(pi^(2))/(8)`
B. `(pi^(2))/(16)`
C. `(pi^(2))/(4)`
D. `(pi^(2))/(32)`

Answer

To solve the integral \( I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx \), we will 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 apply this property to our integral:

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

### 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 gives:

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

This simplifies to:

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

### Step 4: Solve for \( I \)

Now, we can express \( I \):

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

### Step 5: Change of variable

Let \( t = \cos x \), then \( dt = -\sin x \, dx \). The limits change as follows:
- When \( x = 0 \), \( t = 1 \)
- When \( x = \pi \), \( t = -1 \)

Thus, we have:

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

### Step 6: Evaluate the integral

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

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

Since \( \tan^{-1}(1) = \frac{\pi}{4} \) and \( \tan^{-1}(-1) = -\frac{\pi}{4} \):

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

### Final Result

Thus, the value of the integral is:

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