In=∫0∞e−x(sin x)ndx(n∈N,>1), then

In=(e−x(sinx)n−1)0∞+n∫0∞(sinx)n−1cosxe−xdxIn=n∫0∞(−(sinx)n+(n−1)(1−sin2x)(sinx)n−2)e−xdx=n(n−1)n2+1In−2Hence, I10I8=90101

In=(e−x(sinx)n−1)0∞+n∫0∞(sinx)n−1cosxe−xdxIn=n∫0∞(−(sinx)n+(n−1)(1−sin2x)(sinx)n−2)e−xdx=n(n−1)n2+1In−2Hence, I10I8=90101

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$I_{n} = \int_{0}^{\infty} e^{- x} {(sin x)}^{n} d x (n \in N, > 1), t h e n$

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$82 \frac{I_{10}}{I_{9}} = 90$

$I_{n} = \frac{n (n - 1)}{{(n - 1)}^{2} + 1} I_{n - 1}$

$I_{n} = \frac{n (n - 1)}{n^{2} + 1} I_{n - 2}$

$101 \frac{I_{10}}{I_{8}} = 90$

answer is A.

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$I_{n} = {(\frac{e^{- x} {(sin x)}^{n}}{- 1})}_{0}^{\infty} + n \int_{0}^{\infty} {(sin x)}^{n - 1} cos x e^{- x} d x$

$I_{n} = n \int_{0}^{\infty} (- {(sin x)}^{n} + (n - 1) (1 - {sin}^{2} x) {(sin x)}^{n - 2}) e^{- x} d x$ $= \frac{n (n - 1)}{n^{2} + 1} I_{n - 2}$

$H e n c e, \frac{I_{10}}{I_{8}} = \frac{90}{101}$

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