If $ I_{n}=\int_{0}^{\frac{\pi}{2}} x^{n} \sin x d x $, where $ n1 $, then

Question

If $ I_{n}=\int_{0}^{\frac{\pi}{2}} x^{n} \sin x d x $, where $ n>1 $, then, $ I_{n}+n(n-1) I_{n-2}=n\left(\frac{\pi}{2}\right)^{n} $, $ I_{n}+n(n-1) I_{n-2}=n\left(\frac{\pi}{2}\right)^{n-1} $, $ I_{n}-n(n-1) I_{n-2}=n\left(\frac{\pi}{2}\right)^{n} $, $ I_{n}-n(n-1) I_{n-2}=-n\left(\frac{\pi}{2}\right)^{n} $

MARKS App · Accepted Answer

We have, I n = ∫ 0 π 2 x n sin x d x ⇒ I n = - x n cos x 0 π 2 + ∫ 0 π 2 n x n - 1 cos x d x ⇒ I n = n ∫ 0 π 2 x n - 1 cos x d x ⇒ I n = n x n - 1 sin x 0 π 2 - ∫ 0 π 2 n - 1 x n - 2 sin x d x ⇒ I n = n π 2 n - 1 - n - 1 I n - 2 Hence, I n + n n - 1 I n - 2 = n π 2 n - 1

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