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
≠
0.
{\displaystyle a \neq 0.}
目录
1 被积函数仅为一种三角函数的多项式
1.1 一次（基本公式）
1.2 二次
1.3 高次：递推公式
2 被积函数是幂函数与三角函数的乘积
2.1 幂函数为一次式
2.2 幂函数为高次：递推公式
3 被积函数是三角函数有理式
3.1 含有 a sin x + b cos x 的积分
3.2 含有 a + b sin x 或 cos x 的积分
3.3 含有 a + b sin x cos x 的积分
3.4 含有 a tan x + b cot x 的积分
被积函数仅为一种三角函数的多项式[]
一次（基本公式）[]
∫
sin
a
x
d
x
=
−
1
a
cos
a
x
.
{\displaystyle \int \sin ax \mathrm{d}x = -\dfrac{1}{a} \cos ax.}
∫
cos
a
x
d
x
=
1
a
sin
a
x
.
{\displaystyle \int \cos ax \mathrm{d}x = \dfrac{1}{a} \sin ax.}
∫
tan
a
x
d
x
=
−
1
a
ln
|
cos
a
x
|
.
{\displaystyle \int \tan ax \mathrm{d}x = -\dfrac{1}{a} \ln |\cos ax|.}
∫
cot
a
x
d
x
=
1
a
ln
|
sin
a
x
|
.
{\displaystyle \int \cot ax \mathrm{d}x = \dfrac{1}{a} \ln |\sin ax|.}
∫
sec
a
x
d
x
=
1
a
ln
|
sec
a
x
+
tan
a
x
|
=
1
a
ln
|
tan
a
x
2
|
.
{\displaystyle \int \sec ax \mathrm{d}x = \dfrac{1}{a} \ln |\sec ax + \tan ax| = \dfrac{1}{a} \ln \left| \tan \dfrac{ax}{2} \right|.}
∫
csc
a
x
d
x
=
1
a
ln
|
csc
a
x
−
cot
a
x
|
=
1
a
ln
|
tan
(
a
x
2
+
π
4
)
|
.
{\displaystyle \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[]
∫
sin
2
a
x
d
x
=
−
1
4
a
sin
2
a
x
+
1
2
x
.
{\displaystyle \int \sin^2 ax \mathrm{d}x = -\dfrac{1}{4a} \sin 2ax + \dfrac{1}{2}x.}
∫
cos
2
a
x
d
x
=
1
4
a
sin
2
a
x
+
1
2
x
.
{\displaystyle \int \cos^2 ax \mathrm{d}x = \dfrac{1}{4a} \sin 2ax + \dfrac{1}{2}x.}
∫
tan
2
a
x
d
x
=
1
a
tan
a
x
−
x
.
{\displaystyle \int \tan^2 ax \mathrm{d}x = \dfrac{1}{a} \tan ax - x.}
∫
cot
2
a
x
d
x
=
−
1
a
cot
a
x
−
x
.
{\displaystyle \int \cot^2 ax \mathrm{d}x = -\dfrac{1}{a} \cot ax - x.}
∫
sec
2
a
x
d
x
=
1
a
tan
a
x
.
{\displaystyle \int \sec^2 ax \mathrm{d}x = \dfrac{1}{a} \tan ax.}
∫
csc
2
a
x
d
x
=
−
1
a
cot
a
x
.
{\displaystyle \int \csc^2 ax \mathrm{d}x = -\dfrac{1}{a} \cot ax.}
高次：递推公式[]
以下假设
n
>
1
{\displaystyle n>1}
是正整数。
∫
sin
n
a
x
d
x
=
−
1
n
a
sin
n
−
1
a
x
cos
a
x
+
n
−
1
n
∫
sin
n
−
2
a
x
d
x
.
{\displaystyle \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.}
∫
cos
n
a
x
d
x
=
1
n
a
cos
n
−
1
a
x
sin
a
x
+
n
−
1
n
∫
cos
n
−
2
a
x
d
x
.
{\displaystyle \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.}
∫
tan
n
a
x
d
x
=
1
(
n
−
1
)
a
tan
n
−
1
a
x
−
∫
tan
n
−
2
a
x
d
x
.
{\displaystyle \int \tan^n ax \mathrm{d}x = \dfrac{1}{(n-1)a} \tan^{n-1} ax - \int \tan^{n-2} ax \mathrm{d}x.}
∫
cot
n
a
x
d
x
=
−
1
(
n
−
1
)
a
cot
n
−
1
a
x
−
∫
cot
n
−
2
a
x
d
x
.
{\displaystyle \int \cot^n ax \mathrm{d}x = -\dfrac{1}{(n-1)a} \cot^{n-1} ax - \int \cot^{n-2} ax \mathrm{d}x.}
∫
sec
n
a
x
d
x
=
1
(
n
−
1
)
a
sec
n
−
2
a
x
tan
a
x
+
n
−
2
n
−
1
∫
sec
n
−
2
a
x
d
x
.
{\displaystyle \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.}
∫
csc
n
a
x
d
x
=
−
1
(
n
−
1
)
a
csc
n
−
2
a
x
cot
a
x
+
n
−
2
n
−
1
∫
csc
n
−
2
a
x
d
x
.
{\displaystyle \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[]
∫
x
sin
a
x
d
x
=
1
a
2
sin
a
x
−
1
a
x
cos
a
x
.
{\displaystyle \int x \sin ax \mathrm{d}x = \dfrac{1}{a^2} \sin ax - \dfrac{1}{a} x \cos ax.}
∫
x
cos
a
x
d
x
=
1
a
2
cos
a
x
+
1
a
x
sin
a
x
.
{\displaystyle \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。
∫
x
sin
2
a
x
d
x
=
1
4
x
2
−
1
4
a
x
sin
2
a
x
−
1
8
a
2
cos
2
a
x
.
{\displaystyle \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.}
∫
x
cos
2
a
x
d
x
=
1
4
x
2
+
1
4
a
x
sin
2
a
x
+
1
8
a
2
cos
2
a
x
.
{\displaystyle \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.}
∫
x
tan
2
a
x
d
x
=
−
1
2
x
2
+
1
a
x
tan
a
x
+
1
a
2
ln
|
cos
a
x
|
.
{\displaystyle \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|.}
∫
x
cot
2
a
x
d
x
=
−
1
2
x
2
−
1
a
x
cot
a
x
+
1
a
2
ln
|
sin
a
x
|
.
{\displaystyle \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|.}
∫
x
sec
2
a
x
d
x
=
1
a
x
tan
a
x
+
1
a
2
ln
|
cos
a
x
|
.
{\displaystyle \int x \sec^2 ax \mathrm{d}x = \dfrac{1}{a} x \tan ax + \dfrac{1}{a^2} \ln |\cos ax|.}
∫
x
csc
2
a
x
d
x
=
−
1
a
x
cot
a
x
+
1
a
2
ln
|
sin
a
x
|
.
{\displaystyle \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
{\displaystyle n>1}
是正整数。
∫
x
n
sin
a
x
d
x
=
−
1
a
x
n
cos
a
x
+
n
a
∫
x
n
−
1
cos
a
x
d
x
.
{\displaystyle \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.}
∫
x
n
cos
a
x
d
x
=
1
a
x
n
sin
a
x
−
n
a
∫
x
n
−
1
sin
a
x
d
x
.
{\displaystyle \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.}
∫
x
−
n
sin
a
x
d
x
=
−
1
n
−
1
x
−
n
+
1
sin
a
x
+
a
n
−
1
∫
x
−
n
+
1
cos
a
x
d
x
.
{\displaystyle \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.}
∫
x
−
n
cos
a
x
d
x
=
−
1
n
−
1
x
−
n
+
1
cos
a
x
−
a
n
−
1
∫
x
−
n
+
1
sin
a
x
d
x
.
{\displaystyle \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[]
∫
A
sin
x
+
B
cos
x
a
sin
x
+
b
cos
x
d
x
{\displaystyle \int \dfrac{A\sin x + B\cos x}{a\sin x + b\cos x} \mathrm{d}x}
I
1
:=
∫
sin
x
a
sin
x
+
b
cos
x
d
x
=
b
x
+
a
ln
|
a
sin
x
+
b
cos
x
|
a
2
+
b
2
.
{\displaystyle 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
:=
∫
cos
x
a
sin
x
+
b
cos
x
d
x
=
a
x
−
b
ln
|
a
sin
x
+
b
cos
x
|
a
2
+
b
2
.
{\displaystyle 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}.}
∫
A
sin
x
+
B
cos
x
+
C
a
sin
x
+
b
cos
x
+
c
d
x
:
{\displaystyle \int \dfrac{A \sin x + B \cos x + C}{a\sin x + b\cos x + c} \mathrm{d}x:}
I
1
:=
∫
sin
x
a
sin
x
+
b
cos
x
+
c
d
x
=
a
a
2
+
b
2
x
−
b
a
2
+
b
2
ln
|
a
sin
x
+
b
cos
x
+
c
|
−
a
c
a
2
+
b
2
I
3
.
{\displaystyle 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
:=
∫
cos
x
a
sin
x
+
b
cos
x
+
c
d
x
=
b
a
2
+
b
2
x
−
a
a
2
+
b
2
ln
|
a
sin
x
+
b
cos
x
+
c
|
−
b
c
a
2
+
b
2
I
3
.
{\displaystyle 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
:=
∫
1
a
sin
x
+
b
cos
x
+
c
d
x
=
∫
d
x
c
+
a
2
+
b
2
sin
(
x
+
arctan
(
b
/
a
)
)
.
{\displaystyle 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）
∫
A
sin
x
+
B
cos
x
(
a
sin
x
+
b
cos
x
)
2
d
x
{\displaystyle \int \dfrac{A\sin x + B\cos x}{(a\sin x + b\cos x)^2} \mathrm{d}x}
I
1
:=
∫
sin
x
(
a
sin
x
+
b
cos
x
)
2
d
x
=
a
(
a
2
+
b
2
)
3
/
2
ln
|
tan
(
x
2
+
1
2
arctan
b
a
)
|
+
b
a
2
+
b
2
1
a
sin
x
+
b
cos
x
.
{\displaystyle 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
:=
∫
cos
x
(
a
sin
x
+
b
cos
x
)
2
d
x
=
b
(
a
2
+
b
2
)
3
/
2
ln
|
tan
(
x
2
+
1
2
arctan
b
a
)
|
−
a
a
2
+
b
2
1
a
sin
x
+
b
cos
x
.
{\displaystyle 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}.}
∫
d
x
(
a
sin
x
+
b
cos
x
)
n
=
1
(
n
−
1
)
(
a
2
+
b
2
)
(
b
sin
x
−
a
cos
x
(
a
sin
x
+
b
cos
x
)
n
−
1
+
(
n
−
2
)
∫
d
x
(
a
sin
x
+
b
cos
x
)
n
−
1
)
.
{\displaystyle \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).}
∫
A
sin
x
+
B
cos
x
(
a
sin
x
+
b
cos
x
)
n
d
x
=
A
a
+
B
b
a
2
+
b
2
∫
d
x
(
a
sin
x
+
b
cos
x
)
n
−
1
−
B
a
−
A
b
(
n
−
1
)
(
a
2
+
b
2
)
1
(
a
sin
x
+
b
cos
x
)
n
−
1
.
{\displaystyle \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}}.}
∫
A
sin
2
x
+
2
B
sin
x
cos
x
+
C
cos
2
x
a
sin
x
+
b
cos
x
d
x
{\displaystyle \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
:=
∫
sin
2
x
a
sin
x
+
b
cos
x
d
x
=
b
2
(
a
2
+
b
2
)
3
/
2
ln
|
tan
x
2
+
1
2
arctan
b
a
|
−
a
cos
x
+
b
sin
x
a
2
+
b
2
.
{\displaystyle 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
:=
∫
sin
x
cos
x
a
sin
x
+
b
cos
x
d
x
=
−
a
b
(
a
2
+
b
2
)
3
/
2
ln
|
tan
x
2
+
1
2
arctan
b
a
|
+
a
sin
x
−
b
cos
x
a
2
+
b
2
.
{\displaystyle 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
:=
∫
cos
2
x
a
sin
x
+
b
cos
x
d
x
=
a
2
(
a
2
+
b
2
)
3
/
2
ln
|
tan
x
2
+
1
2
arctan
b
a
|
+
a
cos
x
+
b
sin
x
a
2
+
b
2
.
{\displaystyle 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}.}
∫
A
sin
2
x
+
2
B
sin
x
cos
x
+
C
cos
2
x
(
a
sin
x
+
b
cos
x
)
2
d
x
{\displaystyle \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
:=
∫
sin
2
x
(
a
sin
x
+
b
cos
x
)
2
d
x
=
b
2
(
a
2
+
b
2
)
2
b
sin
x
−
a
cos
x
a
sin
x
+
b
cos
x
+
a
2
−
b
2
(
a
2
+
b
2
)
2
x
−
2
a
b
(
a
2
+
b
2
)
2
ln
|
a
sin
x
+
b
cos
x
|
.
{\displaystyle 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
:=
∫
sin
x
cos
x
(
a
sin
x
+
b
cos
x
)
2
d
x
=
a
2
(
a
2
+
b
2
)
2
b
sin
x
−
a
cos
x
a
sin
x
+
b
cos
x
−
a
2
−
b
2
(
a
2
+
b
2
)
2
x
+
2
a
b
(
a
2
+
b
2
)
2
ln
|
a
sin
x
+
b
cos
x
|
.
{\displaystyle 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|.}
I
3
:=
∫
cos
2
x
(
a
sin
x
+
b
cos
x
)
2
d
x
=
−
b
2
a
I
1
+
a
2
b
I
3
=
−
a
b
(
a
2
+
b
2
)
2
b
sin
x
−
a
cos
x
a
sin
x
+
b
cos
x
−
(
a
2
−
b
2
)
2
2
a
b
(
a
2
+
b
2
)
2
x
+
2
(
a
2
−
b
2
)
(
a
2
+
b
2
)
2
ln
|
a
sin
x
+
b
cos
x
|
.
{\displaystyle \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}}
∫
A
sin
3
x
+
B
cos
3
x
sin
x
+
cos
x
d
x
{\displaystyle \int \dfrac{A\sin^3 x + B\cos^3 x}{\sin x + \cos x} \mathrm{d}x}
I
1
:=
∫
sin
3
x
sin
x
+
cos
x
d
x
=
1
2
x
+
1
4
cos
2
x
+
1
4
cos
x
sin
x
+
1
4
ln
|
sin
x
+
cos
x
|
.
{\displaystyle 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
:=
∫
cos
3
x
sin
x
+
cos
x
d
x
=
1
2
x
+
1
4
cos
2
x
−
1
4
cos
x
sin
x
−
1
4
ln
|
sin
x
+
cos
x
|
.
{\displaystyle 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
∫
1
a
+
b
sin
x
d
x
{\displaystyle \int \dfrac{1}{a + b\sin x} \mathrm{d}x}
和
∫
1
a
−
b
cos
x
d
x
{\displaystyle \int \dfrac{1}{a - b\cos x} \mathrm{d}x}
。若积分为
∫
1
a
−
b
sin
x
d
x
,
|
a
|
>
|
b
|
.
{\displaystyle \int \dfrac{1}{a - b\sin x} \mathrm{d}x, \quad |a| > |b|.}
仅需注意到
sin
(
x
+
π
)
=
−
sin
x
{\displaystyle \sin(x+\pi)=-\sin x}
，若积分为
∫
1
a
+
b
cos
x
d
x
,
|
a
|
>
|
b
|
.
{\displaystyle \int \dfrac{1}{a + b\cos x} \mathrm{d}x, \quad |a| > |b|.}
仅需注意到
cos
(
x
+
π
)
=
−
cos
x
.
{\displaystyle \cos(x+\pi)=-\cos x.}
若分母是
sin
x
,
cos
x
{\displaystyle \sin x,\cos x}
的多项式，可以使用部分分式分解。
当
|
a
|
>
|
b
|
>
0
{\displaystyle |a| > |b| > 0}
∫
1
a
+
b
sin
x
d
x
.
{\displaystyle \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
:=
∫
1
a
+
b
sin
x
d
x
=
1
a
2
−
b
2
(
2
arctan
b
cos
x
+
a
sin
x
+
b
−
a
2
−
b
2
sin
x
b
sin
x
−
a
cos
x
+
a
+
a
2
−
b
2
+
x
)
.
{\displaystyle 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
=
−
2
a
2
−
b
2
arctan
(
a
−
b
a
+
b
tan
(
π
4
−
x
2
)
)
.
{\displaystyle 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).}
∫
1
a
−
b
cos
x
d
x
.
{\displaystyle \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
:=
∫
1
a
−
b
cos
x
d
x
=
1
a
2
−
b
2
(
2
arctan
(
a
2
−
b
2
−
b
−
a
)
sin
x
(
a
+
b
−
a
2
−
b
2
)
cos
x
−
a
−
b
−
a
2
−
b
2
+
x
)
.
{\displaystyle 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
=
∫
1
a
−
b
cos
x
d
x
=
1
a
2
−
b
2
arctan
a
2
−
b
2
sin
x
a
cos
x
−
b
.
{\displaystyle 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}.}
当
|
a
|
=
|
b
|
{\displaystyle |a|=|b|}
时就是下述情形：
∫
1
1
+
sin
x
d
x
=
−
2
1
+
tan
x
2
.
{\displaystyle \int \dfrac{1}{1+\sin x}\mathrm{d}x = - \dfrac{2}{1+\tan \frac{x}{2}}.}
∫
1
1
+
cos
x
d
x
=
tan
x
2
.
{\displaystyle \int \dfrac{1}{1+\cos x}\mathrm{d}x = \tan \dfrac{x}{2}.}
当
0
<
|
a
|
<
|
b
|
{\displaystyle 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。
∫
d
x
a
+
b
sin
x
=
1
b
2
−
a
2
ln
|
a
tan
x
2
+
b
−
b
2
−
a
2
a
tan
x
2
+
b
+
b
2
−
a
2
|
.
{\displaystyle \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|.}
∫
d
x
a
−
b
cos
x
=
1
b
2
−
a
2
ln
|
(
a
+
b
)
tan
x
2
−
b
2
−
a
2
(
a
+
b
)
tan
x
2
+
b
2
−
a
2
|
.
{\displaystyle \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[]
∫
A
cos
x
+
B
sin
x
a
+
b
sin
x
cos
x
d
x
{\displaystyle \int \dfrac{A\cos x + B\sin x}{a + b\sin x \cos x} \mathrm{d}x}
I
1
=
∫
cos
x
a
+
b
sin
x
cos
x
d
x
,
I
2
=
∫
sin
x
a
+
b
sin
x
cos
x
d
x
.
{\displaystyle 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.}
当
2
a
<
b
{\displaystyle 2a < b}
时，注意到
{
I
1
+
I
2
=
1
b
(
b
+
2
a
)
ln
|
b
+
2
a
+
b
(
sin
x
−
cos
x
)
b
+
2
a
−
b
(
sin
x
−
cos
x
)
|
,
I
1
−
I
2
=
1
b
(
b
−
2
a
)
ln
|
b
(
sin
x
+
cos
x
)
−
b
−
2
a
b
(
sin
x
+
cos
x
)
+
b
−
2
a
|
.
{\displaystyle \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}}
当
2
a
>
b
{\displaystyle 2a>b}
时，注意到
{
I
1
+
I
2
=
1
b
(
b
+
2
a
)
ln
|
b
+
2
a
+
b
(
sin
x
−
cos
x
)
b
+
2
a
−
b
(
sin
x
−
cos
x
)
|
,
I
1
−
I
2
=
2
b
(
2
a
−
b
)
arctan
b
(
sin
x
+
cos
x
)
2
a
−
b
.
{\displaystyle \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[]
∫
A
tan
x
+
B
cot
x
a
tan
x
+
b
cot
x
d
x
{\displaystyle \int \dfrac{A\tan x + B\cot x}{a\tan x + b\cot x} \mathrm{d}x}
I
1
:=
∫
tan
x
a
tan
x
+
b
cot
x
d
x
=
1
a
−
b
(
x
−
1
a
b
arctan
(
a
b
tan
x
)
)
{\displaystyle 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
:=
∫
cot
x
a
tan
x
+
b
cot
x
d
x
=
1
b
−
a
(
x
−
1
a
b
arctan
(
a
b
tan
x
)
)
{\displaystyle 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)}
∫
A
tan
2
x
+
B
cot
2
x
a
tan
x
+
b
cot
x
d
x
{\displaystyle \int \dfrac{A\tan^2 x + B\cot^2 x}{a\tan x + b\cot x} \mathrm{d}x}
I
1
:=
∫
tan
2
x
a
tan
x
+
b
cot
x
d
x
=
−
1
2
a
(
2
ln
|
cos
x
|
+
b
a
−
b
ln
|
a
sin
2
x
+
b
cos
2
x
|
)
.
{\displaystyle 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
:=
∫
cot
2
x
a
tan
x
+
b
cot
x
d
x
=
−
1
2
b
(
−
2
ln
|
sin
x
|
+
a
a
−
b
ln
|
a
sin
2
x
+
b
cos
2
x
|
)
.
{\displaystyle 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).}