The value of the integral `int_(-pi//2)^(pi//2) (x^(2) + log" (pi-x)/(pi+x)) cos x dx `

Options

A. 0
B. `(pi^(2))/(2)-4`
C. `(pi^(2))/(2)+4`
D. `(pi^(2))/(2)`

Answer

To solve the integral

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

we will use the property of definite integrals that states:

\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} \left( f(x) + f(-x) \right) \, dx. \]

### Step 1: Rewrite the integral using the property
We can express the integral as:

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

where

\[ f(x) = \left( x^2 + \log \left( \frac{\pi - x}{\pi + x} \right) \right) \cos x. \]

### Step 2: Calculate \( f(-x) \)
Now we need to find \( f(-x) \):

\[ f(-x) = \left( (-x)^2 + \log \left( \frac{\pi - (-x)}{\pi + (-x)} \right) \right) \cos(-x). \]

Since \( (-x)^2 = x^2 \) and \( \cos(-x) = \cos x \), we have:

\[ f(-x) = x^2 + \log \left( \frac{\pi + x}{\pi - x} \right) \cos x. \]

### Step 3: Combine \( f(x) \) and \( f(-x) \)
Now we can add \( f(x) \) and \( f(-x) \):

\[ f(x) + f(-x) = \left( x^2 + \log \left( \frac{\pi - x}{\pi + x} \right) \right) \cos x + \left( x^2 + \log \left( \frac{\pi + x}{\pi - x} \right) \right) \cos x. \]

This simplifies to:

\[ f(x) + f(-x) = 2x^2 \cos x + \left( \log \left( \frac{\pi - x}{\pi + x} \right) + \log \left( \frac{\pi + x}{\pi - x} \right) \right) \cos x. \]

Using the property of logarithms:

\[ \log a + \log b = \log(ab), \]

we find that:

\[ \log \left( \frac{\pi - x}{\pi + x} \cdot \frac{\pi + x}{\pi - x} \right) = \log(1) = 0. \]

Thus, we have:

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

### Step 4: Substitute back into the integral
Now we can substitute back into the integral:

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

### Step 5: Evaluate the integral
Now we need to evaluate:

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

We will use integration by parts. Let:

- \( u = x^2 \) and \( dv = \cos x \, dx \).
- Then, \( du = 2x \, dx \) and \( v = \sin x \).

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we get:

\[ \int x^2 \cos x \, dx = x^2 \sin x \bigg|_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2x \sin x \, dx. \]

Evaluating \( x^2 \sin x \) at the limits:

\[ \left( \frac{\pi^2}{4} \cdot 1 \right) - (0) = \frac{\pi^2}{4}. \]

Now we need to evaluate \( \int_{0}^{\frac{\pi}{2}} 2x \sin x \, dx \) using integration by parts again:

Let \( u = 2x \) and \( dv = \sin x \, dx \):

- Then, \( du = 2 \, dx \) and \( v = -\cos x \).

Thus:

\[ \int 2x \sin x \, dx = -2x \cos x \bigg|_{0}^{\frac{\pi}{2}} + 2 \int_{0}^{\frac{\pi}{2}} \cos x \, dx. \]

Evaluating:

\[ -2 \left( \frac{\pi}{2} \cdot 0 - 0 \cdot 1 \right) + 2 \left( \sin x \bigg|_{0}^{\frac{\pi}{2}} \right) = 0 + 2(1 - 0) = 2. \]

Putting it all together:

\[ \int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx = \frac{\pi^2}{4} - 2. \]

### Step 6: Final value of \( I \)
Thus,

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

### Final Answer
The value of the integral is

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