The value of `int_(-pi//2)^(pi//2)(x^(2)cosx)/(1+e^(x))` d x is equal to

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 solve the integral

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

we can use the property of definite integrals and the substitution \( x \to -x \).

### Step 1: Substitute \( x \) with \( -x \)

We start 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 2: Simplify the integral

We can simplify the denominator:

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

Thus, we have:

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

### Step 3: 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 equations gives:

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

### Step 4: Simplify the expression inside the integral

Notice that:

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

Thus, we have:

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

### Step 5: Evaluate the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx \)

Now we need to evaluate:

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

Using 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:

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

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

At \( x = \frac{\pi}{2} \):
\[ \left( \frac{\pi}{2} \right)^2 \cdot 1 = \frac{\pi^2}{4}. \]

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

Thus,

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

### Step 6: Evaluate the remaining integral

Now we need to evaluate:

\[ -2 \int \sin x \cdot x \, dx. \]

Using integration by parts again:

Let \( u = x \) and \( dv = \sin x \, dx \), then \( du = dx \) and \( v = -\cos x \):

\[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x. \]

Evaluating this from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):

At \( x = \frac{\pi}{2} \):
\[ -\frac{\pi}{2} \cdot 0 + 1 = 1. \]

At \( x = -\frac{\pi}{2} \):
\[ -\left(-\frac{\pi}{2}\right) \cdot 0 - 1 = -1. \]

Thus,

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \sin x \, dx = 1 - (-1) = 2. \]

### Step 7: Combine results

Putting it all together:

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

So,

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

### Final Result

Thus, the value of the integral is:

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