Question

Evaluate \( \int_{-\pi}^{\pi} \frac{\cos^{2022}x}{1 + (2022)^x} \, dx \).

• A. \( \frac{(2022)!}{2^{2022}((1011)!)^2} \pi \)
• B. \( \binom{2022}{1011} \pi \)
• C. \( \binom{2022}{1011} \frac{\pi}{2^{1011}} \)
• D. \( \frac{(2022)!}{(1011)! \cdot 2^{2022}} \pi \)

Hint:

For integrals of the form \( \int_{-\pi}^{\pi} f(x) dx \), check if the integrand is even or odd. Even functions simplify the computation to: \[ \int_{-\pi}^{\pi} f(x) dx = 2 \int_{0}^{\pi} f(x) dx \] Use symmetry and standard binomial integral identities.

Answer 1

The Correct Option is A

This is a definite integral of the form: \[ \int_{-\pi}^{\pi} \frac{f(x)}{g(x)} \, dx \] We note that: - The function \( \cos^{2022}x \) is even - The denominator \( 1 + (2022)^x \) is neither even nor odd But the whole integrand is even, because: \[ f(-x) = \frac{\cos^{2022}(-x)}{1 + (2022)^{-x}} = \frac{\cos^{2022}x}{1 + \frac{1}{(2022)^x}} = \frac{\cos^{2022}x \cdot (2022)^x}{(2022)^x + 1} \] Use standard results of definite integrals of even functions over symmetric intervals. Given the final answer evaluates to: \[ \frac{(2022)!}{2^{2022}((1011)!)^2} \pi \]