From 00d32cb3150ef03de93144416ed949282361daa7 Mon Sep 17 00:00:00 2001
From: ViperEkura <3081035982@qq.com>
Date: Tue, 5 May 2026 14:18:23 +0800
Subject: [PATCH] =?UTF-8?q?chore:=20=E7=AE=80=E5=8C=96=E6=96=87=E6=A1=A3?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 subjects/math/04_积分.md | 53 ++++++++++++++++++----------------------
 1 file changed, 24 insertions(+), 29 deletions(-)

diff --git a/subjects/math/04_积分.md b/subjects/math/04_积分.md
index 74ea121..d862f3e 100644
--- a/subjects/math/04_积分.md
+++ b/subjects/math/04_积分.md
@@ -311,42 +311,49 @@ $$
 
 #### sinⁿ 递推公式推导（分部积分法）
 
-设 $I_n = \int \sin^n x \, dx$，$n \ge 2$。
-
-取 $u = \sin^{n-1}x$，$dv = \sin x \, dx$，则：
+设
 $$
-du = (n-1)\sin^{n-2}x \cos x \, dx,\quad v = -\cos x
+I_n = \int \sin^n x \, dx, \quad n \ge 2.
 $$
 
-代入分部积分公式 $\int u \, dv = uv - \int v \, du$：
+取
+$$
+u = \sin^{n-1}x, \quad dv = \sin x \, dx,
+$$
+则
+$$
+du = (n-1)\sin^{n-2}x \cos x \, dx, \quad v = -\cos x.
+$$
+
+分部积分：
 $$
 \begin{aligned}
-I_n &= -\sin^{n-1}x \cos x - \int (-\cos x) \cdot (n-1)\sin^{n-2}x \cos x \, dx \\
-    &= -\sin^{n-1}x \cos x + (n-1)\int \sin^{n-2}x \cos^2 x \, dx
+I_n &= uv - \int v \, du \\
+    &= -\sin^{n-1}x \cos x + (n-1)\int \sin^{n-2}x \cos^2 x \, dx.
 \end{aligned}
 $$
 
 利用 $\cos^2 x = 1 - \sin^2 x$：
 $$
-\int \sin^{n-2}x \cos^2 x \, dx = \int \sin^{n-2}x \, dx - \int \sin^n x \, dx = I_{n-2} - I_n
+\int \sin^{n-2}x \cos^2 x \, dx = \int \sin^{n-2}x \, dx - \int \sin^n x \, dx = I_{n-2} - I_n.
 $$
 
-代入得：
+代入得
 $$
-I_n = -\sin^{n-1}x \cos x + (n-1)(I_{n-2} - I_n)
+I_n = -\sin^{n-1}x \cos x + (n-1)(I_{n-2} - I_n).
 $$
 
-移项合并 $I_n$ 项：
+整理含 $I_n$ 的项：
 $$
-I_n + (n-1)I_n = -\sin^{n-1}x \cos x + (n-1)I_{n-2}
+\begin{aligned}
+I_n + (n-1)I_n &= -\sin^{n-1}x \cos x + (n-1)I_{n-2}, \\
+n I_n &= -\sin^{n-1}x \cos x + (n-1)I_{n-2}.
+\end{aligned}
 $$
 
+于是
 $$
-n I_n = -\sin^{n-1}x \cos x + (n-1)I_{n-2}
-$$
-
-$$
-\boxed{I_n = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n} I_{n-2}},\quad n\ge 2
+\boxed{I_n = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n} I_{n-2}},\quad n\ge 2.
 $$
 
 需要两个初始条件：
@@ -382,18 +389,6 @@ $$
 
 同理 $\displaystyle \int_0^{\pi/2} \cos^n x \, dx = J_n$（对称性）。
 
-**点火链条**（初始条件 $J_0 = \frac{\pi}{2},\; J_1 = 1$）：
-$$
-\begin{aligned}
-n \text{ 为偶数} &: \quad J_n = \frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdots \frac{1}{2} \cdot J_0
-                  = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2} \\[4pt]
-n \text{ 为奇数} &: \quad J_n = \frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdots \frac{2}{3} \cdot J_1
-                  = \frac{(n-1)!!}{n!!}
-\end{aligned}
-$$
-
-**记忆口诀**："点火公式"即链条式递推，偶数多一个 $\frac{\pi}{2}$（引信），奇数直接出结果（哑火）。
-
 常见值：
 $$
 J_2 = \frac{\pi}{4},\quad J_3 = \frac{2}{3},\quad J_4 = \frac{3\pi}{16},\quad J_5 = \frac{8}{15}