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Today's calculation of Integral 872

Source: 2013 Tsukuba University entrance exam

March 15, 2013
calculusintegrationtrigonometrygeometrylimitfunctioncalculus computations

Problem Statement

Let nn be a positive integer.
(1) For a positive integer kk such that 1≤k≤n1\leq k\leq n, Show that : ∫k−12nπk2nπsin⁡2ntcos⁡t dt=(−1)k+12n4n2−1(cos⁡k2nπ+cos⁡k−12nπ).\int_{\frac{k-1}{2n}\pi}^{\frac{k}{2n}\pi} \sin 2nt\cos t\ dt=(-1)^{k+1}\frac{2n}{4n^2-1}(\cos \frac{k}{2n}\pi +\cos \frac{k-1}{2n}\pi).
(2) Find the area SnS_n of the part expressed by a parameterized curve Cn:x=sin⁡t, y=sin⁡2nt (0≤t≤π).C_n: x=\sin t,\ y=\sin 2nt\ (0\leq t\leq \pi).
If necessary, you may use ∑k=1n−1cos⁡k2nπ=12(1tan⁡π4n−1) (n≥2).{\sum_{k=1}^{n-1} \cos \frac{k}{2n}\pi =\frac 12(\frac{1}{\tan \frac{\pi}{4n}}-1})\ (n\geq 2).
(3) Find lim⁡n→∞Sn.\lim_{n\to\infty} S_n.
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