From 9d430991c3735b87a59ba179dc92226e0d14ad88 Mon Sep 17 00:00:00 2001
From: ViperEkura <3081035982@qq.com>
Date: Sun, 26 Apr 2026 21:51:00 +0800
Subject: [PATCH] =?UTF-8?q?feat:=20=E4=BF=AE=E6=94=B9=E7=A7=AF=E5=88=86?=
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---
 mistakes/math/03_中值定理.md |   2 +-
 subjects/math/04_积分.md     | 255 ++++++++++++++++++++++++++++-------
 2 files changed, 209 insertions(+), 48 deletions(-)

diff --git a/mistakes/math/03_中值定理.md b/mistakes/math/03_中值定理.md
index 69e7cf9..6a9791e 100644
--- a/mistakes/math/03_中值定理.md
+++ b/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  
 $$
 f(\xi) = \frac{b - \xi}{a} f'(\xi).
 $$
diff --git a/subjects/math/04_积分.md b/subjects/math/04_积分.md
index 69a8e63..6841754 100644
--- a/subjects/math/04_积分.md
+++ b/subjects/math/04_积分.md
@@ -27,18 +27,41 @@ $$
 **推导**（换元法）：令 $x = a \tan t$，则 $dx = a \sec^2 t \, dt$
 
 $$
-\int \frac{dx}{x^2 + a^2} = \int \frac{a \sec^2 t}{a^2 + a^2 \tan^2 t} \, dt = \int \frac{a \sec^2 t}{a^2 \sec^2 t} \, dt = \int \frac{dt}{a} = \frac{t}{a} + C = \frac{1}{a} \arctan \frac{x}{a} + C
+\int \frac{dx}{x^2 + a^2} = \int \frac{a \sec^2 t}{a^2 + a^2 \tan^2 t} \, dt
+$$
+
+$$
+= \int \frac{a \sec^2 t}{a^2 \sec^2 t} \, dt
+$$
+
+$$
+= \int \frac{dt}{a}
+$$
+
+$$
+= \frac{t}{a} + C
+$$
+
+$$
+= \frac{1}{a} \arctan \frac{x}{a} + C
 $$
 
 #### 推广形式
 
 $$
-\begin{aligned}
-\int \frac{dx}{x^2 + a^2} &= \frac{1}{a} \arctan \frac{x}{a} + C \\
-\int \frac{dx}{b^2 + (x + c)^2} &= \frac{1}{b} \arctan \frac{x + c}{b} + C \\
-\int \frac{x \, dx}{x^2 + a^2} &= \frac{1}{2} \ln(x^2 + a^2) + C \\
-\int \frac{dx}{(x^2 + a^2)^2} &= \frac{x}{2a^2(x^2 + a^2)} + \frac{1}{2a^3} \arctan \frac{x}{a} + C
-\end{aligned}
+\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan \frac{x}{a} + C
+$$
+
+$$
+\int \frac{dx}{b^2 + (x + c)^2} = \frac{1}{b} \arctan \frac{x + c}{b} + C
+$$
+
+$$
+\int \frac{x \, dx}{x^2 + a^2} = \frac{1}{2} \ln(x^2 + a^2) + C
+$$
+
+$$
+\int \frac{dx}{(x^2 + a^2)^2} = \frac{x}{2a^2(x^2 + a^2)} + \frac{1}{2a^3} \arctan \frac{x}{a} + C
 $$
 
 ---
@@ -48,35 +71,59 @@ $$
 #### 基本公式
 
 $$
-\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C = \operatorname{arsinh} \frac{x}{a} + C
+\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C
 $$
 
 $$
-\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C = \operatorname{arcosh} \frac{x}{a} + C \quad (|x| > |a|)
+\int \frac{dx}{\sqrt{x^2 + a^2}} = \operatorname{arsinh} \frac{x}{a} + C
 $$
 
 $$
-\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \frac{x}{a} + C = -\arccos \frac{x}{a} + C \quad (|x| < |a|)
+\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C \quad (|x| > |a|)
+$$
+
+$$
+\int \frac{dx}{\sqrt{x^2 - a^2}} = \operatorname{arcosh} \frac{x}{a} + C \quad (|x| > |a|)
+$$
+
+$$
+\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \frac{x}{a} + C \quad (|x| < |a|)
+$$
+
+$$
+\int \frac{dx}{\sqrt{a^2 - x^2}} = -\arccos \frac{x}{a} + C \quad (|x| < |a|)
 $$
 
 #### 推导方法
 
-**$\sqrt{x^2 + a^2}$ 型**：令 $x = a \sinh t$，则 $dx = a \cosh t \, dt$，$\sqrt{x^2 + a^2} = a \cosh t$
-
 $$
-\int \frac{dx}{\sqrt{x^2 + a^2}} = \int \frac{a \cosh t}{a \cosh t} \, dt = \int dt = t + C = \ln\left|x + \sqrt{x^2 + a^2}\right| + C
+\begin{align}
+\int \frac{dx}{\sqrt{x^2 + a^2}}
+    &= \int \frac{a \cosh t}{a \cosh t} \, dt && (x = a \sinh t) \\
+    &= \int dt \\
+    &= t + C \\
+    &= \ln\left|x + \sqrt{x^2 + a^2}\right| + C
+\end{align}
 $$
 
-**$\sqrt{x^2 - a^2}$ 型**：令 $x = a \cosh t$（$x > a$），则 $dx = a \sinh t \, dt$，$\sqrt{x^2 - a^2} = a \sinh t$
-
 $$
-\int \frac{dx}{\sqrt{x^2 - a^2}} = \int \frac{a \sinh t}{a \sinh t} \, dt = \int dt = t + C = \ln\left|x + \sqrt{x^2 - a^2}\right| + C
+\begin{align}
+\int \frac{dx}{\sqrt{x^2 - a^2}}
+    &= \int \frac{a \sinh t}{a \sinh t} \, dt && (x = a \cosh t) \\
+    &= \int dt \\
+    &= t + C \\
+    &= \ln\left|x + \sqrt{x^2 - a^2}\right| + C
+\end{align}
 $$
 
-**$\sqrt{a^2 - x^2}$ 型**：令 $x = a \sin t$，则 $dx = a \cos t \, dt$，$\sqrt{a^2 - x^2} = a \cos t$
-
 $$
-\int \frac{dx}{\sqrt{a^2 - x^2}} = \int \frac{a \cos t}{a \cos t} \, dt = \int dt = t + C = \arcsin \frac{x}{a} + C
+\begin{align}
+\int \frac{dx}{\sqrt{a^2 - x^2}}
+    &= \int \frac{a \cos t}{a \cos t} \, dt && (x = a \sin t) \\
+    &= \int dt \\
+    &= t + C \\
+    &= \arcsin \frac{x}{a} + C
+\end{align}
 $$
 
 #### 等效形式
@@ -88,55 +135,168 @@ $$
 #### 推广形式
 
 $$
-\begin{aligned}
-\int \frac{dx}{\sqrt{(x + b)^2 + a^2}} &= \ln\left|x + b + \sqrt{(x + b)^2 + a^2}\right| + C \\
-\int \frac{dx}{\sqrt{(x + b)^2 - a^2}} &= \ln\left|x + b + \sqrt{(x + b)^2 - a^2}\right| + C \quad (|x + b| > |a|) \\
-\int \frac{x \, dx}{\sqrt{x^2 + a^2}} &= \sqrt{x^2 + a^2} + C \\
-\int \frac{x \, dx}{\sqrt{x^2 - a^2}} &= \sqrt{x^2 - a^2} + C \\
-\int \sqrt{x^2 + a^2} \, dx &= \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C \\
-\int \sqrt{x^2 - a^2} \, dx &= \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
-\end{aligned}
+\int \frac{dx}{\sqrt{(x + b)^2 + a^2}} = \ln\left|x + b + \sqrt{(x + b)^2 + a^2}\right| + C
+$$
+
+$$
+\int \frac{dx}{\sqrt{(x + b)^2 - a^2}} = \ln\left|x + b + \sqrt{(x + b)^2 - a^2}\right| + C \quad (|x + b| > |a|)
+$$
+
+$$
+\int \frac{x \, dx}{\sqrt{x^2 + a^2}} = \sqrt{x^2 + a^2} + C
+$$
+
+$$
+\int \frac{x \, dx}{\sqrt{x^2 - a^2}} = \sqrt{x^2 - a^2} + C
+$$
+
+$$
+\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
+$$
+
+$$
+\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
 $$
 
 ---
 
-### 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）：
+$$
+\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C \quad (|x| > |a|)
+$$
 
 $$
-\int \sec x \, dx = \int \sec x \cdot \frac{\sec x + \tan x}{\sec x + \tan x} \, dx = \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, dx = \ln|\sec x + \tan x| + C
+\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C \quad (|x| < |a|)
+$$
+
+#### 推导方法
+
+$$
+\begin{align}
+\int \sqrt{x^2 + a^2} \, dx
+    &= \int a \cosh t \cdot a \cosh t \, dt = a^2 \int \cosh^2 t \, dt && (x = a \sinh t) \\
+    &= a^2 \int \frac{\cosh 2t + 1}{2} \, dt = \frac{a^2}{2}\left(\frac{\sinh 2t}{2} + t\right) + C \\
+    &= \frac{a^2}{2}(\sinh t \cosh t + t) + C \\
+    &= \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
+\end{align}
+$$
+
+$$
+\begin{align}
+\int \sqrt{x^2 - a^2} \, dx
+    &= \int a \sinh t \cdot a \sinh t \, dt = a^2 \int \sinh^2 t \, dt && (x = a \cosh t) \\
+    &= a^2 \int \frac{\cosh 2t - 1}{2} \, dt = \frac{a^2}{2}\left(\frac{\sinh 2t}{2} - t\right) + C \\
+    &= \frac{a^2}{2}(\sinh t \cosh t - t) + C \\
+    &= \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln\left|x + \sqrt{x^2 - a^2}\right| + C
+\end{align}
+$$
+
+$$
+\begin{align}
+\int \sqrt{a^2 - x^2} \, dx
+    &= \int a \cos t \cdot a \cos t \, dt = a^2 \int \cos^2 t \, dt && (x = a \sin t) \\
+    &= \frac{a^2}{2}\left(t + \frac{\sin 2t}{2}\right) + C \\
+    &= \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C
+\end{align}
+$$
+
+#### 推广形式
+
+$$
+\int (x + b)\sqrt{x^2 + a^2} \, dx = \frac{1}{3}(x + b)(x^2 + a^2)^{3/2} - \frac{b}{2}x\sqrt{x^2 + a^2} - \frac{ab^2}{2}\ln\left|x + \sqrt{x^2 + a^2}\right| + C
+$$
+
+$$
+\int x\sqrt{x^2 + a^2} \, dx = \frac{1}{3}(x^2 + a^2)^{3/2} + C
+$$
+
+$$
+\int x\sqrt{x^2 - a^2} \, dx = \frac{1}{3}(x^2 - a^2)^{3/2} + C
+$$
+
+$$
+\int x\sqrt{a^2 - x^2} \, dx = -\frac{1}{3}(a^2 - x^2)^{3/2} + C
 $$
 
 ---
 
-$$
-\int \csc x \, dx = -\ln|\csc x + \cot x| + C
-$$
+### 要点 05 - 三角函数积分
 
-**推导方法**（类似地）：
+#### 降幂公式
 
 $$
-\int \csc x \, dx = \int \csc x \cdot \frac{\csc x - \cot x}{\csc x - \cot x} \, dx = \int \frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x} \, dx = -\ln|\csc x + \cot x| + C
+\sin^2 x = \frac{1 - \cos 2x}{2}
 $$
 
-#### 其他常用积分
+$$
+\cos^2 x = \frac{1 + \cos 2x}{2}
+$$
 
 $$
-\begin{aligned}
-\int \sec^2 x \, dx &= \tan x + C \\
-\int \csc^2 x \, dx &= -\cot x + C \\
-\int \sec x \tan x \, dx &= \sec x + C \\
-\int \csc x \cot x \, dx &= -\csc x + C \\
-\int \sec^3 x \, dx &= \frac{1}{2}(\sec x \tan x + \ln|\sec x + \tan x|) + C \\
-\int \csc^3 x \, dx &= \frac{1}{2}(-\csc x \cot x + \ln|\csc x + \cot x|) + C
-\end{aligned}
+\sin^3 x = \frac{3\sin x - \sin 3x}{4}
+$$
+
+$$
+\cos^3 x = \frac{3\cos x + \cos 3x}{4}
+$$
+
+#### 基本积分
+
+$$
+\int \sin x \, dx = -\cos x + C
+$$
+
+$$
+\int \cos x \, dx = \sin x + C
+$$
+
+$$
+\int \tan x \, dx = -\ln|\cos x| + C
+$$
+
+$$
+\int \cot x \, dx = \ln|\sin x| + C
+$$
+
+#### 万能代换
+
+令 $t = \tan\frac{x}{2}$，则：
+
+$$
+\sin x = \frac{2t}{1 + t^2}, \quad \cos x = \frac{1 - t^2}{1 + t^2}, \quad dx = \frac{2 \, dt}{1 + t^2}
+$$
+
+适用类型：$R(\sin x, \cos x)$（有理函数形式）
+
+#### 常用结论
+
+$$
+\int \sin^n x \, dx = -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n}\int \sin^{n-2} x \, dx
+$$
+
+$$
+\int \cos^n x \, dx = \frac{\cos^{n-1} x \sin x}{n} + \frac{n-1}{n}\int \cos^{n-2} x \, dx
+$$
+
+#### 积化和差
+
+$$
+\sin A \cos B = \frac{1}{2}\sin(A+B) + \frac{1}{2}\sin(A-B)
+$$
+
+$$
+\cos A \cos B = \frac{1}{2}\cos(A+B) + \frac{1}{2}\cos(A-B)
+$$
+
+$$
+\sin A \sin B = \frac{1}{2}\cos(A-B) - \frac{1}{2}\cos(A+B)
 $$
 
 ---
@@ -147,4 +307,5 @@ $$
 - 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
+- 三角函数积分（降幂、万能代换、积化和差）