T18_sqrt_1-x_2_三角换元

T18 $\sqrt{1-x^2}$ 三角换元

三角函数递推变体 华里士公式

$$\begin{array}{rl}{b}_n& ={\int }_0^{\frac{\pi }{2}}{\sin }^{n}x\mathrm{d}x\\ a_n& ={\int }_0^1{x}^n\sqrt{1-x^2}\mathrm{d}x\\ & ={\int }_0^{\frac{\pi }{2}}{\sin }^{n}x-{\sin }^{n+2}\mathrm{d}x\\ & =b_n-b_{n+2}\\ & =(1-\frac{n-1}{n})b_n\end{array}$$