feat: 修改积分方法
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## 错题记录
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### 题目 01
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设 $ f(x) $ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，$f(a) = 0$，$a > 0$ ，证明：存在 $\xi \in (a, b)$，使得
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设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，$f(a) = 0$，$a > 0$ ，证明：存在 $\xi \in (a, b)$，使得
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$$
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f(\xi) = \frac{b - \xi}{a} f'(\xi).
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$$
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**推导**（换元法）：令 $x = a \tan t$，则 $dx = a \sec^2 t \, dt$
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$$
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\int \frac{dx}{x^2 + a^2} = \int \frac{a \sec^2 t}{a^2 + a^2 \tan^2 t} \, dt = \int \frac{a \sec^2 t}{a^2 \sec^2 t} \, dt = \int \frac{dt}{a} = \frac{t}{a} + C = \frac{1}{a} \arctan \frac{x}{a} + C
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\int \frac{dx}{x^2 + a^2} = \int \frac{a \sec^2 t}{a^2 + a^2 \tan^2 t} \, dt
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$$
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$$
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= \int \frac{a \sec^2 t}{a^2 \sec^2 t} \, dt
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$$
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$$
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= \int \frac{dt}{a}
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$$
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$$
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= \frac{t}{a} + C
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$$
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$$
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= \frac{1}{a} \arctan \frac{x}{a} + C
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$$
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#### 推广形式
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$$
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\begin{aligned}
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\int \frac{dx}{x^2 + a^2} &= \frac{1}{a} \arctan \frac{x}{a} + C \\
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\int \frac{dx}{b^2 + (x + c)^2} &= \frac{1}{b} \arctan \frac{x + c}{b} + C \\
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\int \frac{x \, dx}{x^2 + a^2} &= \frac{1}{2} \ln(x^2 + a^2) + C \\
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\int \frac{dx}{(x^2 + a^2)^2} &= \frac{x}{2a^2(x^2 + a^2)} + \frac{1}{2a^3} \arctan \frac{x}{a} + C
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\end{aligned}
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\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan \frac{x}{a} + C
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$$
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$$
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\int \frac{dx}{b^2 + (x + c)^2} = \frac{1}{b} \arctan \frac{x + c}{b} + C
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$$
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$$
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\int \frac{x \, dx}{x^2 + a^2} = \frac{1}{2} \ln(x^2 + a^2) + C
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$$
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$$
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\int \frac{dx}{(x^2 + a^2)^2} = \frac{x}{2a^2(x^2 + a^2)} + \frac{1}{2a^3} \arctan \frac{x}{a} + C
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$$
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---
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@ -48,35 +71,59 @@ $$
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#### 基本公式
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$$
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\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C = \operatorname{arsinh} \frac{x}{a} + C
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\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C
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$$
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$$
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\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C = \operatorname{arcosh} \frac{x}{a} + C \quad (|x| > |a|)
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\int \frac{dx}{\sqrt{x^2 + a^2}} = \operatorname{arsinh} \frac{x}{a} + C
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$$
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$$
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\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \frac{x}{a} + C = -\arccos \frac{x}{a} + C \quad (|x| < |a|)
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\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C \quad (|x| > |a|)
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$$
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$$
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\int \frac{dx}{\sqrt{x^2 - a^2}} = \operatorname{arcosh} \frac{x}{a} + C \quad (|x| > |a|)
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$$
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$$
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\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \frac{x}{a} + C \quad (|x| < |a|)
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$$
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$$
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\int \frac{dx}{\sqrt{a^2 - x^2}} = -\arccos \frac{x}{a} + C \quad (|x| < |a|)
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$$
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#### 推导方法
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**$\sqrt{x^2 + a^2}$ 型**：令 $x = a \sinh t$，则 $dx = a \cosh t \, dt$，$\sqrt{x^2 + a^2} = a \cosh t$
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$$
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\int \frac{dx}{\sqrt{x^2 + a^2}} = \int \frac{a \cosh t}{a \cosh t} \, dt = \int dt = t + C = \ln\left|x + \sqrt{x^2 + a^2}\right| + C
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\begin{align}
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\int \frac{dx}{\sqrt{x^2 + a^2}}
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&= \int \frac{a \cosh t}{a \cosh t} \, dt && (x = a \sinh t) \\
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&= \int dt \\
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&= t + C \\
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&= \ln\left|x + \sqrt{x^2 + a^2}\right| + C
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\end{align}
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$$
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**$\sqrt{x^2 - a^2}$ 型**：令 $x = a \cosh t$（$x > a$），则 $dx = a \sinh t \, dt$，$\sqrt{x^2 - a^2} = a \sinh t$
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$$
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\int \frac{dx}{\sqrt{x^2 - a^2}} = \int \frac{a \sinh t}{a \sinh t} \, dt = \int dt = t + C = \ln\left|x + \sqrt{x^2 - a^2}\right| + C
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\begin{align}
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\int \frac{dx}{\sqrt{x^2 - a^2}}
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&= \int \frac{a \sinh t}{a \sinh t} \, dt && (x = a \cosh t) \\
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&= \int dt \\
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&= t + C \\
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&= \ln\left|x + \sqrt{x^2 - a^2}\right| + C
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\end{align}
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$$
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**$\sqrt{a^2 - x^2}$ 型**：令 $x = a \sin t$，则 $dx = a \cos t \, dt$，$\sqrt{a^2 - x^2} = a \cos t$
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$$
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\int \frac{dx}{\sqrt{a^2 - x^2}} = \int \frac{a \cos t}{a \cos t} \, dt = \int dt = t + C = \arcsin \frac{x}{a} + C
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\begin{align}
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\int \frac{dx}{\sqrt{a^2 - x^2}}
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&= \int \frac{a \cos t}{a \cos t} \, dt && (x = a \sin t) \\
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&= \int dt \\
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&= t + C \\
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&= \arcsin \frac{x}{a} + C
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\end{align}
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$$
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#### 等效形式
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#### 推广形式
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$$
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\begin{aligned}
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\int \frac{dx}{\sqrt{(x + b)^2 + a^2}} &= \ln\left|x + b + \sqrt{(x + b)^2 + a^2}\right| + C \\
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\int \frac{dx}{\sqrt{(x + b)^2 - a^2}} &= \ln\left|x + b + \sqrt{(x + b)^2 - a^2}\right| + C \quad (|x + b| > |a|) \\
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\int \frac{x \, dx}{\sqrt{x^2 + a^2}} &= \sqrt{x^2 + a^2} + C \\
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\int \frac{x \, dx}{\sqrt{x^2 - a^2}} &= \sqrt{x^2 - a^2} + C \\
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\int \sqrt{x^2 + a^2} \, dx &= \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C \\
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\int \sqrt{x^2 - a^2} \, dx &= \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
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\end{aligned}
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\int \frac{dx}{\sqrt{(x + b)^2 + a^2}} = \ln\left|x + b + \sqrt{(x + b)^2 + a^2}\right| + C
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$$
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$$
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\int \frac{dx}{\sqrt{(x + b)^2 - a^2}} = \ln\left|x + b + \sqrt{(x + b)^2 - a^2}\right| + C \quad (|x + b| > |a|)
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$$
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$$
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\int \frac{x \, dx}{\sqrt{x^2 + a^2}} = \sqrt{x^2 + a^2} + C
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$$
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$$
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\int \frac{x \, dx}{\sqrt{x^2 - a^2}} = \sqrt{x^2 - a^2} + C
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$$
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$$
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\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
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$$
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$$
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\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
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$$
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---
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### 要点 04 - sec x、csc x 的积分
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### 要点 04 - $\sqrt{x^2 \pm a^2}$ 型积分
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#### 基本积分公式
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#### 基本公式
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$$
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\int \sec x \, dx = \ln|\sec x + \tan x| + C
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\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
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$$
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**推导方法**（分子分母策略）：
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$$
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\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C \quad (|x| > |a|)
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$$
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$$
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\int \sec x \, dx = \int \sec x \cdot \frac{\sec x + \tan x}{\sec x + \tan x} \, dx = \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, dx = \ln|\sec x + \tan x| + C
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\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C \quad (|x| < |a|)
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$$
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#### 推导方法
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$$
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\begin{align}
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\int \sqrt{x^2 + a^2} \, dx
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&= \int a \cosh t \cdot a \cosh t \, dt = a^2 \int \cosh^2 t \, dt && (x = a \sinh t) \\
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&= a^2 \int \frac{\cosh 2t + 1}{2} \, dt = \frac{a^2}{2}\left(\frac{\sinh 2t}{2} + t\right) + C \\
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&= \frac{a^2}{2}(\sinh t \cosh t + t) + C \\
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&= \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
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\end{align}
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$$
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$$
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\begin{align}
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\int \sqrt{x^2 - a^2} \, dx
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&= \int a \sinh t \cdot a \sinh t \, dt = a^2 \int \sinh^2 t \, dt && (x = a \cosh t) \\
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&= a^2 \int \frac{\cosh 2t - 1}{2} \, dt = \frac{a^2}{2}\left(\frac{\sinh 2t}{2} - t\right) + C \\
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&= \frac{a^2}{2}(\sinh t \cosh t - t) + C \\
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&= \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
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\end{align}
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$$
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$$
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\begin{align}
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\int \sqrt{a^2 - x^2} \, dx
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&= \int a \cos t \cdot a \cos t \, dt = a^2 \int \cos^2 t \, dt && (x = a \sin t) \\
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&= \frac{a^2}{2}\left(t + \frac{\sin 2t}{2}\right) + C \\
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&= \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C
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\end{align}
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$$
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#### 推广形式
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$$
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\int (x + b)\sqrt{x^2 + a^2} \, dx = \frac{1}{3}(x + b)(x^2 + a^2)^{3/2} - \frac{b}{2}x\sqrt{x^2 + a^2} - \frac{ab^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
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$$
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$$
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\int x\sqrt{x^2 + a^2} \, dx = \frac{1}{3}(x^2 + a^2)^{3/2} + C
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$$
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$$
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\int x\sqrt{x^2 - a^2} \, dx = \frac{1}{3}(x^2 - a^2)^{3/2} + C
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$$
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$$
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\int x\sqrt{a^2 - x^2} \, dx = -\frac{1}{3}(a^2 - x^2)^{3/2} + C
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$$
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---
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$$
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\int \csc x \, dx = -\ln|\csc x + \cot x| + C
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$$
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### 要点 05 - 三角函数积分
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**推导方法**（类似地）：
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#### 降幂公式
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$$
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\int \csc x \, dx = \int \csc x \cdot \frac{\csc x - \cot x}{\csc x - \cot x} \, dx = \int \frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x} \, dx = -\ln|\csc x + \cot x| + C
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\sin^2 x = \frac{1 - \cos 2x}{2}
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$$
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#### 其他常用积分
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$$
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\cos^2 x = \frac{1 + \cos 2x}{2}
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$$
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$$
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\begin{aligned}
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\int \sec^2 x \, dx &= \tan x + C \\
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\int \csc^2 x \, dx &= -\cot x + C \\
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\int \sec x \tan x \, dx &= \sec x + C \\
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\int \csc x \cot x \, dx &= -\csc x + C \\
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\int \sec^3 x \, dx &= \frac{1}{2}(\sec x \tan x + \ln|\sec x + \tan x|) + C \\
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\int \csc^3 x \, dx &= \frac{1}{2}(-\csc x \cot x + \ln|\csc x + \cot x|) + C
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\end{aligned}
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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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$$
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#### 万能代换
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令 $t = \tan\frac{x}{2}$，则：
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$$
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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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$$
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#### 积化和差
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$$
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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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$$
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---
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@ -147,4 +307,5 @@ $$
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- 均匀分割技巧
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- $\frac{1}{x^2 + a^2}$ 型积分公式
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- $\frac{1}{\sqrt{x^2 \pm a^2}}$ 型积分公式
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- sec x、csc x 的积分公式
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- $\sqrt{x^2 \pm a^2}$ 型积分公式
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- 三角函数积分（降幂、万能代换、积化和差）
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