定积分的积分方法

说明

\begin{array}{r} \int_{a}^{b} f (x) d x = \int_{α}^{β} f [φ (t)] φ^{'} (t) d t \end{array}

注意换元要换积分限

\begin{aligned} \int_{0}^{π / 2} f (\sin x) d x & = - \int_{π / 2}^{0} f [\sin (\frac{π}{2} - t)] d t = \int_{0}^{π / 2} f (\cos t) d t \\ = \int_{0}^{π / 2} f (\cos x) d x \end{aligned}

\begin{aligned} \int_{0}^{π} x f (\sin x) d x & = - \int_{π}^{0} (π - t) f [\sin (π - t)] d t = \int_{0}^{π} (π - t) f (\sin t) d t \\ = π \int_{0}^{π} f (\sin x) d x - \int_{0}^{π} x f (\sin x) d x \\ = \frac{π}{2} \int_{0}^{π} f (\sin x) d x \end{aligned}

\begin{array}{r} \int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x \end{array}

\begin{array}{r} \int_{- a}^{a} f (x) d x = 0 \end{array}

\begin{array}{r} \int_{a}^{a + T} f (x) d x = \int_{0}^{T} f (x) d x \\ \int_{a}^{a + n T} f (x) d x = n \int_{0}^{T} f (x) d x \end{array}

\begin{array}{r} \int_{a}^{b} u d v = {[u v]}_{a}^{b} - \int_{a}^{b} v d u \end{array}

瓦里斯公式：点火公式

\begin{aligned} I_{n} & = \int_{0}^{π / 2} \sin^{n} x d x = \int_{0}^{π / 2} \cos^{n} x d x \\ = {\begin{cases} \frac{(n - 1)!!}{n!!} \cdot \frac{π}{2} & = \frac{n - 1}{n} \cdot \frac{n - 3}{n - 2} \dots \frac{1}{2} \cdot \frac{π}{2} n为正偶数 \\ \frac{(n - 1)!!}{n!!} & = \frac{n - 1}{n} \cdot \frac{n - 3}{n - 2} \dots \frac{4}{5} \cdot \frac{2}{3} n为正奇数 \end{cases} \end{aligned}

证明过程：

\begin{aligned} I_{n} & = \int_{0}^{π / 2} \sin^{n} x d x = - \int_{0}^{π / 2} \sin^{n - 1} x d (\cos x) \\ = {[- \cos x \sin^{n - 1} x]}_{0}^{π / 2} + (n - 1) \int_{0}^{π / 2} \sin^{n - 2} x \cos^{2} x d x \\ = 0 + (n - 1) \int_{0}^{π / 2} \sin^{n - 2} x d x - (n - 1) \int_{0}^{π / 2} \sin^{n} x d x \\ = (n - 1) I_{n - 2} - (n - 1) I_{n} \\ = \frac{n - 1}{n} I_{n - 2} \end{aligned}

{\begin{cases} I_{2 m} = \frac{2 m - 1}{2 m} \cdot \frac{2 m - 3}{2 m - 5} \dots \frac{3}{4} \cdot \frac{1}{2} I_{0} \\ I_{2 m + 1} = \frac{2 m}{2 m + 1} \cdot \frac{2 m - 2}{2 m - 1} \dots \frac{4}{5} \cdot \frac{2}{3} I_{1} \end{cases}

\begin{array}{r} I_{0} = \int_{0}^{π / 2} d x = \frac{π}{2} \end{array}

\begin{array}{r} I_{1} = \int_{0}^{π / 2} \sin x d x = 1 \end{array}