Definite Integrals

This problem tests the power-reduction identity together with the symmetry property that the integral of sine-squared equals the integral of cosine-squared over the quarter period, written as \int_0^{\pi/2} \sin^2 x,dx = \int_0^{\pi/2} \cos^2 x,dx. The core idea is that a squared sine has no elementary antiderivative until we lower its power, so we rewrite \sin^2 x = (1 - \cos 2x)/2. The integral then becomes \frac{1}{2}\int_0^{\pi/2}(1 - \cos 2x),dx = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_0^{\pi/2}. Substituting the limits gives \frac{1}{2}\left[(\pi/2 - 0) - (0 - 0)\right] = \pi/4, since \sin \pi and \sin 0 both vanish. The symmetry confirmation follows because adding the sine and cosine integrals yields \int_0^{\pi/2} 1,dx = \pi/2, and the reflection substitution x \to \pi/2 - x interchanges sine and cosine, forcing each integral to equal \pi/4. Option \pi/2 is the sum of both integrals, not a single one. Option 1 ignores the angular scaling and the averaging factor entirely. Option \pi mistakes the full-period result for the quarter-period value. As a final plausibility check, the answer is positive and strictly less than \pi/2, which is exactly what we expect when averaging a bounded squared sine over a quarter period.