设 $n\in\bbN^+$, 计算积分 $\dps{\int_0^{\pi/2} \cfrac{\sin nx}{\sin x}\rd x}.$
解答: (1) 由 $$\beex \bea 2\sin x\cdot \cfrac{1}{2}&=\sin x,\\ 2\sin x\cdot \cos 2x&=\sin 3x-\sin x,\\ 2\sin x\cdot \cos 4x&=\sin 5x-\sin 3x,\\ \cdots&=\cdots,\\ 2\sin x\cdot \cos 2nx&=\sin (2n+1)x-\sin(2n-1)x \eea \eeex$$ 知 $$\bex 2\sin x\sex{\cfrac{1}{2}+\sum_{k=1}^n \cos 2kx}=\sin (2n+1)x. \eex$$ 于是 $$\bex \int_0^{\pi/2}\cfrac{\sin (2n+1)x}{\sin x}\rd x =\int_0^{\pi/2} \sex{1+2\sum_{k=1}^n \cos 2kx}\rd x =\cfrac{\pi}{2}. \eex$$ (2) 由 $$\beex \bea 2\sin x\cos x&=\sin 2x,\\ 2\sin x\cos 3x&=\sin 4x-\sin 2x,\\ 2\sin x\cos 5x&=\sin 6x-\sin 4x,\\ \cdots&=\cdots,\\ 2\sin x\cos(2n-1)x&=\sin 2nx-\sin(2n-2)x \eea \eeex$$ 知 $$\bex 2\sin x\sum_{k=1}^n \cos (2k-1)x=\sin 2nx. \eex$$ 于是 $$\bex \int_0^{\pi/2} \cfrac{\sin 2nx}{\sin x}\rd x =2\int_0^{\pi/2} \sum_{k=1}^n \cos(2k-1)x\rd x =2\sum_{k=1}^n \cfrac{(-1)^{k-1}}{2k-1}. \eex$$
[再寄小读者之数学篇](2014-06-20 求积分),布布扣,bubuko.com
原文地址：http://www.cnblogs.com/zhangzujin/p/3798619.html
