<p>Evaluate <span>\(\int_{0}^{\pi}\frac{\cos 2x}{\cos^2 x + 1}\,dx\)</span></p>
Step-by-Step Solution
Key Concept: Use the property of even/odd functions or split the integral based on the behavior of trigonometric functions.
<p><strong>Solution:</strong> Using properties of definite integrals and the fact that $\cos 2x = \cos^2 x - \sin^2 x$:</p><p>$\int_{0}^{\pi}\cos 2x\,dx = \int_{0}^{\pi/2}\cos x\,dx - \int_{\pi/2}^{\pi}\cos x\,dx = 2[1-\cos\pi] = 2(1-(-1)) = 0$</p><p>∴ Answer is 0</p>
Correct Answer: 0