If `I=int_(-pi)^(pi) (sin^(2))/(1+a^(x))dx, a gt 0`, then I equals

Options

A. `pi`
B. `(pi)/(2)`
C. `a pi`
D. `a(pi)/(2)`

Answer

To solve the integral \( I = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1 + a^x} \, dx \), where \( a > 0 \), we can use the property of definite integrals.

### Step 1: Use the Symmetry Property of Integrals
We know that:
\[ I = \int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(a + b - x) \, dx \]
In our case, we can express \( I \) as:
\[ I = \int_{-\pi}^{\pi} \frac{\sin^2(-x)}{1 + a^{-x}} \, dx \]
Since \( \sin(-x) = -\sin(x) \), we have \( \sin^2(-x) = \sin^2(x) \). Thus:
\[ I = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1 + \frac{1}{a^x}} \, dx \]

### Step 2: Simplify the Integral
We can rewrite \( \frac{1}{a^x} \) as \( \frac{1}{a^x} = a^{-x} \), hence:
\[ I = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1 + \frac{1}{a^x}} \, dx = \int_{-\pi}^{\pi} \frac{\sin^2 x \cdot a^x}{a^x + 1} \, dx \]

### Step 3: Add the Two Expressions for \( I \)
Now, we have two expressions for \( I \):
1. \( I = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1 + a^x} \, dx \) (Equation 1)
2. \( I = \int_{-\pi}^{\pi} \frac{a^x \sin^2 x}{1 + a^x} \, dx \) (Equation 2)

Adding these two equations gives:
\[ 2I = \int_{-\pi}^{\pi} \left( \frac{\sin^2 x}{1 + a^x} + \frac{a^x \sin^2 x}{1 + a^x} \right) \, dx \]
\[ 2I = \int_{-\pi}^{\pi} \frac{\sin^2 x (1 + a^x)}{1 + a^x} \, dx = \int_{-\pi}^{\pi} \sin^2 x \, dx \]

### Step 4: Evaluate the Integral of \( \sin^2 x \)
We know that:
\[ \int_{-\pi}^{\pi} \sin^2 x \, dx = 2 \int_{0}^{\pi} \sin^2 x \, dx \]
Using the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \):
\[ \int_{0}^{\pi} \sin^2 x \, dx = \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_{0}^{\pi} \]
Calculating this, we find:
\[ = \frac{1}{2} \left[ \pi - 0 \right] = \frac{\pi}{2} \]
Thus:
\[ \int_{-\pi}^{\pi} \sin^2 x \, dx = 2 \cdot \frac{\pi}{2} = \pi \]

### Step 5: Solve for \( I \)
Now substituting back, we have:
\[ 2I = \pi \implies I = \frac{\pi}{2} \]

### Final Answer
Therefore, the value of \( I \) is:
\[ \boxed{\frac{\pi}{2}} \]