The value of `int_(-(pi)/(2))^(pi/2)(x^(2)cosx)/(1+e^(x))dx` 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)`

Answer

To solve the integral

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

we will use the property of definite integrals and a substitution.

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

We first consider the integral with the substitution \( x \to -x \):

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

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

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

This simplifies to:

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

### Step 2: Combine the two integrals

Now we have two expressions for \( I \):

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

Now, let's add these two equations:

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

### Step 3: Simplify the integrand

The integrand can be simplified:

\[ \frac{x^2 \cos x}{1 + e^x} + \frac{x^2 \cos x \cdot e^x}{e^x + 1} = x^2 \cos x \left( \frac{1 + e^x}{1 + e^x} \right) = x^2 \cos x. \]

Thus, we have:

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

### Step 4: Calculate the integral

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 have:

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

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

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

Now we need to compute \( -2 \int x \sin x \, dx \).

### Step 5: Another integration by parts

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

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

Thus,

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

Evaluating \( -x \cos x \) at the limits gives:

\[ -\left( \frac{\pi}{2} \cdot 0 - \left(-\frac{\pi}{2} \cdot 0\right)\right) = 0, \]

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

Thus,

\[ \int x \sin x \, dx = 0 + 2 = 2. \]

### Step 6: Putting it all together

Now we can substitute back:

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

Thus,

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

### Final Answer

The value of the integral is:

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