Question

Evaluate the integral \[ \int_{0}^{\pi} \frac{x \cos x \sin x}{\cos^3 x + \cos x}\,dx \]

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

Hint:

For definite integrals involving \(x\) and trigonometric functions over symmetric limits, always try the substitution \(x \to a - x\) to simplify.

Answer 1

The Correct Option is B

Step 1: Simplify the integrand.
The denominator can be factorized as \[ \cos^3 x + \cos x = \cos x(\cos^2 x + 1) \] Thus, the integrand becomes \[ \frac{x \cos x \sin x}{\cos x(\cos^2 x + 1)} = \frac{x \sin x}{1 + \cos^2 x} \] Step 2: Use symmetry property.
Let \[ I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x}\,dx \] Replace \(x\) by \(\pi - x\): \[ I = \int_{0}^{\pi} \frac{(\pi - x)\sin x}{1 + \cos^2 x}\,dx \] Step 3: Add the two integrals.
\[ 2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x}\,dx \] Step 4: Evaluate the remaining integral.
Let \(u = \cos x\), then \(du = -\sin x\,dx\). \[ \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x}\,dx = \int_{1}^{-1} \frac{-du}{1 + u^2} = \int_{-1}^{1} \frac{du}{1 + u^2} = \left[\tan^{-1}u\right]_{-1}^{1} = \frac{\pi}{2} \] Step 5: Final calculation.
\[ 2I = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2} \Rightarrow I = \frac{\pi^2}{4} \]