Question

The integral \( I = \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} \, dx \) is equal to:

• A. \( \frac{8\pi \sqrt{3}}{5} \)
• B. \( \frac{2\pi \sqrt{3}}{9} \)
• C. \( \frac{4\pi^2 \sqrt{3}}{9} \)
• D. \( \frac{\pi^2}{6\sqrt{3}} \)

Hint:

For definite integrals involving trigonometric functions, consider using symmetry properties and substitution techniques like \( t = \tan x \).

Answer 1

The Correct Option is D

Given,\[I = \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} \, dx\]Step 1: Split the integral\[I = \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x}{2 - \cos 2x} \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\pi}{4}}{2 - \cos 2x} \, dx\]Let\[I_1 = \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x}{2 - \cos 2x} \, dx\]and \[I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\pi}{4}}{2 - \cos 2x} \, dx.\]
Step 2: Evaluate \( I_1 \)Define \( f(x) = \frac{x}{2 - \cos 2x} \).\[f(-x) = \frac{-x}{2 - \cos 2(-x)} = -f(x).\]As \( f(x) \) is an odd function, \[I_1 = \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} f(x) \, dx = 0.\]
Step 3: Evaluate \( I_2 \)\[I_2 = 2 \times \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{2 - \cos 2x}.\]Using the identity,\[2 - \cos 2x = 1 + \tan^2 x,\]substitute \( t = \tan x \), so \( dt = \sec^2 x \, dx \):\[I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sec^2 x \, dx}{1 + \tan^2 x} = \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{dt}{1 + t^2}.\]Since \[\int \frac{dt}{1 + t^2} = \tan^{-1} t,\]calculate:\[I_2 = \tan^{-1} (\sqrt{3}) - \tan^{-1} (0) = \frac{\pi}{3}.\]
Step 4: Final Calculation\[I = \frac{\pi}{2\sqrt{3}} \left( \tan^{-1}(\sqrt{3}) - \tan^{-1}(0) \right) = \frac{\pi^2}{6\sqrt{3}}.\]The correct answer is \(\boxed{(d)}\).