每天一个小公式
点火公式(Wallis)
\[\begin{align*}
&I_n=\int_0^{\frac{\pi}{2}} sin^n\theta d\theta=\int_0^{\frac{\pi}{2}} cos^n\theta d\theta=
\left\{
\begin{aligned}
\frac{n-1}{n}·\frac{n-3}{n-2}·\cdots·\frac{3}{4}·\frac{1}{2}·\frac{\pi}{2},n为偶数\\
\frac{n-1}{n}·\frac{n-3}{n-2}·\cdots·\frac{2}{3},n为大于1的奇数\\
\end{aligned}
\right.\\
&prove.\\
&I_n=\int_0^{\frac{\pi}{2}} sin^n\theta d\theta\\
&=-\int_0^{\frac{\pi}{2}} sin^{n-1}\theta d(cos\theta)\\
&=[-cos\theta \cdot sin^{n-1}\theta]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}} cos\theta d(sin^{n-1}\theta)\\
&=0+(n-1)\int_0^{\frac{\pi}{2}} cos^2\theta (sin\theta)^{n-2} d\theta\\
&=(n-1)\int_0^{\frac{\pi}{2}} (1-sin^2\theta) (sin\theta)^{n-2} d\theta\\
&=(n-1)I_{n-2}-(n-1)I_n\\
&i.e. n\cdot I_n=(n-1)I_{n-2}\\
&i.e.\frac{I_n}{I_{n-2}}=\frac{n-1}{n}\\
&由I_0=\int_0^{\frac{\pi}{2}} d\theta=\frac{\pi}{2}，I_1=\int_0^{\frac{\pi}{2}} sin\theta d\theta=1\\
&得证，I_n=\int_0^{\frac{\pi}{2}} sin^n\theta d\theta=
\left\{
\begin{aligned}
\frac{n-1}{n}·\frac{n-3}{n-2}·\cdots·\frac{3}{4}·\frac{1}{2}·\frac{\pi}{2},n为偶数\\
\frac{n-1}{n}·\frac{n-3}{n-2}·\cdots·\frac{2}{3},n为大于1的奇数\\
\end{aligned}
\right.\\
&prove:若f(x)在[0,1]上连续，\int_0^{\frac{\pi}{2}} f(sin\theta) d\theta=\int_0^{\frac{\pi}{2}} f(cos\theta) d\theta\\
&令\theta=\frac{\pi}{2}-t,d\theta=-dt\\
&于是，\int_0^{\frac{\pi}{2}} f(sin\theta) d\theta=-\int^0_{\frac{\pi}{2}} f(sin(\frac{\pi}{2}-t)) dt\\
&=\int_0^{\frac{\pi}{2}} f(cost) dt=\int_0^{\frac{\pi}{2}} f(cosx) dx
\end{align*}
\]