Question

If \[ I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 \frac{3}{2}x}{\sin^2 x + \cos^2 x} \, dx, \] then \[ \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx \] equals:

• A. \( \frac{\pi^2}{16} \)
• B. \( \frac{\pi^2}{4} \)
• C. \( \frac{\pi^2}{8} \)
• D. \( \frac{\pi^2}{12} \)

Hint:

When solving integrals involving trigonometric functions, use identities to simplify and check for symmetry between integrals to find relationships between them.

Answer 1

The Correct Option is A

We are given the first integral: \[ I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 \frac{3}{2}x}{\sin^2 x + \cos^2 x} \, dx. \] Since \( \sin^2 x + \cos^2 x = 1 \), the integral becomes: \[ I = \int_0^{\frac{\pi}{2}} \sin^2 \frac{3}{2}x \, dx. \] Using standard trigonometric identities, we solve the integral and obtain: \[ I = \frac{\pi^2}{16}. \] Now, for the second integral: \[ \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx, \] using symmetry and the known result from the first integral, we conclude: \[ \int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx = \frac{\pi^2}{16}. \]