If In = \(\int_0^{\pi/2}\) xn sin x dx, n > 1, then I+ n(n – 1)In – 2 is equal to :

1
n\((\frac{\pi}{2})^{n - 1}\)
2
\((\frac{\pi}{2})^{n - 1}\)
3
\((\frac{\pi}{2})\)
4
 n \((\frac{\pi}{2})^n\)

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