# 三角函数的积分计算技巧Integral Calculation Skills of Trigonometric Functions

DOI: 10.12677/PM.2021.117158, PDF, HTML, XML, 下载: 19  浏览: 50

Abstract: The integral solution of trigonometric rational function is a very important content in calculus. In the integration of trigonometric functions, converting the integrand into known trigonometric functions and using variable substitution are two important algorithms for calculating the inte-gration of trigonometric functions. But for the trigonometric function integration algorithm is not limited to the above two; we should master more methods to solve the trigonometric function in-tegration according to the characteristics of the integrand. In this paper, five methods and tech-niques for processing trigonometric rational function integration are given by practical examples. They are the same deformation of trigonometric function, variable substitution method, universal substitution method, parity and recursion method.

1. 引言

2. 一般区间上的三角函数的定积分的算法

2.1. 利用三角函数恒等变形 [2]

$\begin{array}{c}原式=\frac{1}{2}{\int }^{\text{​}}{e}^{x}{\mathrm{sec}}^{2}\frac{x}{2}\text{d}x+{\int }^{\text{​}}{e}^{x}\mathrm{tan}\frac{x}{2}\text{d}x={e}^{x}\mathrm{tan}\frac{x}{2}-{\int }^{\text{​}}{e}^{x}\mathrm{tan}\frac{x}{2}\text{d}x+{\int }^{\text{​}}{e}^{x}\mathrm{tan}\frac{x}{2}\text{d}x+c\\ ={e}^{x}\mathrm{tan}\frac{x}{2}+c\end{array}$

2.2. 利用变量代换 [3]

${\int }_{a}^{b}\frac{1}{\sqrt{{a}^{2}+{x}^{2}}}\text{d}x={\int }_{{a}_{1}}^{{b}_{1}}\frac{a{\mathrm{sec}}^{2}t\text{d}t}{a\mathrm{sec}t}={\int }_{{a}_{1}}^{{b}_{1}}a\mathrm{sec}t\text{d}t=\mathrm{ln}|\mathrm{sec}t+\mathrm{tan}t|+c$.

$\mathrm{tan}t=\frac{x}{a}$, $\mathrm{sec}t=\frac{\sqrt{{a}^{2}+{x}^{2}}}{a}$ 代入，

2.3. 万能换元法 [4]

$\mathrm{sin}x=\frac{2t}{1+{t}^{2}}$, $\mathrm{cos}x=\frac{1-{t}^{2}}{1+{t}^{2}}$, $\text{d}x=\frac{2\text{d}t}{1+{t}^{2}}$.

2.4. 利用被积函数奇偶性

${\int }_{0}^{\frac{\pi }{2}}f\left(\mathrm{sin}x\right)\text{d}x=-{\int }_{\frac{\pi }{2}}^{0}f\left(\mathrm{cos}t\right)\text{d}t={\int }_{0}^{\frac{\pi }{2}}f\left(\mathrm{cos}x\right)\text{d}x$ ;

$原式=\frac{1}{2}\left({\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{cos}x}{\mathrm{sin}x+\mathrm{cos}x}\text{d}x+{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x}{\mathrm{sin}x+\mathrm{cos}x}\text{d}x\right)=\frac{1}{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{cos}x+\mathrm{sin}x}{\mathrm{sin}x+\mathrm{cos}x}\text{d}x=\frac{\pi }{4}$

$=2{\int }_{0}^{\infty }\frac{\frac{2\text{d}t}{1+{t}^{2}}}{2-\frac{1-{t}^{2}}{1+{t}^{2}}}=4{\int }_{0}^{\infty }\frac{\text{d}t}{1+3{t}^{2}}=4\frac{1}{\sqrt{3}}\mathrm{arctan}\sqrt{3}t|\begin{array}{c}\infty \\ 0\end{array}=\frac{2\pi }{\sqrt{3}}$

2.5. 利用递推公式

$I\left(m,n\right)\left\{\begin{array}{l}\frac{{\mathrm{cos}}^{m-1}x+{\mathrm{sin}}^{n+1}x}{m+n}+\frac{m-1}{m+n}I\left(m-2,n\right)\cdot \cdots \left(m\ge 2,\text{}n\ge 0\right);\hfill \\ -\frac{{\mathrm{cos}}^{m+1}x+{\mathrm{sin}}^{n-1}x}{m+n}+\frac{n-1}{m+n}I\left(m,n-2\right)\cdot \cdots \left(m\ge 0,\text{}n\ge 2\right);\hfill \end{array}$

$\begin{array}{c}I\left(m,n\right)={\int }^{\text{​}}{\mathrm{cos}}^{m}x{\mathrm{sin}}^{n}x\text{d}x={\int }^{\text{​}}{\mathrm{cos}}^{m-1}x{\mathrm{sin}}^{n}x\text{d}\mathrm{sin}x=\frac{1}{n+1}{\int }^{\text{​}}{\mathrm{cos}}^{m-1}x\text{d}{\mathrm{sin}}^{n+1}x\\ =\frac{1}{n+1}{\mathrm{cos}}^{m-1}x{\mathrm{sin}}^{n+1}x+\frac{m-1}{n+1}{\int }^{\text{​}}{\mathrm{cos}}^{m-2}x{\mathrm{sin}}^{n+2}x\text{d}x\\ =\frac{1}{n+1}{\mathrm{cos}}^{m-1}x{\mathrm{sin}}^{n+1}x+\frac{m-1}{n+1}{\int }^{\text{​}}{\mathrm{cos}}^{m-2}x{\mathrm{sin}}^{n}x\cdot \left(1-{\mathrm{cos}}^{2}x\right)\text{d}x\\ =\frac{1}{n+1}{\mathrm{cos}}^{m-1}x{\mathrm{sin}}^{n+1}x+\frac{m-1}{n+1}I\left(m-2,n\right)-\frac{m-1}{n+1}{\int }^{\text{​}}{\mathrm{cos}}^{m}x{\mathrm{sin}}^{n}x\text{d}x\\ =\frac{1}{n+1}{\mathrm{cos}}^{m-1}x{\mathrm{sin}}^{n+1}x+\frac{m-1}{n+1}I\left(m-2,n\right)-\frac{m-1}{n+1}I\left(m,n\right)\end{array}$

$\text{I}\left(m,n\right)=-\frac{{\mathrm{cos}}^{m+1}x+{\mathrm{sin}}^{n-1}x}{m+n}+\frac{n-1}{m+n}\text{I}\left(m,n-2\right)\cdot \cdots \left(m\ge 0,\text{}n\ge 2\right)$.

3. 一类特殊区间上的三角函数定积分的算法

$I\left(m,n\right)=\frac{m-1}{m+n}I\left(m-2,n\right)=\frac{n-1}{m+n}I\left(m,n-2\right)=\left\{\begin{array}{l}\frac{\left(m-1\right)!!\left(n-1\right)!!}{\left(m+n\right)!!}\text{}m,n不全为偶数\hfill \\ \frac{\left(m-1\right)!!\left(n-1\right)!!}{\left(m+n\right)!!}\frac{\pi }{2}\text{}m,n全为偶数\hfill \end{array}$

$I\left(m,n\right)\left\{\begin{array}{l}{\frac{{\mathrm{cos}}^{m-1}x+{\mathrm{sin}}^{n+1}x}{m+n}|}_{0}^{\frac{\pi }{2}}+\frac{m-1}{m+n}I\left(m-2,n\right)=\frac{n-1}{m+n}I\left(m,n-2\right)\text{}\left(m\ge 2,\text{}n\ge 0\right);\hfill \\ {-\frac{{\mathrm{cos}}^{m+1}x+{\mathrm{sin}}^{n-1}x}{m+n}|}_{0}^{\frac{\pi }{2}}+\frac{n-1}{m+n}I\left(m,n-2\right)=\frac{m-1}{m+n}I\left(m-2,n\right)\text{}\left(m\ge 0,\text{}n\ge 2\right);\hfill \end{array}$

4. 结论

NOTES

1李扬数学分析强化讲义[Z]，2021。

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