The value of `int_(-pi//2)^(pi//2) (x^(2)cos x )/(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 can use the property of definite integrals that states:

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

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

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 have:

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

### Step 2: Simplify the denominator

Now, we simplify the denominator \( 1 + e^{-x} \):

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

Thus, we can rewrite the integral:

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

### 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:

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

### Step 4: Factor out the common term

Factoring out \( x^2 \cos x \):

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

### Step 5: Evaluate the integral

Since \( x^2 \cos x \) is an even function, we can simplify the integral:

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

Thus,

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

### Step 6: Integration by parts

Using integration by parts, let:

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

Applying integration by parts:

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

we have:

\[ \int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx. \]

### Step 7: Apply integration by parts again

Now we need to integrate \( \int 2x \sin x \, dx \) again by parts:

Let:

- \( u = 2x \) → \( du = 2 \, dx \)
- \( dv = \sin x \, dx \) → \( v = -\cos x \)

Thus,

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

### Step 8: Combine results

Putting it all together:

\[ \int x^2 \cos x \, dx = x^2 \sin x + 2x \cos x - 2 \sin x. \]

Evaluating from \( 0 \) to \( \frac{\pi}{2} \):

At \( x = \frac{\pi}{2} \):

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

At \( x = 0 \):

\[ 0. \]

Thus,

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

### Final Answer

Therefore, the value of the integral is:

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