按牛顿二项式展开及代换 $x=\sin t$ 两种方法计算积分 $\dps{\int_0^1 (1-x^2)^n\rd x}$ ($n$ 为正整数). 并由此说明: $$\bex \sum_{k=0}^n C_n^k(-1)^k \frac{1}{2k+1}=\frac{(2n)!!}{(2n+1)!!}. \eex$$
证明: $$\beex \bea \int_0^1 (1-x^2)^n\rd x &=\int_0^1 \sum_{k=0}^n C_n^k (-x^2)^k\rd x =\sum_{k=0}^n C_n^k(-1)^k \frac{1}{2k+1},\\ \int_0^1 (1-x^2)^n\rd x &=\int_0^\frac{\pi}{2} \cos^{2n}t \cdot \cos t\rd t\quad\sex{x=\sin t}\\ &\equiv I_{2n+1}, \eea \eeex$$ 而由 $$\beex \bea I_{2n+1}&=\int_0^\frac{\pi}{2} \cos^{2n-1}t\cdot (1-\sin^2t)\rd t\\ &=I_{2n-1}+\frac{1}{2n}\int_0^\frac{\pi}{2} \sin t\rd \cos^{2n}t\\ &=I_{2n-1}-\frac{1}{2n}\int_0^\frac{\pi}{2}\cos^{2n+1}t\rd t\quad\sex{\mbox{分部积分}} \eea \eeex$$ 知 $$\bex I_{2n+1}=\frac{2n}{2n+1}I_{2n-1}=\cdots=\frac{(2n)!!}{(2n+1)!!}I_1=\frac{(2n)!!}{(2n+1)!!}. \eex$$
