2021-04-12 积分题

题目

求积分 ${\int }_0^{\frac{\pi }{2}}{\left(\frac{\sin nx}{\sin x}\right)}^2\mathrm{d}x$ .

题目已经明显暗示了，构造差分即可.

$$\begin{array}{rl}{I}_{n+1}-I_n& ={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^2\left(n+1\right)x-{\sin }^{2}nx}{{\sin }^{2}x}\mathrm{d}x\\ & ={\int }_0^{\frac{\pi }{2}}\frac{\sin \left(2n+1\right)x\sin x}{{\sin }^{2}x}\mathrm{d}x\\ & =\frac{1}{2}{\int }_0^{\pi }\frac{\sin \left(2n+1\right)x}{\sin x}\mathrm{d}x\end{array}$$

可得

$${\int }_0^{\pi }\frac{\sin (2n+1)x}{\sin x}\mathrm{d}x=\pi$$

故

$$I_n=I_1+\sum _{k=1}^{n-1}\left(I_{k+1}-I_k\right)=\frac{\pi }{2}+\frac{\left(n-1\right)\pi }{2}=\frac{n\pi }{2}$$