feat: 修改积分方法

mistakes/math/03_中值定理.md

@@ -1,7 +1,7 @@
 ## 错题记录
 
 ### 题目 01
-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  
+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  
 $$
 f(\xi) = \frac{b - \xi}{a} f'(\xi).
 $$

subjects/math/04_积分.md

@@ -100,42 +100,66 @@ $$
 
 ---
 
-### 要点 04 - sec x、csc x 的积分
+### 要点 04 - $\sqrt{x^2 \pm a^2}$ 型积分
 
-#### 基本积分公式
+#### 基本公式
 
 $$
-\int \sec x \, dx = \ln|\sec x + \tan x| + C
+\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
 $$
 
-**推导方法**（分子分母策略）：
-
 $$
-\int \sec x \, dx = \int \sec x \cdot \frac{\sec x + \tan x}{\sec x + \tan x} \, dx = \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, dx = \ln|\sec x + \tan x| + C
+\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C \quad (|x| > |a|)
 $$
 
----
-
 $$
-\int \csc x \, dx = -\ln|\csc x + \cot x| + C
+\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C \quad (|x| < |a|)
 $$
 
-**推导方法**（类似地）：
+#### 推导方法
 
-$$
+**$\sqrt{x^2 + a^2}$ 型**：令 $x = a \tan t$，则 $dx = a \sec^2 t \, dt$，$\sqrt{x^2 + a^2} = a \sec t$
-\int \csc x \, dx = \int \csc x \cdot \frac{\csc x - \cot x}{\csc x - \cot x} \, dx = \int \frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x} \, dx = -\ln|\csc x + \cot x| + C
-$$
-
-#### 其他常用积分
 
 $$
 \begin{aligned}
-\int \sec^2 x \, dx &= \tan x + C \\
+\int \sqrt{x^2 + a^2} \, dx &= \int a \sec t \cdot a \sec^2 t \, dt = a^2 \int \sec^3 t \, dt \\
-\int \csc^2 x \, dx &= -\cot x + C \\
+&= a^2 \cdot \frac{1}{2}(\sec t \tan t + \ln|\sec t + \tan t|) + C \\
-\int \sec x \tan x \, dx &= \sec x + C \\
+&= \frac{a^2}{2}\left(\frac{x}{a}\sqrt{1 + \frac{x^2}{a^2}} + \ln\left|\frac{x}{a} + \sqrt{1 + \frac{x^2}{a^2}}\right|\right) + C \\
-\int \csc x \cot x \, dx &= -\csc x + C \\
+&= \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
-\int \sec^3 x \, dx &= \frac{1}{2}(\sec x \tan x + \ln|\sec x + \tan x|) + C \\
+\end{aligned}
-\int \csc^3 x \, dx &= \frac{1}{2}(-\csc x \cot x + \ln|\csc x + \cot x|) + C
+$$
+
+**$\sqrt{x^2 - a^2}$ 型**：令 $x = a \sec t$（$x > a$），则 $dx = a \sec t \tan t \, dt$，$\sqrt{x^2 - a^2} = a \tan t$
+
+$$
+\begin{aligned}
+\int \sqrt{x^2 - a^2} \, dx &= \int a \tan t \cdot a \sec t \tan t \, dt = a^2 \int \sec t \tan^2 t \, dt \\
+&= a^2 \int \sec t (\sec^2 t - 1) \, dt = a^2 \int (\sec^3 t - \sec t) \, dt \\
+&= a^2 \left[\frac{1}{2}(\sec t \tan t + \ln|\sec t + \tan t|) - \ln|\sec t + \tan t|\right] + C \\
+&= \frac{a^2}{2}\sec t \tan t - \frac{a^2}{2}\ln|\sec t + \tan t| + C \\
+&= \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
+\end{aligned}
+$$
+
+**$\sqrt{a^2 - x^2}$ 型**：令 $x = a \sin t$，则 $dx = a \cos t \, dt$，$\sqrt{a^2 - x^2} = a \cos t$
+
+$$
+\begin{aligned}
+\int \sqrt{a^2 - x^2} \, dx &= \int a \cos t \cdot a \cos t \, dt = a^2 \int \cos^2 t \, dt \\
+&= a^2 \int \frac{1 + \cos 2t}{2} \, dt = \frac{a^2}{2}\left(t + \frac{\sin 2t}{2}\right) + C \\
+&= \frac{a^2}{2}\left(\arcsin\frac{x}{a} + \frac{x}{a}\sqrt{1 - \frac{x^2}{a^2}}\right) + C \\
+&= \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C
+\end{aligned}
+$$
+
+#### 推广形式
+
+$$
+\begin{aligned}
+\int (x + b)\sqrt{x^2 + a^2} \, dx &= \frac{1}{3}(x + b)(x^2 + a^2)^{3/2} - \frac{b}{2}x\sqrt{x^2 + a^2} - \frac{ab^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C \\
+\int x\sqrt{x^2 + a^2} \, dx &= \frac{1}{3}(x^2 + a^2)^{3/2} + C \\
+\int x\sqrt{x^2 - a^2} \, dx &= \frac{1}{3}(x^2 - a^2)^{3/2} + C \\
+\int x\sqrt{a^2 - x^2} \, dx &= -\frac{1}{3}(a^2 - x^2)^{3/2} + C
 \end{aligned}
 $$
 
@@ -147,4 +171,4 @@ $$
 - 均匀分割技巧
 - $\frac{1}{x^2 + a^2}$ 型积分公式
 - $\frac{1}{\sqrt{x^2 \pm a^2}}$ 型积分公式
-- sec x、csc x 的积分公式
+- $\sqrt{x^2 \pm a^2}$ 型积分公式