Question

The value of the integral \(\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x\)is :

• A. \(\frac{\pi^2}{12 \sqrt{3}}\)
• B. \(\frac{\pi^2}{6}\)
• C. \(\frac{\pi^2}{3 \sqrt{3}}\)
• D. \(\frac{\pi^2}{6 \sqrt{3}}\)

Hint:

When solving definite integrals involving symmetric functions, consider substitution techniques and symmetry properties to simplify calculations.

Answer 1

The Correct Option is D

Let: \[ I = \int_{-\pi/3}^{\pi/3} \frac{x + \pi/4}{2 - \cos 2x} \, dx. \tag{1} \] Using the substitution\(x \to -x\), the integral becomes: \[ I = \int_{-\pi/3}^{\pi/3} \frac{-x + \pi/4}{2 - \cos 2x} \, dx. \tag{2} \] Adding equations (1) and (2): \[ 2I = \int_{-\pi/3}^{\pi/3} \frac{\pi/2}{2 - \cos 2x} \, dx. \] Simplify: \[ I = \frac{\pi}{4} \int_{-\pi/3}^{\pi/3} \frac{1}{2 - \cos 2x} \, dx. \] Since \(\cos 2x\) is an even function, the integral can be written as: \[ I = \frac{\pi}{4} \cdot 2 \int_{0}^{\pi/3} \frac{1}{2 - \cos 2x} \, dx. \] \[ I = \frac{\pi}{2} \int_{0}^{\pi/3} \frac{1}{2 - \cos 2x} \, dx. \] Simplify the Integral: Using the trigonometric identity \(\cos 2x = \frac{1 - t^2}{1 + t^2}\), let \(t = \tan x\), so \(dt = \sec^2 x \, dx\). Then: \[ \cos 2x = \frac{1 - t^2}{1 + t^2}, \quad \sec^2 x \, dx = dt, \quad \text{and } t = 0 \text{ to } t = 1. \] Substituting: \[ I = \frac{\pi}{2} \int_{0}^{1} \frac{1 + t^2}{2(1 + t^2) - (1 - t^2)} \cdot \frac{dt}{1 + t^2}. \] \[ I = \frac{\pi}{2} \int_{0}^{1} \frac{1}{3t^2 + 1} \, dt. \] Let \(u = \sqrt{3}t, so  \ du = \sqrt{3} \, dt\). The limits change as \(t = 0 \to u = 0\)and \(t = 1 \to u = \sqrt{3}\) The integral becomes: \[ I = \frac{\pi}{2} \cdot \frac{1}{\sqrt{3}} \int_{0}^{\sqrt{3}} \frac{1}{u^2 + 1} \, du. \] \[ I = \frac{\pi}{2\sqrt{3}} \left[ \tan^{-1}(u) \right]_0^{\sqrt{3}}. \] \[ I = \frac{\pi}{2\sqrt{3}} \left[ \tan^{-1}(\sqrt{3}) - \tan^{-1}(0) \right]. \] \[ I = \frac{\pi}{2\sqrt{3}} \cdot \frac{\pi}{3}. \] \[ I = \frac{\pi^2}{6\sqrt{3}}. \]  Conclusion: The value of the integral is \(\frac{\pi^2}{6\sqrt{3}}\)(Option 4).