The value of the integral `int_0^pi(x^2sinx)/((2x-pi)(1+cos^2x))\ dx`

Options

A. `(pi^(2))/(4)`
B. `(pi^(2))/(2)`
C. `(pi^(2))/(6)`
D. None of these

Answer

To solve the integral

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

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 substitute \( x \) with \( \pi - x \):

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

### Step 2: Simplify the expression

Using the identities \( \sin(\pi - x) = \sin x \) and \( \cos(\pi - x) = -\cos x \):

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

which simplifies to:

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

### Step 3: Expand the numerator

Now, expand \((\pi - x)^2\):

\[ (\pi - x)^2 = \pi^2 - 2\pi x + x^2. \]

Thus, we can rewrite the integral as:

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

### Step 4: Split the integral

Now we can split \( I \) into three separate integrals:

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

### Step 5: Combine the integrals

Notice that the second integral can be rewritten in terms of \( I \):

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

### Step 6: Solve for \( I \)

Rearranging gives:

\[ 0 = \int_0^\pi \frac{\pi^2 \sin x}{(\pi - 2x)(1 + \cos^2 x)} \, dx - 2\pi \int_0^\pi \frac{x \sin x}{(\pi - 2x)(1 + \cos^2 x)} \, dx. \]

### Step 7: Evaluate the integrals

Let’s denote:

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

Then we can express \( I \) in terms of \( J \):

\[ 2I = \pi J. \]

### Step 8: Final calculation

Now, we can evaluate \( J \):

Using the substitution \( t = \cos x \), we find:

\[ J = \int_1^{-1} \frac{-dt}{1 + t^2} = \int_{-1}^{1} \frac{dt}{1 + t^2} = 2 \tan^{-1}(1) = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}. \]

Thus,

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

### Conclusion

The value of the integral is

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