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

Options

A. 0
B. `(pi^(2))/(2)-4`
C. `(pi^(2))/(2)-6`
D. `(pi^(2))/(2)`

Explanation

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 can break it into two parts:

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

### Step 1: Evaluate the first integral

Let

\[ I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx. \]

Since \(x^2\) is an even function and \(\cos x\) is also an even function, their product \(x^2 \cos x\) is even. Therefore, we can simplify the integral:

\[ I_1 = 2 \int_0^{\frac{\pi}{2}} x^2 \cos x \, dx. \]

### Step 2: Evaluate the second integral

Now consider

\[ I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log\left(\frac{\pi + x}{\pi - x}\right) \cos x \, dx. \]

To determine whether this integral is odd or even, we can evaluate \(I_2\) at \(-x\):

\[ I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log\left(\frac{\pi - x}{\pi + x}\right) \cos(-x) \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log\left(\frac{\pi - x}{\pi + x}\right) \cos x \, dx. \]

Notice that

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

Thus, we have:

\[ I_2 = -I_2 \implies 2I_2 = 0 \implies I_2 = 0. \]

### Step 3: Combine the results

Now we can combine the results:

\[ I = I_1 + I_2 = I_1 + 0 = I_1. \]

### Step 4: Evaluate \(I_1\)

To evaluate \(I_1 = 2 \int_0^{\frac{\pi}{2}} x^2 \cos x \, dx\), we can use integration by parts. Let:

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

Applying 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 the boundary term:

\[ x^2 \sin x \bigg|_0^{\frac{\pi}{2}} = \left(\frac{\pi^2}{4} \cdot 1\right) - (0) = \frac{\pi^2}{4}. \]

Now we need to evaluate

\[ \int_0^{\frac{\pi}{2}} 2x \sin x \, dx. \]

Using integration by parts again with:

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

We have:

\[ \int 2x \sin x \, dx = -2x \cos x \bigg|_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} 2 \cos x \, dx. \]

Evaluating the boundary term:

\[ -2x \cos x \bigg|_0^{\frac{\pi}{2}} = 0 + 2(0) = 0. \]

Now we evaluate

\[ \int_0^{\frac{\pi}{2}} 2 \cos x \, dx = 2 \sin x \bigg|_0^{\frac{\pi}{2}} = 2(1 - 0) = 2. \]

Thus,

\[ \int_0^{\frac{\pi}{2}} 2x \sin x \, dx = 0 + 2 = 2. \]

### Final Calculation

So we have:

\[ I_1 = \frac{\pi^2}{4} - 2. \]

Thus,

\[ I = 2I_1 = 2\left(\frac{\pi^2}{4} - 2\right) = \frac{\pi^2}{2} - 4. \]

### Conclusion

The value of the integral is:

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