三角积分表

这个页面列举了一些含有三角函数的不定积分公式。以下的积分在被积函数的定义域内点的一个邻域上进行，且均省略积分常数且均假设伸缩系数$a \neq 0.$

被积函数仅为一种三角函数的多项式

一次（基本公式）

1. $\int \sin ax \mathrm{d}x = -\dfrac{1}{a} \cos ax.$
2. $\int \cos ax \mathrm{d}x = \dfrac{1}{a} \sin ax.$
3. $\int \tan ax \mathrm{d}x = -\dfrac{1}{a} \ln |\cos ax|.$
4. $\int \cot ax \mathrm{d}x = \dfrac{1}{a} \ln |\sin ax|.$
5. $\int \sec ax \mathrm{d}x = \dfrac{1}{a} \ln |\sec ax + \tan ax| = \dfrac{1}{a} \ln \left| \tan \dfrac{ax}{2} \right|.$
6. $\int \csc ax \mathrm{d}x = \dfrac{1}{a} \ln |\csc ax - \cot ax| = \dfrac{1}{a} \ln \left| \tan \left( \dfrac{ax}{2} + \dfrac{\pi}{4}\right) \right|.$

二次

1. $\int \sin^2 ax \mathrm{d}x = -\dfrac{1}{4a} \sin 2ax + \dfrac{1}{2}x.$
2. $\int \cos^2 ax \mathrm{d}x = \dfrac{1}{4a} \sin 2ax + \dfrac{1}{2}x.$
3. $\int \tan^2 ax \mathrm{d}x = \dfrac{1}{a} \tan ax - x.$
4. $\int \cot^2 ax \mathrm{d}x = -\dfrac{1}{a} \cot ax - x.$
5. $\int \sec^2 ax \mathrm{d}x = \dfrac{1}{a} \tan ax.$
6. $\int \csc^2 ax \mathrm{d}x = -\dfrac{1}{a} \cot ax.$

高次：递推公式

以下假设$n>1$是正整数。

1. $\int \sin^n ax \mathrm{d}x = -\dfrac{1}{na} \sin^{n-1} ax \cos ax + \dfrac{n-1}{n} \int \sin^{n-2} ax \mathrm{d}x.$
2. $\int \cos^n ax \mathrm{d}x = \dfrac{1}{na} \cos^{n-1} ax \sin ax + \dfrac{n-1}{n} \int \cos^{n-2} ax \mathrm{d}x.$
3. $\int \tan^n ax \mathrm{d}x = \dfrac{1}{(n-1)a} \tan^{n-1} ax - \int \tan^{n-2} ax \mathrm{d}x.$
4. $\int \cot^n ax \mathrm{d}x = -\dfrac{1}{(n-1)a} \cot^{n-1} ax - \int \cot^{n-2} ax \mathrm{d}x.$
5. $\int \sec^n ax \mathrm{d}x = \dfrac{1}{(n-1)a} \sec^{n-2} ax \tan ax + \dfrac{n-2}{n-1} \int \sec^{n-2} ax \mathrm{d}x.$
6. $\int \csc^n ax \mathrm{d}x = -\dfrac{1}{(n-1)a} \csc^{n-2} ax \cot ax + \dfrac{n-2}{n-1} \int \csc^{n-2} ax \mathrm{d}x.$

被积函数是幂函数与三角函数的乘积

幂函数为一次式

1. $\int x \sin ax \mathrm{d}x = \dfrac{1}{a^2} \sin ax - \dfrac{1}{a} x \cos ax.$
2. $\int x \cos ax \mathrm{d}x = \dfrac{1}{a^2} \cos ax + \dfrac{1}{a} x \sin ax.$
   其它四类不是初等函数。
3. $\int x \sin^2 ax \mathrm{d}x = \dfrac{1}{4} x^2 - \dfrac{1}{4a} x \sin 2ax - \dfrac{1}{8a^2} \cos 2ax.$
4. $\int x \cos^2 ax \mathrm{d}x = \dfrac{1}{4} x^2 + \dfrac{1}{4a} x \sin 2ax + \dfrac{1}{8a^2} \cos 2ax.$
5. $\int x \tan^2 ax \mathrm{d}x = -\dfrac{1}{2} x^2 + \dfrac{1}{ a} x \tan ax + \dfrac{1}{ a^2} \ln|\cos ax|.$
6. $\int x \cot^2 ax \mathrm{d}x = -\dfrac{1}{2} x^2 - \dfrac{1}{ a} x \cot ax + \dfrac{1}{ a^2} \ln|\sin ax|.$
7. $\int x \sec^2 ax \mathrm{d}x = \dfrac{1}{a} x \tan ax + \dfrac{1}{a^2} \ln |\cos ax|.$
8. $\int x \csc^2 ax \mathrm{d}x = -\dfrac{1}{a} x \cot ax + \dfrac{1}{a^2} \ln |\sin ax|.$

幂函数为高次：递推公式

以下假设$n>1$是正整数。

1. $\int x^n \sin ax \mathrm{d}x = -\dfrac{1}{a} x^n \cos ax + \dfrac{n}{a} \int x^{n-1} \cos ax \mathrm{d}x.$
2. $\int x^n \cos ax \mathrm{d}x = \dfrac{1}{a} x^n \sin ax - \dfrac{n}{a} \int x^{n-1} \sin ax \mathrm{d}x.$
3. $\int x^{-n} \sin ax \mathrm{d}x = -\dfrac{1}{n-1} x^{-n+1} \sin ax + \dfrac{a}{n-1} \int x^{-n+1} \cos ax \mathrm{d}x.$
4. $\int x^{-n} \cos ax \mathrm{d}x = -\dfrac{1}{n-1} x^{-n+1} \cos ax - \dfrac{a}{n-1} \int x^{-n+1} \sin ax \mathrm{d}x.$

被积函数是三角函数有理式

以下均假设被积函数的分母有定义，且结果中的函数表达式在有极限的地方均有定义。这类问题的通用做法是半角代换，但是往往在一些问题中半角代换的结果很复杂，甚至使得原函数不再连续，其它可考虑的方法主要有三角恒等变形化简和配对积分法。

含有 a sin x + b cos x 的积分

1. $\int \dfrac{A\sin x + B\cos x}{a\sin x + b\cos x} \mathrm{d}x$

   $I_1 := \int \dfrac{\sin x}{a\sin x + b\cos x} \mathrm{d}x = \dfrac{bx + a\ln|a\sin x + b\cos x|}{a^2+b^2}.$

   $I_2 := \int \dfrac{\cos x}{a\sin x + b\cos x} \mathrm{d}x = \dfrac{ax - b\ln|a\sin x + b\cos x|}{a^2+b^2}.$

2. $\int \dfrac{A \sin x + B \cos x + C}{a\sin x + b\cos x + c} \mathrm{d}x:$

   $I_1 := \int \dfrac{\sin x}{a\sin x + b\cos x + c} \mathrm{d}x = \dfrac{a}{a^2+b^2}x - \dfrac{b}{a^2+b^2} \ln |a\sin x + b\cos x + c| - \dfrac{ac}{a^2+b^2} I_3.$

   $I_2 := \int \dfrac{\cos x}{a\sin x + b\cos x + c} \mathrm{d}x = \dfrac{b}{a^2+b^2}x - \dfrac{a}{a^2+b^2} \ln |a\sin x + b\cos x + c| - \dfrac{bc}{a^2+b^2} I_3.$

   $I_3 := \int \dfrac{1}{a\sin x + b\cos x + c} \mathrm{d}x = \int \dfrac{\mathrm{d}x}{c + \sqrt{a^2+b^2} \sin(x + \arctan(b/a))}.$（见下文）

3. $\int \dfrac{A\sin x + B\cos x}{(a\sin x + b\cos x)^2} \mathrm{d}x$

   $I_1 := \int \dfrac{\sin x}{(a\sin x + b\cos x)^2} \mathrm{d}x = \dfrac{a}{(a^2+b^2)^{3/2}} \ln \left| \tan \left( \dfrac{x}{2} + \dfrac{1}{2} \arctan \dfrac{b}{a} \right) \right| + \dfrac{b}{a^2+b^2} \dfrac{1}{a\sin x + b\cos x}.$

   $I_2 := \int \dfrac{\cos x}{(a\sin x + b\cos x)^2} \mathrm{d}x = \dfrac{b}{(a^2+b^2)^{3/2}} \ln \left| \tan \left( \dfrac{x}{2} + \dfrac{1}{2} \arctan \dfrac{b}{a} \right) \right| - \dfrac{a}{a^2+b^2} \dfrac{1}{a\sin x + b\cos x}.$

4. $\int \dfrac{\mathrm{d}x}{(a\sin x + b\cos x)^n} = \dfrac{1}{(n-1)(a^2+b^2)} \left( \dfrac{b\sin x - a\cos x}{(a\sin x + b\cos x)^{n-1}} + (n-2) \int \dfrac{\mathrm{d}x}{(a\sin x + b\cos x)^{n-1}} \right).$
5. $\int \dfrac{A\sin x + B\cos x}{(a\sin x+b\cos x)^n} \mathrm{d}x = \dfrac{Aa+Bb}{a^2+b^2} \int \dfrac{\mathrm{d}x}{(a\sin x + b\cos x)^{n-1}} - \dfrac{Ba - Ab}{(n-1)(a^2+b^2)} \dfrac{1}{(a\sin x + b\cos x)^{n-1}}.$
6. $\int \dfrac{A\sin^2 x + 2B \sin x \cos x + C\cos^2 x}{a\sin x + b\cos x} \mathrm{d}x$

   $I_1 := \int \dfrac{\sin^2 x}{a\sin x + b\cos x} \mathrm{d}x = \dfrac{b^2}{(a^2+b^2)^{3/2}} \ln \left| \tan \dfrac{x}{2} + \dfrac{1}{2} \arctan \dfrac{b}{a} \right| - \dfrac{a\cos x + b\sin x}{a^2+b^2}.$

   $I_2 := \int \dfrac{\sin x \cos x}{a\sin x + b\cos x} \mathrm{d}x = -\dfrac{ab}{(a^2+b^2)^{3/2}} \ln \left| \tan \dfrac{x}{2} + \dfrac{1}{2} \arctan \dfrac{b}{a} \right| + \dfrac{a\sin x - b\cos x}{a^2+b^2}.$

   $I_3 := \int \dfrac{\cos^2 x}{a\sin x + b\cos x} \mathrm{d}x = \dfrac{a^2}{(a^2+b^2)^{3/2}} \ln \left| \tan \dfrac{x}{2} + \dfrac{1}{2} \arctan \dfrac{b}{a} \right| + \dfrac{a\cos x + b\sin x}{a^2+b^2}.$

7. $\int \dfrac{A\sin^2 x + 2B \sin x \cos x + C\cos^2 x}{(a\sin x + b\cos x)^2} \mathrm{d}x$

   $I_1 := \int \dfrac{\sin^2 x}{(a\sin x + b\cos x)^2} \mathrm{d}x = \dfrac{b^2}{(a^2+b^2)^2} \dfrac{b\sin x - a\cos x}{a\sin x + b\cos x} + \dfrac{a^2-b^2}{(a^2+b^2)^2} x - \dfrac{2ab}{(a^2+b^2)^2} \ln |a\sin x + b\cos x|.$

   $I_2 := \int \dfrac{\sin x \cos x}{(a\sin x + b\cos x)^2} \mathrm{d}x = \dfrac{a^2}{(a^2+b^2)^2} \dfrac{b\sin x - a\cos x}{a\sin x + b\cos x} - \dfrac{a^2-b^2}{(a^2+b^2)^2} x + \dfrac{2ab}{(a^2+b^2)^2} \ln |a\sin x + b\cos x|.$

   $\begin{align}I_3 := \int \dfrac{\cos^2 x}{(a\sin x + b\cos x)^2} \mathrm{d}x & = -\dfrac{b}{2a} I_1 + \dfrac{a}{2b} I_3 \\ & = - \dfrac{ab}{(a^2+b^2)^2} \dfrac{b\sin x - a\cos x}{a\sin x + b\cos x} - \dfrac{(a^2-b^2)^2}{2ab(a^2+b^2)^2} x + \dfrac{2(a^2-b^2)}{(a^2+b^2)^2} \ln |a\sin x + b\cos x|.\end{align}$

8. $\int \dfrac{A\sin^3 x + B\cos^3 x}{\sin x + \cos x} \mathrm{d}x$

   $I_1 := \int \dfrac{\sin^3 x}{\sin x + \cos x} \mathrm{d}x = \dfrac{1}{2}x + \dfrac{1}{4} \cos^2 x + \dfrac{1}{4} \cos x \sin x + \dfrac{1}{4} \ln |\sin x + \cos x|.$

   $I_2 := \int \dfrac{\cos^3 x}{\sin x + \cos x} \mathrm{d}x = \dfrac{1}{2}x + \dfrac{1}{4} \cos^2 x - \dfrac{1}{4} \cos x \sin x - \dfrac{1}{4} \ln |\sin x + \cos x|.$

含有 a + b sin x 或 cos x 的积分

主要考虑$\int \dfrac{1}{a + b\sin x} \mathrm{d}x$和$\int \dfrac{1}{a - b\cos x} \mathrm{d}x$。若积分为$\int \dfrac{1}{a - b\sin x} \mathrm{d}x, \quad |a| > |b|.$仅需注意到$\sin(x+\pi)=-\sin x$，若积分为$\int \dfrac{1}{a + b\cos x} \mathrm{d}x, \quad |a| > |b|.$仅需注意到$\cos(x+\pi)=-\cos x.$若分母是$\sin x,\cos x$的多项式，可以使用部分分式分解。

1. 当$|a| > |b| > 0$
   1. $\int \dfrac{1}{a + b\sin x} \mathrm{d}x.$用半角代换的结果可能不是连续的，下述结果中第一个是连续的，第二个不连续。

      $I_1 := \int \dfrac{1}{a + b \sin x} \mathrm{d}x = \dfrac{1}{\sqrt{a^2-b^2}} \left( 2\arctan \dfrac{b\cos x + a\sin x + b - \sqrt{a^2-b^2} \sin x}{b\sin x - a\cos x + a + \sqrt{a^2-b^2}} + x \right).$

      $I_1 = -\dfrac{2}{\sqrt{a^2-b^2}} \arctan \left( \sqrt{\dfrac{a-b}{a+b}} \tan \left( \dfrac{\pi}{4} - \dfrac{x}{2} \right) \right).$

   2. $\int \dfrac{1}{a - b\cos x} \mathrm{d}x.$用半角代换的结果可能不是连续的，下述结果中第一个是连续的，第二个不连续。

      $I_1 := \int \dfrac{1}{a - b \cos x} \mathrm{d}x = \dfrac{1}{\sqrt{a^2-b^2}} \left( 2 \arctan \dfrac{(\sqrt{a^2-b^2}-b-a) \sin x}{(a+b-\sqrt{a^2-b^2})\cos x - a - b - \sqrt{a^2-b^2}} + x \right).$

      $I_1 = \int \dfrac{1}{a - b\cos x} \mathrm{d}x = \dfrac{1}{\sqrt{a^2-b^2}} \arctan \dfrac{\sqrt{a^2-b^2} \sin x}{a\cos x - b}.$

2. 当$|a|=|b|$时就是下述情形：

   $\int \dfrac{1}{1+\sin x}\mathrm{d}x = - \dfrac{2}{1+\tan \frac{x}{2}}.$

   $\int \dfrac{1}{1+\cos x}\mathrm{d}x = \tan \dfrac{x}{2}.$

3. 当$0 < |a| < |b|$时用半角代换容易求得，这时被积函数存在无穷间断点。

   $\int \dfrac{\mathrm{d}x}{a + b\sin x} = \dfrac{1}{\sqrt{b^2-a^2}} \ln \left| \dfrac{a\tan\frac{x}{2}+b-\sqrt{b^2-a^2}}{a\tan\frac{x}{2}+b+\sqrt{b^2-a^2}} \right|.$

   $\int \dfrac{\mathrm{d}x}{a - b\cos x} = \dfrac{1}{\sqrt{b^2-a^2}} \ln \left| \dfrac{(a+b)\tan\frac{x}{2}-\sqrt{b^2-a^2}}{(a+b)\tan\frac{x}{2}+\sqrt{b^2-a^2}} \right|.$

含有 a + b sin x cos x 的积分

1. $\int \dfrac{A\cos x + B\sin x}{a + b\sin x \cos x} \mathrm{d}x$

   $I_1 = \int \dfrac{\cos x}{a + b\sin x \cos x} \mathrm{d}x, \quad I_2 = \int \dfrac{\sin x}{a + b\sin x \cos x} \mathrm{d}x.$

   1. 当$2a < b$时，注意到
      $\begin{cases} I_1 + I_2 & = \dfrac{1}{\sqrt{b(b+2a)}} \ln \left| \dfrac{\sqrt{b+2a} + \sqrt{b} (\sin x - \cos x)}{\sqrt{b+2a} - \sqrt{b} (\sin x - \cos x)} \right|, \\ I_1 - I_2 & = \dfrac{1}{\sqrt{b(b-2a)}} \ln \left|\dfrac{\sqrt{b} (\sin x + \cos x) - \sqrt{b-2a}}{\sqrt{b} (\sin x + \cos x) + \sqrt{b-2a}} \right|. \end{cases}$
   2. 当$2a>b$时，注意到
      $\begin{cases} I_1 + I_2 & = \dfrac{1}{\sqrt{b(b+2a)}} \ln \left| \dfrac{\sqrt{b+2a} + \sqrt{b} (\sin x - \cos x)}{\sqrt{b+2a} - \sqrt{b} (\sin x - \cos x)} \right|, \\ I_1 - I_2 & = \dfrac{2}{\sqrt{b(2a-b)}} \arctan \dfrac{\sqrt{b}(\sin x + \cos x)}{\sqrt{2a-b}}. \end{cases}$

含有 a tan x + b cot x 的积分

1. $\int \dfrac{A\tan x + B\cot x}{a\tan x + b\cot x} \mathrm{d}x$

   $I_1 := \int \dfrac{\tan x}{a\tan x + b\cot x}\mathrm{d}x = \dfrac{1}{a-b} \left( x - \dfrac{1}{\sqrt{ab}} \arctan \left( \sqrt{\dfrac{a}{b}} \tan x \right) \right)$

   $I_2 := \int \dfrac{\cot x}{a\tan x + b\cot x}\mathrm{d}x = \dfrac{1}{b-a} \left( x - \dfrac{1}{\sqrt{ab}} \arctan \left( \sqrt{\dfrac{a}{b}} \tan x \right) \right)$

2. $\int \dfrac{A\tan^2 x + B\cot^2 x}{a\tan x + b\cot x} \mathrm{d}x$

   $I_1 := \int \dfrac{\tan^2 x}{a\tan x + b\cot x} \mathrm{d}x = -\dfrac{1}{2a} \left( 2\ln|\cos x| + \dfrac{b}{a-b} \ln|a\sin^2 x + b\cos^2 x| \right).$

   $I_{2}:=\int {\dfrac {\cot ^{2}x}{a\tan x+b\cot x}}\mathrm {d} x=-{\dfrac {1}{2b}}\left(-2\ln |\sin x|+{\dfrac {a}{a-b}}\ln |a\sin ^{2}x+b\cos ^{2}x|\right).$