313 lines
6.9 KiB
Markdown
313 lines
6.9 KiB
Markdown
## 笔记记录
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### 要点 05 - 三角函数积分
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#### 降幂公式
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$$
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\sin^2 x = \frac{1 - \cos 2x}{2}
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$$
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$$
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\cos^2 x = \frac{1 + \cos 2x}{2}
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$$
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$$
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\sin^3 x = \frac{3\sin x - \sin 3x}{4}
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$$
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$$
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\cos^3 x = \frac{3\cos x + \cos 3x}{4}
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$$
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#### 基本积分
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$$
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\int \sin x \, dx = -\cos x + C
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$$
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$$
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\int \cos x \, dx = \sin x + C
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$$
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$$
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\int \tan x \, dx = -\ln|\cos x| + C
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$$
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$$
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\int \cot x \, dx = \ln|\sin x| + C
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$$
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$$
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\int \sec x \, dx = \ln|\sec x + \tan x| + C
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$$
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$$
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\int \csc x \, dx = -\ln|\csc x + \cot x| + C
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$$
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$$
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\int \sec^2 x \, dx = \tan x + C
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$$
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$$
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\int \csc^2 x \, dx = -\cot x + C
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$$
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$$
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\int \sec x \tan x \, dx = \sec x + C
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$$
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$$
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\int \csc x \cot x \, dx = -\csc x + C
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$$
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#### 万能代换
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令 $t = \tan\frac{x}{2}$，则：
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$$
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\sin x = \frac{2t}{1 + t^2}, \quad \cos x = \frac{1 - t^2}{1 + t^2}, \quad dx = \frac{2 \, dt}{1 + t^2}
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$$
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适用类型：$R(\sin x, \cos x)$（有理函数形式）
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#### 常用结论
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$$
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\int \sin^n x \, dx = -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n}\int \sin^{n-2} x \, dx
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$$
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$$
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\int \cos^n x \, dx = \frac{\cos^{n-1} x \sin x}{n} + \frac{n-1}{n}\int \cos^{n-2} x \, dx
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$$
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#### sinⁿ 递推公式推导（分部积分法）
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设
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$$
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I_n = \int \sin^n x \, dx, \quad n \ge 2.
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$$
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取
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$$
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u = \sin^{n-1}x, \quad dv = \sin x \, dx,
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$$
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则
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$$
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du = (n-1)\sin^{n-2}x \cos x \, dx, \quad v = -\cos x.
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$$
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分部积分：
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$$
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\begin{aligned}
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I_n &= uv - \int v \, du \\
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&= -\sin^{n-1}x \cos x + (n-1)\int \sin^{n-2}x \cos^2 x \, dx.
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\end{aligned}
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$$
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利用 $\cos^2 x = 1 - \sin^2 x$：
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$$
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\int \sin^{n-2}x \cos^2 x \, dx = \int \sin^{n-2}x \, dx - \int \sin^n x \, dx = I_{n-2} - I_n.
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$$
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代入得
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$$
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I_n = -\sin^{n-1}x \cos x + (n-1)(I_{n-2} - I_n).
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$$
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整理含 $I_n$ 的项：
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$$
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\begin{aligned}
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I_n + (n-1)I_n &= -\sin^{n-1}x \cos x + (n-1)I_{n-2}, \\
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n I_n &= -\sin^{n-1}x \cos x + (n-1)I_{n-2}.
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\end{aligned}
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$$
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于是
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$$
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\boxed{I_n = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n} I_{n-2}},\quad n\ge 2.
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$$
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需要两个初始条件：
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- $I_0 = \int dx = x + C$
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- $I_1 = \int \sin x \, dx = -\cos x + C$
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#### cosⁿ 递推公式推导
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类似地，设 $J_n = \int \cos^n x \, dx$，取 $u = \cos^{n-1}x$，$dv = \cos x \, dx$，利用 $\sin^2 x = 1 - \cos^2 x$，可得：
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$$
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\boxed{J_n = \frac{\cos^{n-1}x \sin x}{n} + \frac{n-1}{n} J_{n-2}},\quad n\ge 2
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$$
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需要两个初始条件：
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- $J_0 = \int dx = x + C$
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- $J_1 = \int \cos x \, dx = \sin x + C$
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#### 点火公式（Wallis 公式）
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利用 $\sin^n$ 递推公式在 $[0, \frac{\pi}{2}]$ 上积分，边界项为零：
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$$
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J_n = \int_0^{\pi/2} \sin^n x \, dx = \frac{n-1}{n} J_{n-2}
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$$
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**递推过程**：
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$$
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\begin{aligned}
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J_n &= \int_0^{\pi/2} \sin^n x \, dx
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= \left[-\frac{\sin^{n-1} x \cos x}{n}\right]_0^{\pi/2} + \frac{n-1}{n} J_{n-2} \\
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&= \frac{n-1}{n} J_{n-2}
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\end{aligned}
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$$
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同理 $\displaystyle \int_0^{\pi/2} \cos^n x \, dx = J_n$（对称性）。
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常见值：
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$$
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J_2 = \frac{\pi}{4},\quad J_3 = \frac{2}{3},\quad J_4 = \frac{3\pi}{16},\quad J_5 = \frac{8}{15}
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$$
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---
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$$
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\int \sec^n x \, dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1}\int \sec^{n-2} x \, dx \quad (n \neq 1)
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$$
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$$
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\int \csc^n x \, dx = -\frac{\csc^{n-2} x \cot x}{n-1} + \frac{n-2}{n-1}\int \csc^{n-2} x \, dx \quad (n \neq 1)
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$$
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**secⁿ 递推公式推导**（分部积分法）：
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设 $I_n = \int \sec^n x \, dx$，改写为 $\int \sec^{n-2} x \cdot \sec^2 x \, dx$。
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令 $u = \sec^{n-2} x$，$dv = \sec^2 x \, dx$，则：
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$$
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du = (n-2)\sec^{n-2} x \tan x \, dx,\quad v = \tan x
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$$
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代入分部积分公式 $\int u \, dv = uv - \int v \, du$：
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$$
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\begin{aligned}
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I_n &= \sec^{n-2} x \tan x - \int \tan x \cdot (n-2)\sec^{n-2} x \tan x \, dx \\
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&= \sec^{n-2} x \tan x - (n-2)\int \sec^{n-2} x \tan^2 x \, dx
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\end{aligned}
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$$
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利用 $\tan^2 x = \sec^2 x - 1$：
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$$
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I_n = \sec^{n-2} x \tan x - (n-2)\int \sec^{n-2} x (\sec^2 x - 1) \, dx
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$$
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$$
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I_n = \sec^{n-2} x \tan x - (n-2) I_n + (n-2) I_{n-2}
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$$
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移项合并 $I_n$ 项：
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$$
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(n-1) I_n = \sec^{n-2} x \tan x + (n-2) I_{n-2}
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$$
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$$
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\boxed{I_n = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} I_{n-2}}
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$$
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需要两个初始条件：
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- $I_1 = \int \sec x \, dx = \ln|\sec x + \tan x| + C$
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- $I_2 = \int \sec^2 x \, dx = \tan x + C$
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**cscⁿ 递推公式推导**类似，利用 $\cot^2 x = \csc^2 x - 1$。
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---
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#### 递推式的完全展开
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递推公式重复代入即可展开为有限项和（初等函数的闭式表达）。
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**sinⁿ / cosⁿ 的完全展开**（不定积分，设 $I_n = \int \sin^n x \, dx$）：
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边界条件 $I_0 = x + C,\; I_1 = -\cos x + C$
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$$
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\begin{aligned}
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I_n &= -\frac{\sin^{n-1} x \cos x}{n}
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+ \frac{n-1}{n} \left(-\frac{\sin^{n-3} x \cos x}{n-2}\right)
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+ \frac{n-1}{n}\cdot\frac{n-3}{n-2} \left(-\frac{\sin^{n-5} x \cos x}{n-4}\right) + \cdots \\[4pt]
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&= -\cos x \sum_{k=1}^{n} \frac{(n-1)!!}{(n-k)!!} \cdot \frac{\sin^{n-k} x}{(n-k+1)!!} \quad \text{（示意模式）}
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\end{aligned}
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$$
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更清晰地，按奇偶展开：
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**$n$ 为偶数**（$n=2m$）：
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$$
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\begin{aligned}
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\int \sin^{2m} x \, dx &= -\cos x \sum_{k=1}^{m} \frac{(2m-1)(2m-3)\cdots(2m-2k+1)}{(2m-2)(2m-4)\cdots(2m-2k+2)}
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\cdot \frac{\sin^{2m-2k+1} x}{2m-2k+1} + C
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\end{aligned}
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$$
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其中 $C$ 为常数项（来自 $I_0$）。
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**$n$ 为奇数**（$n=2m+1$）：
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$$
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\begin{aligned}
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\int \sin^{2m+1} x \, dx &= -\cos x \sum_{k=1}^{m} \frac{(2m)(2m-2)\cdots(2m-2k+2)}{(2m-1)(2m-3)\cdots(2m-2k+1)}
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\cdot \frac{\sin^{2m-2k+1} x}{2m-2k+2} + C
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\end{aligned}
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$$
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其中 $C$ 来自 $I_1 = -\cos x$ 项。
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**实际记忆**：通常直接用递推公式比记忆展开式更实用，考试中一般只需求特定 $n$ 的值或用到递推关系。
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**secⁿ 的完全展开**（设 $I_n = \int \sec^n x \, dx$）：
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递推式同样可展开，以奇数/偶数分界：
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**$n$ 为偶数**（$n=2m$，终止于 $I_2 = \tan x + C$）：
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$$
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I_{2m} = \sum_{k=1}^{m} \frac{(2m-2)(2m-4)\cdots(2m-2k+2)}{(2m-1)(2m-3)\cdots(2m-2k+1)} \cdot \frac{\sec^{2m-2k} x \tan x}{2m-2k+1} + C
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$$
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例如：
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$$
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\begin{aligned}
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I_2 &= \tan x + C \\[2pt]
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I_4 &= \frac{1}{3}\sec^2 x \tan x + \frac{2}{3}\tan x + C \\[2pt]
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I_6 &= \frac{1}{5}\sec^4 x \tan x + \frac{4}{15}\sec^2 x \tan x + \frac{8}{15}\tan x + C
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\end{aligned}
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$$
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**$n$ 为奇数**（$n=2m+1$，终止于 $I_1 = \ln|\sec x + \tan x| + C$）：
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$$
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\begin{aligned}
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I_{2m+1} &= \sum_{k=1}^{m} \frac{(2m-1)(2m-3)\cdots(2m-2k+1)}{(2m)(2m-2)\cdots(2m-2k+2)} \cdot \frac{\sec^{2m-2k+1} x \tan x}{2m-2k+2}
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+ \frac{(2m-1)!!}{(2m)!!} \ln|\sec x + \tan x| + C
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\end{aligned}
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$$
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例如：
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$$
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\begin{aligned}
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I_1 &= \ln|\sec x + \tan x| + C \\[2pt]
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I_3 &= \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C \\[2pt]
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I_5 &= \frac{1}{4}\sec^3 x \tan x + \frac{3}{8}\sec x \tan x + \frac{3}{8}\ln|\sec x + \tan x| + C
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\end{aligned}
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$$
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---
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#### 积化和差
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$$
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\sin A \cos B = \frac{1}{2}\sin(A+B) + \frac{1}{2}\sin(A-B)
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$$
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$$
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\cos A \cos B = \frac{1}{2}\cos(A+B) + \frac{1}{2}\cos(A-B)
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$$
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$$
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\sin A \sin B = \frac{1}{2}\cos(A-B) - \frac{1}{2}\cos(A+B)
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$$
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