If $A_n = \int_0^{\pi/2} \frac{\sin(2n-1)x}{\sin x} dx$; $B_n = \int_0^{\pi/2} \left(\frac{\sin nx}{\sin x}\right)^2 dx$, for $n \in \mathbb{N}$, then:
Step-by-Step Solution
Key Concept: Product-to-sum trigonometric identities simplify difference integrals and reveal recurrence patterns.
For the integrals $A_n = \int_0^{\pi/2} \frac{\sin(2n+1)x - \sin(2n-1)x}{\sin x}\,dx$, using the identity $\sin(2n+1)x - \sin(2n-1)x = 2\cos(2nx)\sin x$, we get $A_{n+1} - A_n = \int_0^{\pi/2} 2\cos(2nx)\,dx = 0$. Similarly, $B_{n+1} - B_n = A_{n+1}$, establishing the recurrence relation.
Correct Answer: 1,4