计算积分 $\int_0^{\frac{\pi}{2}} \frac{\sin \frac{2 n+1}{2} x}{\sin \frac{x}{2}} \mathrm{~d} x$ ,其中 $n$ 为正整数.
【答案】 【参考解答】由三角函数变换公式,有
$$
\begin{aligned}
\sum_{k=1}^n \cos 2 k x & =\frac{\sin 4 x+\ldots+\sin (2 n+2) x-\sin (2 n-2) x}{2 \sin 2 x} \\
& =\frac{\sin (2 n+2) x+\sin 2 n x-\sin 2 x}{2 \sin 2 x} \\
& =\frac{2 \sin (2 n+1) x \cos x+\sin 2 x}{4 \sin x \cos x} \\
& =\frac{\sin (2 n+1) x}{2 \sin x}-\frac{1}{2}
\end{aligned}
$$
所以
$$
\begin{gathered}
\int_0^\pi \frac{\sin (2 n+1) x}{\sin x} \mathrm{~d} x=2 \int_0^\pi\left(\frac{1}{2}+\sum_{k=1}^n \cos 2 k x\right) d x \\
=\pi+2 \sum_{k=1}^n \int_0^\pi(\cos 2 k x) d x=\pi
\end{gathered}
$$


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