Question

Evaluate the integral: \[ \int_0^{\pi} \frac{x \sin x}{\sin^2 x + 2\cos^2 x} \, dx \]

• A. \(\dfrac{\pi}{2}\)
• B. \(\dfrac{\pi^2}{2}\)
• C. \(\dfrac{\pi^2}{4}\)
• D. \(\dfrac{\pi}{4}\)

Hint:

For integrals involving expressions like \( f(x) \) and \( f(a - x) \), try using the symmetry property: \[ \int_0^a f(x) dx = \int_0^a f(a - x) dx \] It often simplifies the integration.

Answer 1

The Correct Option is C

Step 1: Use the property of definite integrals For any continuous function \( f(x) \), \[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \] Let \[ I = \int_0^{\pi} \frac{x \sin x}{\sin^2 x + 2\cos^2 x} \, dx \] Using the property: \[ I = \int_0^{\pi} \frac{(\pi - x) \sin(\pi - x)}{\sin^2(\pi - x) + 2\cos^2(\pi - x)} \, dx \] Recall: - \( \sin(\pi - x) = \sin x \) - \( \cos(\pi - x) = -\cos x \) - So, \( \sin^2(\pi - x) = \sin^2 x \), \( \cos^2(\pi - x) = \cos^2 x \) Then, \[ I = \int_0^{\pi} \frac{(\pi - x) \sin x}{\sin^2 x + 2\cos^2 x} \, dx \] Step 2: Add the two forms Add both representations of \( I \): \[ 2I = \int_0^{\pi} \frac{x \sin x + (\pi - x) \sin x}{\sin^2 x + 2\cos^2 x} \, dx = \int_0^{\pi} \frac{\pi \sin x}{\sin^2 x + 2\cos^2 x} \, dx \] \[ I = \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{\sin^2 x + 2\cos^2 x} \, dx \] Step 3: Simplify the denominator Note: \[ \sin^2 x + 2\cos^2 x = 1 - \cos^2 x + 2\cos^2 x = 1 + \cos^2 x \] So the integral becomes: \[ I = \frac{\pi}{2} \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx \] Let \( u = \cos x \), then \( du = -\sin x \, dx \) Limits change: - When \( x = 0 \Rightarrow u = \cos 0 = 1 \) - When \( x = \pi \Rightarrow u = \cos \pi = -1 \) So, \[ I = \frac{\pi}{2} \int_1^{-1} \frac{-1}{1 + u^2} \, du = \frac{\pi}{2} \int_{-1}^{1} \frac{1}{1 + u^2} \, du \] \[ = \frac{\pi}{2} \left[ \tan^{-1}(u) \right]_{-1}^{1} = \frac{\pi}{2} \left( \tan^{-1}(1) - \tan^{-1}(-1) \right) = \frac{\pi}{2} \left( \frac{\pi}{4} + \frac{\pi}{4} \right) = \frac{\pi^2}{4} \]