Source: Moldova National MO 2008, 12 Grade, Problem 2
Tags: integration, trigonometry, real analysis, real analysis unsolved
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Find the exact value of $ E=\displaystyle\int_0^{\frac\pi2}\cos^{1003}x\text{d}x\cdot\int_0^{\frac\pi2}\cos^{1004}x\text{d}x\cdot$.
$ E=\frac{\pi}{2008}$.
We have that
$ E=\frac{1}{2}\beta(502, 1/2)\frac{1}{2}\beta(1005/2,1/2)$
$ =\frac{1}{4}\frac{\Gamma(502)\Gamma(1/2)}{\Gamma(1005/2)}\frac{\Gamma(1005/2)\Gamma(1/2)}{\Gamma(503)}$
$ =\frac{\pi}{4\cdot 502}=\frac{\pi}{2008}.$
$ \int{\cos^{n}{x} dx}=\frac{\cos^{n-1}{x}\sin {x}}{n}+\frac{n-1}{n}\int{\cos^{n-2}{x} dx}$
So we have:
$ \int_{0}^{\frac{\pi}{2}}{\cos^{n}{x} dx}=\frac{n-1}{n}\int_{0}^{\frac{\pi}{2}}{\cos^{n-2}{x} dx}$
We have:
$ \int_{0}^{\frac{\pi}{2}}{\cos^{1003}{x} dx}=\frac{2\cdot 4\cdot 6\cdot ... \cdot 1002}{3\cdot 5\cdot 7\cdot ...\cdot 1003}\int_{0}^{\frac{\pi}{2}}{\cos{x} dx}=\frac{2\cdot 4\cdot 6\cdot ...\cdot 1002}{3\cdot 5\cdot 7\cdot ... \cdot 1003}$
$ \int_{0}^{\frac{\pi}{2}}{\cos^{1004}{x} dx}=\frac{3\cdot 5\cdot 7\cdot ... \cdot 1003}{2\cdot 4\cdot 5\cdot ...\cdot 1004}\int_{0}^{\frac{\pi}{2}}{\cos^{0}{x} dx}=\frac{3\cdot 5\cdot 7\cdot ... \cdot 1003}{2\cdot 4\cdot 5\cdot ...\cdot 1004}\cdot\frac{\pi}{2}$
We will have:
$ E=\frac{2\cdot 4\cdot 6\cdot ...\cdot 1002}{3\cdot 5\cdot 7\cdot ... \cdot 1003}\cdot \frac{3\cdot 5\cdot 7\cdot ... \cdot 1003}{2\cdot 4\cdot 5\cdot ...\cdot 1004}\cdot\frac{\pi}{2}=\frac{\pi}{2008}$