Question

Evaluate : \( \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{2^x + 1} \, dx \)

Hint:

An integrand containing a factor of \( \frac{1}{a^x + 1} \) alongside symmetric bounds \( [-L, L] \) is a classic indicator that applying King's property will eliminate the exponential tracking variable entirely.

Answer 1

Concept: This definite integral can be resolved using King's Property of definite integrals, which states that: \[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

Step 1: Write down the primary integral expression.
Let our initial given integral be represented as equation (1): \[ I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{2^x + 1} \, dx \quad \text{--- (1)} \]

Step 2: Apply King's Property to generate a companion equation.
Here, lower bound \( a = -\frac{\pi}{2} \) and upper bound \( b = \frac{\pi}{2} \). The variable substitute parameter is: \[ a + b - x = -\frac{\pi}{2} + \frac{\pi}{2} - x = -x \] Replacing \( x \) with \( -x \) inside the integrand: \[ I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2(-x)}{2^{-x} + 1} \, dx \] We know that \( \cos(-x) = \cos x \), so \( \cos^2(-x) = \cos^2 x \). Also rewrite the exponential index: \[ 2^{-x} = \frac{1}{2^x} \] Substituting these simplifications back into the definite integral: \[ I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{\frac{1}{2^x} + 1} \, dx = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{\frac{1 + 2^x}{2^x}} \, dx = \int_{-\pi/2}^{\pi/2} \frac{2^x \cos^2 x}{2^x + 1} \, dx \quad \text{--- (2)} \]

Step 3: Add equations (1) and (2) together.
Since both equations have matching limits of integration, we can merge their integrands: \[ 2I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{2^x + 1} \, dx + \int_{-\pi/2}^{\pi/2} \frac{2^x \cos^2 x}{2^x + 1} \, dx \] \[ 2I = \int_{-\pi/2}^{\pi/2} \frac{\cos^2 x (1 + 2^x)}{2^x + 1} \, dx \] The matching factor \( (2^x + 1) \) cancels out cleanly from the numerator and denominator: \[ 2I = \int_{-\pi/2}^{\pi/2} \cos^2 x \, dx \]

Step 4: Integrate the simplified function using even/odd properties.
Since \( \cos^2 x \) is an even function (\( f(-x) = f(x) \)), we can change the integration baseline using the property \( \int_{-a}^{a} f(x)dx = 2\int_{0}^{a} f(x)dx \): \[ 2I = 2 \int_{0}^{\pi/2} \cos^2 x \, dx \quad \Rightarrow \quad I = \int_{0}^{\pi/2} \cos^2 x \, dx \] Using the trigonometric identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \): \[ I = \int_{0}^{\pi/2} \left( \frac{1 + \cos 2x}{2} \right) dx = \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right]_{0}^{\pi/2} \]

Step 5: Compute boundaries to isolate the final numeric answer.
Evaluate at the upper limit \( \frac{\pi}{2} \) and lower limit 0: \[ I = \frac{1}{2} \left[ \left( \frac{\pi}{2} + \frac{\sin(\pi)}{2} \right) - \left( 0 + \frac{\sin(0)}{2} \right) \right] \] Since \( \sin(\pi) = 0 \) and \( \sin(0) = 0 \): \[ I = \frac{1}{2} \left[ \frac{\pi}{2} \right] = \frac{\pi}{4} \]