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Today's calculation of Integral 883

Source: 2013 Fukushima University entrance exam

June 29, 2013
calculusintegrationtrigonometrylogarithmscalculus computationsIntegral inequality

Problem Statement

Prove that for each positive integer nn
4n2+14n2−1∫0π(ex−e−x)cos⁡2nx dx>eπ−e−π−24ln⁡(2n+1)2(2n−1)(n+3).\frac{4n^2+1}{4n^2-1}\int_0^{\pi} (e^{x}-e^{-x})\cos 2nx\ dx>\frac{e^{\pi}-e^{-\pi}-2}{4}\ln \frac{(2n+1)^2}{(2n-1)(n+3)}.
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