Question

The value of \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + 2^{-x}} \, dx \] is equal to:

• A. \( \frac{\pi}{3} \)
• B. \( \frac{\pi}{4} \)
• C. 1
• D. \( \frac{1}{2} \)
• E. \( \frac{\pi}{2} \)

Hint:

When faced with integrals involving symmetric limits, use substitution to exploit the symmetry and simplify the computation. In this case, symmetry allowed us to reduce the problem.

Answer 1

The Correct Option is B

We are asked to evaluate the following integral: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + 2^{-x}} \, dx. \] Step 1: First, let's check if we can use symmetry. We notice that the limits of integration are symmetric, from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).
Consider the substitution \( x = -t \), so that \( dx = -dt \). 
The limits of integration will change as follows: when \( x = -\frac{\pi}{2} \), \( t = \frac{\pi}{2} \), and when \( x = \frac{\pi}{2} \), \( t = -\frac{\pi}{2} \). The integral becomes:
\[ I = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{\cos^2(-t)}{1 + 2^{t}} (-dt). \] Since \( \cos^2(-t) = \cos^2(t) \), the integral simplifies to: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2 t}{1 + 2^{t}} \, dt. \] This is exactly the same as the original integral.
Step 2: Let's now add the original integral and the transformed integral. By symmetry, we have:
\[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{\cos^2 x}{1 + 2^{-x}} + \frac{\cos^2 x}{1 + 2^{x}} \right) dx. \] Notice that: \[ \frac{\cos^2 x}{1 + 2^{-x}} + \frac{\cos^2 x}{1 + 2^{x}} = \cos^2 x \left( \frac{1}{1 + 2^{-x}} + \frac{1}{1 + 2^{x}} \right). \] Step 3: Simplify the sum inside the parentheses: \[ \frac{1}{1 + 2^{-x}} + \frac{1}{1 + 2^{x}} = \frac{(1 + 2^x) + (1 + 2^{-x})}{(1 + 2^{-x})(1 + 2^x)} = \frac{2 + 2^x + 2^{-x}}{(1 + 2^{-x})(1 + 2^x)}. \] Using the identity \( 2^x + 2^{-x} = 2 \cosh(x \ln 2) \), we get: \[ \frac{2 + 2^x + 2^{-x}}{(1 + 2^{-x})(1 + 2^x)} = 2. \] Step 4: The integral becomes: \[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2 \cos^2 x \, dx. \] Now, use the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \), so the integral becomes: \[ 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( 1 + \cos 2x \right) \, dx. \] Step 5: Evaluate the integral: \[ 2I = \left[ x + \frac{\sin 2x}{2} \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}. \] The sine term vanishes at both limits, so we are left with: \[ 2I = \left( \frac{\pi}{2} - \left( -\frac{\pi}{2} \right) \right) = \pi. \] Thus: \[ I = \frac{\pi}{4}. \] Thus, the correct answer is option (B).