Question

The integral \[ \int_0^\pi \frac{8x}{4\cos^2 x + \sin^2 x} \, dx \text{ is equal to:} \]

• A. \( 2\pi^2 \)
• B. \( \pi^2 \)
• C. \( \frac{3\pi^2}{2} \)
• D. \( 4\pi^2 \)

Hint:

When solving integrals with trigonometric expressions in the denominator, try simplifying or using known integration techniques like substitution or symmetry.

Answer 1

The Correct Option is C

The integral to be evaluated is:\[I = \int_0^\pi \frac{8x}{4\cos^2 x + \sin^2 x} \, dx\]The denominator is simplified as:\[4\cos^2 x + \sin^2 x = 4\left( \cos^2 x \right) + \left( \sin^2 x \right)\]Integration is performed using substitution, yielding:\[I = \frac{3\pi^2}{2}\]The value of the integral is \( \frac{3\pi^2}{2} \).