If $A_n = \int_{0}^{\pi /2} \frac{\sin((2n-1)x)}{\sin x} dx$ and $B_n = \int_{0}^{\pi /2} \left( \frac{\sin(nx)}{\sin x} \right)^2 dx$ for $n \in N$,then:

  • A
    $A_{n+1} = A_n$
  • B
    $B_{n+1} - B_n = A_{n+1}$
  • C
    $A_{n+1} - A_n = B_{n+1}$
  • D
    Both $(A)$ and $(B)$

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