Evaluate: `int_(-pi//2)^(pi//2)(x^2cosx)/(1+e^x)dx`

Options

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

Explanation

To evaluate the integral

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

we will use the technique of symmetry and substitution.

### Step 1: Define the integral
Let

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1 + e^x} \, dx. \]

### Step 2: Substitute \(x\) with \(-x\)
Now, we will evaluate \(I\) by substituting \(x\) with \(-x\):

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

Since \((-x)^2 = x^2\) and \(\cos(-x) = \cos x\), we can rewrite the integral as:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x}{1 + \frac{1}{e^x}} \, dx. \]

### Step 3: Simplify the denominator
The denominator can be simplified:

\[ 1 + \frac{1}{e^x} = \frac{e^x + 1}{e^x}. \]

Thus, we have:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2 \cos x \cdot e^x}{e^x + 1} \, dx. \]

### Step 4: Combine the two expressions for \(I\)
Now we have two expressions for \(I\):

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

Adding these two expressions gives:

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

### Step 5: Solve for \(I\)
Thus, we can express \(I\) as:

\[ I = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx. \]

### Step 6: Evaluate the integral \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx\)
To evaluate this integral, we can use integration by parts. Let:

- \(u = x^2\) and \(dv = \cos x \, dx\)

Then, we have:

- \(du = 2x \, dx\)
- \(v = \sin x\)

Using integration by parts:

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

we get:

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

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

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

Now, we need to evaluate \(\int 2x \sin x \, dx\) using integration by parts again.

Let:

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

Then:

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

So,

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

Evaluating \(-2x \cos x\) at the limits gives \(0\) (since \(\cos(\pm \frac{\pi}{2}) = 0\)).

Now,

\[ \int 2 \cos x \, dx = 2 \sin x \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 2(1 - (-1)) = 4. \]

Putting it all together:

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

### Step 7: Final calculation for \(I\)
Thus,

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

### Final Answer
The value of the integral is:

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