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

Options

A. `pi//4`
B. `pi//8`
C. `pi//2`
D. `4pi`

Explanation

To solve the integral

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

we will use the property of definite integrals and some substitutions.

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

We start by substituting \( x \) with \( -x \):

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

Since \( \sin(-x) = -\sin(x) \) and \( \sin^2(-x) = \sin^2(x) \), we have:

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

### Step 2: Simplify the denominator

The expression \( 1 + \frac{1}{2^x} \) can be rewritten as:

\[ 1 + \frac{1}{2^x} = \frac{2^x + 1}{2^x}. \]

Thus, we can rewrite the integral as:

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

### Step 3: Combine the two expressions for \( I \)

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

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

Let's denote the second integral as \( I_2 \):

\[ I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x \cdot 2^x}{2^x + 1} \, dx. \]

### Step 4: Add the two integrals

Now, we add the two integrals:

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

The expression simplifies to:

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

### Step 5: Evaluate the integral of \( \sin^2 x \)

Using the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \):

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

Now, we evaluate:

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

Calculating the limits:

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

\[ \frac{\frac{\pi}{2}}{2} - \frac{\sin(\pi)}{4} = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

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

\[ \frac{-\frac{\pi}{2}}{2} - \frac{\sin(-\pi)}{4} = -\frac{\pi}{4} - 0 = -\frac{\pi}{4}. \]

Thus,

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

### Step 6: Solve for \( I \)

Now substituting back:

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

### Final Answer

Thus, the value of the integral is:

\[ \boxed{\frac{\pi}{4}}. \]