Question

The value of \( \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x} dx \) is:

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

Hint:

For definite integrals from 0 to \(a\) with an \(x f(\sin x, \cos^2 x)\) structure, always try the property \( \int_0^a g(x)dx = \int_0^a g(a-x)dx \). It often helps to eliminate the \(x\) factor, leaving a simpler integral to solve, as demonstrated in this problem.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a definite integral involving trigonometric functions. A standard technique for integrals of the form \( \int_0^a x f(x) dx \) where \(f(a-x)=f(x)\) is to use the property \( \int_0^a x g(x) dx = \frac{a}{2} \int_0^a g(x) dx \).

Step 2: Key Formula or Approach:
Let the integral be \( I \). \[ I = \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x} dx \] We use the property \( \int_0^a h(x) dx = \int_0^a h(a-x) dx \) with \( a=\pi \): \[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1+\cos^2(\pi - x)} dx \] Since \( \sin(\pi - x) = \sin x \) and \( \cos(\pi - x) = -\cos x \), we have \( \cos^2(\pi - x) = (-\cos x)^2 = \cos^2 x \). \[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1+\cos^2 x} dx = \int_{0}^{\pi} \frac{\pi \sin x}{1+\cos^2 x} dx - \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x} dx \] The second term is just \(I\). \[ I = \pi \int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x} dx - I \]
Step 3: Detailed Explanation:
From the equation above, we get: \[ 2I = \pi \int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x} dx \] To evaluate this integral, let \( u = \cos x \). Then \( du = -\sin x dx \). The limits of integration change as well: When \( x=0 \), \( u = \cos 0 = 1 \). When \( x=\pi \), \( u = \cos \pi = -1 \). \[ 2I = \pi \int_{1}^{-1} \frac{-du}{1+u^2} = \pi \int_{-1}^{1} \frac{du}{1+u^2} \] \[ 2I = \pi [\tan^{-1}(u)]_{-1}^{1} = \pi (\tan^{-1}(1) - \tan^{-1}(-1)) \] \[ 2I = \pi \left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right) = \pi \left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \pi \left(\frac{2\pi}{4}\right) = \frac{\pi^2}{2} \] \[ I = \frac{1}{2} \left(\frac{\pi^2}{2}\right) = \frac{\pi^2}{4} \]
Step 4: Final Answer:
The value of the integral is \( \frac{\pi^2}{4} \).