n次三角函数不定积分的探究

Investigation of Indefinite Integrals of Trigonometric Functions of the n Degree

摘要: 结合具体的实例，通过低次方三角函数不定积分的研究，总结，逐步深入到研究n次三角。最终探究，${\displaystyle \int {\sin }^{m}x}\text{d}x,{\displaystyle \int {\cos }^{n}x}\text{d}x$ 以及${\displaystyle \int {\sin }^{m}x\cdot {\cos }^{n}x}\text{d}x$ 的显式结果及其证明。

Abstract: Combined with concrete examples, through the study of the indefinite integral of low power trigonometric function, summary, and step by step in-depth study of n trigonometry. The explicit results of ${\displaystyle \int {\sin }^{m}x}\text{d}x$ , ${\displaystyle \int {\cos }^{n}x}\text{d}x$ and ${\displaystyle \int {\sin }^{m}x\cdot {\cos }^{n}x}\text{d}x$ and their proofs are discussed.

1. 引言

三角函数的不定积分在众多理工专业中，应用广泛。在波动理论里，三角函数可精确描述波的各种特性，帮助分析波的传播方向和速度。简谐振动中，物体位移与时间关系用正弦或余弦函数表示，速度、加速度是其导数，借此可分析振动的周期、频率等特征，为深入研究波动现象及相关技术应用提供有力数学工具。金融领域，可用于计算期权定价模型中的积分。同时也是不少学生困扰的问题，其中高次三角函数的问题更甚。高等数学教材文献[1]的附录中，给出了众多三角函数相关的积分公式以及两相邻偶数或奇数次的积分关联式。文献[2]中宋显花研究${\displaystyle \int {\sec }^{n}x}\text{d}x,{\displaystyle \int {\csc }^{n}x}\text{d}x$ 等的关联公式，并给出了证明和几个在解题上的应用实例。文献[3]，李茜在此基础上给出了众多含三角函数的不定积分解法和例子。但是上述文章，均只给出了积分关联式以及含三角函数积分的解法与例子，没有给出具体的高次方积分结果式。在解决复杂的乘积函数积分如${\displaystyle \int {\text{e}}^x{\sin }^{n}x}\text{d}x$ ，上面两位作者的研究内容显然是不好解决的。在傅里叶级数等理论中，常需要计算高次方三角函数来确定函数展开式的系数，以将复杂的周期函数用三角函数表示，用于分析函数性质。由此思考，能否利用关联式和归纳法更进一步的探究${\displaystyle \int {\sin }^{m}x}\text{d}x,{\displaystyle \int {\cos }^{n}x}\text{d}x\cdots {\displaystyle \int {\sin }^{m}x{\cos }^{n}x}\text{d}x$ 的积分结果成为本文的核心。同时，为了探索出n次积分的结果，在预备知识也详尽的给出一般情形5次方及其以下的不定积分过程与结果。不仅能帮助总结归纳出n次积分，也能对学生学习基本积分方法，积分策略起到一定的积极作用。

2. 预备知识

这里给出5次方以内的正余弦三角函数的不定积分式子，其中较为简单的就直接给出，复杂的再给出具体的过程。

基础公式：

${\displaystyle \int \sin x}\text{d}x=-\cos x+c{\displaystyle \int \cos x}\text{d}x=\sin x+c$

$\begin{array}{l}{\displaystyle \int {\sin }^{2}x}\text{d}x=\frac{1}{2}x-\frac{1}{2}\sin x\cos x+c{\displaystyle \int {\cos }^{2}x}\text{d}x=\frac{1}{2}x+\frac{1}{2}\sin x\cos x+c\\ {\displaystyle \int {\sin }^{3}x}\text{d}x=-\frac{2}{3}\cos x\left(1+\frac{1}{2}{\sin }^{2}x\right)+c{\displaystyle \int {\cos }^{3}x}\text{d}x=\frac{2}{3}\sin x+\left(1+\frac{1}{2}{\cos }^{2}x\right)+c\end{array}$

下面阐明4次与5次的正余弦函数的不定积分。为了方便，在求解高次不定积分时，文章省略掉后面的常数c。

${\displaystyle \int {\sin }^{4}x\text{d}x}\underset{\_}{\underset{\_}{�形}}{\displaystyle \int \left(1-{\cos }^{2}x\right)}{\sin }^{2}x\text{d}x\underset{\_}{\underset{\_}{�分性�}}{\displaystyle \int {\sin }^{2}x}\text{d}x-\frac{1}{4}{\displaystyle \int 4{\sin }^{2}x{\cos }^{2}x\text{d}x}$

$\underset{\_}{\underset{\_}{��}}{\displaystyle \int {\sin }^{2}x\text{d}x}-\frac{1}{4}{\displaystyle \int {\left(2\sin x\cos x\right)}^2\text{d}x}$

$\underset{\_}{\underset{\_}{�微分}}{\displaystyle \int {\sin }^{2}x\text{d}x}-\frac{1}{8}{\displaystyle \int {\sin }^{2}2x}\text{d}\left(2x\right)$

$\underset{\_}{\underset{\_}{�,�}}\frac{1}{2}x-\frac{1}{2}\sin x\cos x-\frac{1}{8}\left(\frac{1}{2}\cdot 2x-\frac{1}{2}\sin 2x\cos 2x\right)$

$\underset{\_}{\underset{\_}{化�}}\frac{3}{8}x-\frac{1}{2}\sin x\cos x+\frac{1}{16}\left(2\sin x\cos x\right)\left(1-2{\sin }^{2}x\right)$

$\underset{\_}{\underset{\_}{整理}}\frac{3}{8}x-\frac{3}{8}\cos x\left(\sin x+\frac{2}{3}{\sin }^{3}x\right)$

$\begin{array}{c}{\displaystyle \int {\cos }^4}x\text{d}x={\displaystyle \int \left(1-{\sin }^{2}x\right)}{\cos }^{2}x\text{d}x\\ ={\displaystyle \int {\cos }^2}x\text{d}x-\frac{1}{4}{\displaystyle \int 4{\sin }^2}x{\cos }^{2}x\text{d}x\\ ={\displaystyle \int {\cos }^2}x\text{d}x-\frac{1}{8}{\displaystyle \int {\sin }^{2}2x}\text{d}\left(2x\right)\\ =\frac{1}{2}x+\frac{1}{2}\sin x\cos x-\frac{1}{8}\left(\frac{1}{2}\cdot 2x-\frac{1}{2}\sin 2x\cos 2x\right)\\ =\frac{3}{8}x+\frac{3}{8}\sin x\left(\cos x+\frac{2}{3}{\cos }^{3}x\right)\end{array}$

求正弦4次方积分时，每一步的做法做了说明，由于已经默认读者很熟悉第一换元积分，加之内容简单，中间就没有写出换元的过程。求余弦时，就略去这一步。另：在求4次方的正余弦积分，还可以考虑另一种求方程思想。在此详细的介绍并且做简易的推演。

$\begin{array}{c}{\displaystyle \int {\sin }^4}x\text{d}x={\displaystyle \int {\left(1-{\cos }^{2}x\right)}^2\text{d}x}\\ ={\displaystyle \int \left(1-2{\cos }^{2}x+{\cos }^{4}x\right)\text{d}x}\\ ={\displaystyle \int \left(-\cos 2x\right)\text{d}x}+{\displaystyle \int {\cos }^{4}x\text{d}x}\\ =-\frac{1}{2}\sin 2x+{\displaystyle \int {\cos }^{4}x}\text{d}x\\ =-\sin x\cos x+{\displaystyle \int {\cos }^{4}x}\text{d}x\end{array}$

可得：${\displaystyle \int {\sin }^4}x\text{d}x-{\displaystyle \int {\cos }^4}x\text{d}x=-\sin x\cos x$

另：

$\begin{array}{c}{\displaystyle \int {\sin }^4}x\text{d}x+{\displaystyle \int {\cos }^4}x\text{d}x={\displaystyle \int \left(1-2{\sin }^{2}x{\cos }^{2}x\right)\text{d}x}\\ =x-\frac{1}{2}{\displaystyle \int \frac{1-\cos 4x}{2}\text{d}x}\\ =\frac{3}{4}x+\frac{1}{16}\sin 4x\end{array}$

将上面两式联立，可求出4次方正余弦函数的积分。但求出答案的形式与前面直接计算所得略有不同，这都是正常的。由于很多三角函数之间形式可以自由变换，两种答案本质是相同的。有兴趣的读者，可以自行验证，这里不再赘述。本文给出的结果中，以包含$\sin x,\cos x$ 以及它们的n次方为主，到后面就可以发现，这样更有利于总结出规律以及证明。

$\begin{array}{c}{\displaystyle \int {\text{sin}}^{5}x\text{d}x}={\displaystyle \int {\left(1-{\cos }^{2}x\right)}^2\text{d}\left(-\cos x\right)}\\ =-{\displaystyle \int \left(1-2{\cos }^{2}x+{\cos }^{4}x\right)\text{d}\left(\cos x\right)}\\ =-{\displaystyle \int \left(1-2t^2+t^4\right)\text{d}t}=-t+\frac{2}{3}{t}^3-\frac{1}{5}{t}^5\\ =-\frac{1}{5}{\cos }^{5}x+\frac{2}{3}{\cos }^{3}x-\cos x\\ =-\frac{1}{5}{\left(1-{\sin }^{2}x\right)}^2\cos x+\frac{2}{3}\left(1-{\sin }^{2}x\right)\cos x-\cos x\\ =-\frac{8}{15}\cos x-\frac{4}{15}\cos x{\sin }^{2}x-\frac{1}{5}\cos x{\sin }^{4}x\\ =-\frac{8}{15}\cos x\left(1+\frac{1}{2}{\sin }^{2}x+\frac{3}{8}{\sin }^{4}x\right)\end{array}$

${\displaystyle \int {\cos }^{5}x\text{d}x}\underset{\_}{\underset{\_}{�形}}{\displaystyle \int {\left(1-{\sin }^{2}x\right)}^2\cos x\text{d}x}\underset{\_}{\underset{\_}{�微分}}{\displaystyle \int \left(1-2{\sin }^{2}x+{\sin }^{4}x\right)\text{d}\left(\sin x\right)}$

$\underset{\_}{\underset{\_}{换元}}{\displaystyle \int \left(1-2t^2+t^4\right)\text{d}t}$

$\underset{\_}{\underset{\_}{�函��分}}t-\frac{2}{3}{t}^3+\frac{1}{5}{t}^5$

$\underset{\_}{\underset{\_}{回代}}\frac{1}{5}{\text{sin}}^{\text{5}}x-\frac{2}{3}{\sin }^{3}x+\sin x$

$\underset{\_}{\underset{\_}{代�}}\frac{1}{5}{\left(1-{\cos }^{2}x\right)}^2\sin x-\frac{2}{3}\left(1-{\cos }^{2}x\right)\sin x+\sin x$

$\underset{\_}{\underset{\_}{化�}}\frac{8}{15}\sin x+\frac{4}{15}\sin x{\cos }^{2}x+\frac{1}{5}\sin x{\cos }^{4}x$

$\underset{\_}{\underset{\_}{整理}}\frac{8}{15}\sin x\left(1+\frac{1}{2}{\cos }^{2}x+\frac{3}{8}{\cos }^{4}x\right)$

主要用到的是第一换元积分法以及三角变换，这里详细写出余弦5次方每一步的操作，求正弦时略去。不难发现：当三角函数的次数逐渐变高时，对应的积分步骤方法以及运算的深度都会相应的加大。这样就面临一个问题，每次求解不同的三角函数不定积分时尤其是高次方，都会很棘手。实际上在通过上面的结果规律，其中的规律已经是有迹可循。下面就是通过这些规律，归纳总结出一般n次方的不定积分结果。

利用分部积分法，可以得到：

$\begin{array}{c}{\displaystyle \int {\sin }^{n}x\text{d}x}={\displaystyle \int {\sin }^{n-1}x}\text{d}\left(-\cos x\right)\\ =-{\sin }^{n-1}x\cos x-{\displaystyle \int \left(-\cos x\right)\text{d}\left({\sin }^{n-1}x\right)}\\ =-{\sin }^{n-1}x\cos x-{\displaystyle \int \left(n-1\right){\sin }^{n-2}x\left(-{\cos }^{2}x\right)\text{d}x}\\ =-{\sin }^{n-1}x\cos x-\left(n-1\right){\displaystyle \int {\sin }^{n-2}x\left({\sin }^{2}x-1\right)\text{d}x}\\ =-{\sin }^{n-1}x\cos x-\left(n-1\right){\displaystyle \int {\sin }^{n}x}\text{d}x+\left(n-1\right){\displaystyle \int {\sin }^{n-2}x\text{d}x}\\ =\frac{-{\sin }^{n-1}x\cos x}{n}+\frac{n-1}{n}{\displaystyle \int {\sin }^{n-2}x\text{d}x}\left(n\in Z^{+}\right)\end{array}$

${\displaystyle \int {\sin }^{n}x\text{d}x}=\frac{-{\sin }^{n-1}x\cos x}{n}+\frac{n-1}{n}{\displaystyle \int {\sin }^{n-2}x\text{d}x}$ (2.1)

类似也可以得到：

${\displaystyle \int {\cos }^{n}x\text{d}x}=\frac{{\cos }^{n-1}x\sin x}{n}+\frac{n-1}{n}{\displaystyle \int {\cos }^{n-2}x\text{d}x}$ (2.2)

称这两式为积分关联公式，这两个结果在参考文献[1]以及其他很多资料中都有详细的说明。

当n取一个大于等于2的正整数时，${\displaystyle \int {\sin }^{n}x}\text{d}x$ 与${\displaystyle \int {\sin }^{n-2}x}\text{d}x$ 有关系，且n与n − 2的奇偶性是保持一致的。当需要求解高次三角函数的积分问题时，一般通过该积分关联公式做降次处理。

3. n次正余弦函数不定积分

定义：当n为自然数时，$n!=n\cdot \left(n-1\right)\cdot \left(n-2\right)\cdots 2\cdot 1$ 称为n的阶乘

当n为偶数时，$n!!=n\cdot \left(n-2\right)\cdot \left(n-4\right)\cdots 4\cdot 2$

当n为奇数时，$n!!=n\cdot \left(n-2\right)\cdot \left(n-4\right)\cdots 3\cdot 1$ 两者均称为n的双阶乘。

特别规定：当$n\le 0$ 时，$n!=0$ 且$n!!=0$

由前面1次到5次方不定积分的情况，总结出n次正弦的不定积分有以下结论。

公式3.1：偶数次的正弦函数积分有

$S_{2n}={\displaystyle \int {\sin }^{2n}x}\text{d}x=\frac{\left(2n-1\right)!!}{\left(2n\right)!!}x-\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\cos x{\displaystyle \sum _{k=1}^n\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\sin }^{2k-1}x$

证明：当取$n=1$ 时，不难验证${\displaystyle \int {\sin }^{2}x\text{d}x}=\frac{1}{2}x-\frac{1}{2}\sin x\cos x$

假设当$n=t$ 时有$S_{2t}={\displaystyle \int {\sin }^{2t}x}\text{d}x=\frac{\left(2t-1\right)!!}{\left(2t\right)!!}x-\frac{\left(2t-1\right)!!}{\left(2t\right)!!}\cos x{\displaystyle \sum _{k=1}^t\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\sin }^{2k-1}x$

当$n=t+1$ 时，由2.1与$n=t$ 时的结论可得

$\begin{array}{c}{S}_{2\left(t+1\right)}={\displaystyle \int {\sin }^{2t+2}x}\text{d}x=-\frac{{\sin }^{2t+1}x\cos x}{2t+2}+\frac{2t+1}{2t+2}{\displaystyle \int {\sin }^{2t}x}\text{d}x\\ =-\frac{{\sin }^{2t+1}x\cos x}{2t+2}+\frac{2t+1}{2t+2}\left[\frac{\left(2t-1\right)!!}{\left(2t\right)!!}x-\frac{\left(2t-1\right)!!}{\left(2t\right)!!}\cos x{\displaystyle \sum _{k=1}^t\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\sin }^{2k-1}x\right]\\ =-\frac{{\sin }^{2t+1}x\cos x}{2t+2}+\left[\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}x-\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}\cos x{\displaystyle \sum _{k=1}^t\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\sin }^{2k-1}x\right]\\ =-\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}\frac{\left(2t\right)!!}{\left(2t+1\right)!!}{\sin }^{2t+1}x\cos x+\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}x-\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}\cos x{\displaystyle \sum _{k=1}^t\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\sin }^{2k-1}x\\ =\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}x-\left[\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}\frac{\left(2t\right)!!}{\left(2t+1\right)!!}{\sin }^{2t+1}x\cos x+\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}\cos x{\displaystyle \sum _{k=1}^t\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\sin }^{2k-1}x\right]\\ =\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}x-\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}\cos x\left[\frac{\left(2t\right)!!}{\left(2t+1\right)!!}{\sin }^{2t+1}x+{\displaystyle \sum _{k=1}^t\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\sin }^{2k-1}x\right]\\ =\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}x-\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}\cos x{\displaystyle \sum _{k=1}^{t+1}\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\sin }^{2k-1}x\end{array}$

由此可得该式成立。

公式3.2：奇数次的正弦函数积分有

$S_{2n-1}={\displaystyle \int {\sin }^{2n-1}x}\text{d}x=-\frac{\left(2n-2\right)!!}{\left(2n-1\right)!!}\cos x{\displaystyle \sum _{k=1}^n\frac{\left(2k-3\right)!!}{\left(2k-2\right)!!}}{\sin }^{2k-2}x$

证明：当$n=1$ 时，易验证：${\displaystyle \int \sin x\text{d}x}=-\cos x$

假设当$n=t$ 时有$S_{2t-1}={\displaystyle \int {\sin }^{2t-1}x}\text{d}x=\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}\cos x{\displaystyle \sum _{k=1}^t\frac{\left(2k-3\right)!!}{\left(2k-2\right)!!}}{\sin }^{2k-2}x$

当$n=t+1$ 时，由2.2与$n=t$ 时的结论可得

$\begin{array}{c}{S}_{2t+1}={\displaystyle \int {\sin }^{2t+1}x}\text{d}x=-\frac{{\sin }^{2t}x\cos x}{2t+1}+\frac{2t}{2t+1}{\displaystyle \int {\sin }^{2t-1}x}\text{d}x\\ =-\frac{{\sin }^{2t}x\cos x}{2t+1}+\frac{2t}{2t+1}\left[\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}\cos x{\displaystyle \sum _{k=1}^t\frac{\left(2k-3\right)!!}{\left(2k-2\right)!!}}{\sin }^{2k-2}x\right]\\ =-{\sin }^{2t}x\cos x\cdot \frac{\left(2t\right)!!}{\left(2t+1\right)!!}\cdot \frac{\left(2t-1\right)!!}{\left(2t\right)!!}+\frac{\left(2t\right)!!}{\left(2t+1\right)!!}\cos x{\displaystyle \sum _{k=1}^t\frac{\left(2k-3\right)!!}{\left(2k-2\right)!!}}{\sin }^{2k-2}x\\ =\frac{\left(2t\right)!!}{\left(2t+1\right)!!}\cdot \left[-{\sin }^{2t}x\cos x\cdot \frac{\left(2t-1\right)!!}{\left(2t\right)!!}+\cos x{\displaystyle \sum _{k=1}^t\frac{\left(2k-3\right)!!}{\left(2k-2\right)!!}}{\sin }^{2k-2}x\right]\\ =\frac{\left(2t\right)!!}{\left(2t+1\right)!!}\cos x\left[-{\sin }^{2t}x\cdot \frac{\left(2t-1\right)!!}{\left(2t\right)!!}+{\displaystyle \sum _{k=1}^t\frac{\left(2k-3\right)!!}{\left(2k-2\right)!!}}{\sin }^{2k-2}x\right]\\ =\frac{\left(2t\right)!!}{\left(2t+1\right)!!}\cos x\left[{\displaystyle \sum _{k=1}^{t+1}\frac{\left(2k-3\right)!!}{\left(2k-2\right)!!}}{\sin }^{2k-2}x\right]\end{array}$

由此可得该式成立。

至此正弦函数的n次不定积分问题解决。

综合上述有：

公式3.3：$S_m=\{\begin{array}{l}\frac{\left(2n-1\right)!!}{\left(2n\right)!!}x-\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\cos x{\displaystyle \sum _{k=1}^n\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\sin }^{2k-1}x\left(m=2n\right)\\ \frac{\left(2n-2\right)!!}{\left(2n-1\right)!!}\cos x{\displaystyle \sum _{k=1}^n\frac{\left(2k-3\right)!!}{\left(2k-2\right)!!}}{\sin }^{2k-2}x\left(m=2n-1\right)\end{array}$

对于余弦三角函数不定积分也分为下面两种情况。当n为正整数时

公式3.4：$C_{2n}={\displaystyle \int {\cos }^{2n}x}\text{d}x=\frac{\left(2n-1\right)!!}{\left(2n\right)!!}x+\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\sin x{\displaystyle \sum _{k=1}^n\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\cos }^{2k-1}x$

证明：当$n=1$ 时，${\displaystyle \int {\text{cos}}^{2}x}\text{d}x=\frac{1}{2}x+\frac{1}{2}\sin x\cos x$ 。上式成立。

假设当$n=t$ 时，原式成立，即

$C_{2t}={\displaystyle \int {\cos }^{2t}x}\text{d}x=\frac{\left(2t-1\right)!!}{\left(2t\right)!!}x+\frac{\left(2t-1\right)!!}{\left(2t\right)!!}\sin x{\displaystyle \sum _{k=1}^t\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\cos }^{2k-1}x$

当$n=t+1$ ，由关联公式2.2和上式得：

$\begin{array}{c}{\displaystyle \int {\cos }^{2\left(t+1\right)}x}\text{d}x=\frac{1}{2t+2}{\cos }^{2t+1}x\sin x+\frac{2t+1}{2t+2}{\displaystyle \int {\cos }^{2t}x}\text{d}x\\ =\frac{{\cos }^{2t+1}x\sin x}{2t+2}+\frac{2t+1}{2t+2}\left(\frac{\left(2t-1\right)!!}{\left(2t\right)!!}x+\frac{\left(2t-1\right)!!}{\left(2t\right)!!}\sin x{\displaystyle \sum _{k=1}^t\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}{\cos }^{2k-1}x}\right)\\ =\frac{\left(2t\right)!!{\cos }^{2t+1}x\sin x}{\left(2t+1\right)!!}\cdot \frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}+\left(\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}x+\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}\sin x{\displaystyle \sum _{k=1}^t\frac{\left(2n-2\right)!!}{\left(2n-1\right)!!}{\cos }^{2k-1}x}\right)\\ =\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}x+\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}\sin x\left(\frac{\left(2t\right)!!{\cos }^{2t+1}x}{\left(2t+1\right)!!}+{\displaystyle \sum _{k=1}^t\frac{\left(2n-2\right)!!}{\left(2n-1\right)!!}{\cos }^{2k-1}x}\right)\\ =\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}x+\frac{\left(2t+1\right)!!}{\left(2t+2\right)!!}\sin x{\displaystyle \sum _{k=1}^{t+1}\frac{\left(2n-2\right)!!}{\left(2n-1\right)!!}{\cos }^{2k-1}x}\end{array}$

由此可得该式成立。

公式3.5：$C_{2n-1}={\displaystyle \int {\cos }^{2n-1}x}\text{d}x=\frac{\left(2n-2\right)!!}{\left(2n-1\right)!!}\sin x{\displaystyle \sum _{k=1}^n\frac{\left(2k-3\right)!!}{\left(2k-2\right)!!}}{\cos }^{2k-2}x$

论证过程略，有兴趣者可尝试自行推演。类似地，对于余弦函数：

公式3.6：$C_m=\{\begin{array}{l}{\displaystyle \int {\cos }^{2n}x}\text{d}x=\frac{\left(2n-1\right)!!}{\left(2n\right)!!}x+\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\sin x{\displaystyle \sum _{k=1}^n\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\cos }^{2k-1}x\left(m=2n\right)\\ {\displaystyle \int {\cos }^{2n-1}x}\text{d}x=\frac{\left(2n-2\right)!!}{\left(2n-1\right)!!}\sin x{\displaystyle \sum _{k=1}^n\frac{\left(2k-3\right)!!}{\left(2k-2\right)!!}}{\cos }^{2k-2}x\left(m=2n-1\right)\end{array}$

4. 正余弦函数乘积${\displaystyle \int {\sin }^{m}x{\cos }^{n}x\text{d}x}$ 的不定积分

此种类型求解时，同样要考虑其中次方m，n的奇偶问题，注意这里只讨论三角函数正整数次方的不定积分。奇偶性有区别时，求解思路方法差别较大。为方便具体论证时分为两种，即至少有一个奇数次${\displaystyle \int {\sin }^{m}x{\cos }^{2n+1}x}\text{d}x$ 与两个偶数次方${\displaystyle \int {\sin }^{2m}x{\cos }^{2n}x}\text{d}x$ 。

$\begin{array}{c}{\displaystyle \int {\sin }^{m}x{\cos }^{2n+1}x}\text{d}x={\displaystyle \int {\sin }^{m}x{\left({\cos }^{2}x\right)}^n\text{d}\left(\sin x\right)}\\ ={\displaystyle \int {\sin }^{m}x{\left(1-{\sin }^{2}x\right)}^n\text{d}\left(\sin x\right)}\\ ={\displaystyle \int {\sin }^{m}x{\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k{\sin }^{2k}x}}\text{d}\left(\sin x\right)\\ ={\displaystyle \int {\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k{t}^{2k+m}}\text{d}t}\\ ={\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k}{\displaystyle \int t^{2k+m}\text{d}t}\\ ={\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k}\frac{t^{2k+m+1}}{2k+m+1}\\ ={\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k}\frac{{\sin }^{2k+m+1}x}{2k+m+1}\end{array}$

综合上述：至少有一个为奇数次方的不定积分为

公式4.1：${\displaystyle \int {\sin }^{m}x}{\cos }^{2n+1}x\text{d}x={\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k}\frac{{\sin }^{2k+m+1}x}{2k+m+1}$

当正弦的次数为奇数或两个次数都是奇数，都可以做类似的处理。下面考虑同为偶数次积分。

$\begin{array}{c}{\displaystyle \int {\sin }^{2m}x{\cos }^{2n}x}\text{d}x={\displaystyle \int {\sin }^{2m}x{\left(1-{\sin }^{2}x\right)}^n\text{d}x}\\ ={\displaystyle \int {\sin }^{2m}x{\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k{\left({\sin }^{2}x\right)}^k}}\text{d}x\\ ={\displaystyle \int {\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k{\sin }^{2m+2k}x}}\text{d}x\\ ={\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k{\displaystyle \int {\sin }^{2\left(m+k\right)}x}\text{d}x}\end{array}$

这里归根结底的问题是${\displaystyle \int {\sin }^{2\left(m+k\right)}x}\text{d}x$ ，其中$2\left(m+k\right)$ 为偶数次方。由此可利用前面的

$S_{2n}\underset{\_}{\underset{\_}{3.1式}}{\displaystyle \int {\sin }^{2n}x}\text{d}x=\frac{\left(2n-1\right)!!}{\left(2n\right)!!}x-\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\cos x{\displaystyle \sum _{k=1}^n\frac{\left(2k-2\right)!!}{\left(2k-1\right)!!}}{\sin }^{2k-1}x$

${\displaystyle \int {\sin }^{2\left(m+k\right)}x}\text{d}x\underset{\_}{\underset{\_}{3.1式}}\frac{\left(2m+2k-1\right)!!}{\left(2m+2k\right)!!}x-\frac{\left(2m+2k-1\right)!!}{\left(2m+2k\right)!!}\cos x{\displaystyle \sum _{t=1}^{m+k}\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x$

$\begin{array}{c}{\displaystyle \int {\sin }^{2m}x{\cos }^{2n}x\text{d}x}={\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k{\displaystyle \int {\sin }^{2\left(m+k\right)}x}\text{d}x}\\ ={\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k}\left[\frac{\left(2m+2k-1\right)!!}{\left(2m+2k\right)!!}x-\frac{\left(2m+2k-1\right)!!}{\left(2m+2k\right)!!}\cos x{\displaystyle \sum _{t=1}^{m+k}\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x\right]\end{array}$

综合上述：正余弦均为偶数次方的不定积分

公式4.2：${\displaystyle \int {\sin }^{2m}x}{\cos }^{2n}x\text{d}x={\displaystyle \sum _{k=0}^n{C}_n^k{\left(-1\right)}^k}\left[\frac{\left(2m+2k-1\right)!!}{\left(2m+2k\right)!!}x-\frac{\left(2m+2k-1\right)!!}{\left(2m+2k\right)!!}\cos x{\displaystyle \sum _{t=1}^{m+k}\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x\right]$

5. 实用举例

求解${\displaystyle \int {\sin }^{4}x{\cos }^{6}x}\text{d}x$ ，常规的办法有两个。第一是凑微分，凑完发现前面的次方有一个变成奇数次方，后续是转化不了的。第二是先利用基本公式进行三角变换，这样会出现正(余)弦10次方，8次方等的不定积分，次方高而且运算量很大，显然不合适。这里直接套用4.2

$\begin{array}{l}{\displaystyle \int {\sin }^{4}x}{\cos }^{6}x\text{d}x\underset{\_}{\underset{\_}{4.2式}}{\displaystyle \sum _{k=0}^3{C}_3^k{\left(-1\right)}^k}\left[\frac{\left(4+2k-1\right)!!}{\left(4+2k\right)!!}x-\frac{\left(4+2k-1\right)!!}{\left(4+2k\right)!!}\cos x{\displaystyle \sum _{t=1}^{2+k}\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x\right]\\ =\frac{3}{8}x-\frac{3}{8}\cos x{\displaystyle \sum _{t=1}^2\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x-\left[\frac{15}{16}x-\frac{15}{16}\cos x{\displaystyle \sum _{t=1}^3\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x\right]\\ +\left[\frac{105}{128}x-\frac{105}{128}\cos x{\displaystyle \sum _{t=1}^4\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x\right]-\left[\frac{945}{3840}x-\frac{945}{3840}\cos x{\displaystyle \sum _{t=1}^5\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x\right]\\ =\frac{3}{256}x-\frac{1}{4}\cos x{\sin }^{3}x+\frac{15}{16}\cos x{\displaystyle \sum _{t=1}^3\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x\\ -\frac{105}{128}\cos x{\displaystyle \sum _{t=1}^4\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x+\frac{945}{3840}\cos x{\displaystyle \sum _{t=1}^5\frac{\left(2t-2\right)!!}{\left(2t-1\right)!!}}{\sin }^{2t-1}x\end{array}$

这个结果显式是可以写出来的，比起上述分析中的两个思路，明显是可行的。这里的求和符号就不再展开。

NOTES

^*通讯作者。