三角函数积分表
以下是部份三角函數的積分表（省略积分常数）：
三角学
历史
三角函数
(
反三角函数
)
广义三角函数
参考
恒等式
精确值
三角表
单位圆
定理
正弦
餘弦
正切
餘切
勾股定理
微积分
三角换元法
积分
(
反三角函数
)
微分
积分只有
sin
的函數
∫
sin
c
x
d
x
=
−
1
c
cos
c
x
{\displaystyle \int \sin cx\;dx=-{\frac {1}{c}}\cos cx\,\!}
∫
sin
n
c
x
d
x
=
−
1
n
c
sin
n
−
1
c
x
cos
c
x
+
n
−
1
n
∫
sin
n
−
2
c
x
d
x
(
{\displaystyle \int \sin ^{n}cx\;dx=-{\frac {1}{nc}}\sin ^{n-1}cx\cos cx+{\frac {n-1}{n}}\int \sin ^{n-2}cx\;dx\qquad (}
其中
n
>
0
)
{\displaystyle n>0\,\!)}
∫
1
−
sin
x
d
x
=
∫
cvs
x
d
x
=
2
cos
x
2
+
sin
x
2
cos
x
2
−
sin
x
2
cvs
x
(
=
2
1
+
sin
x
)
{\displaystyle \int {\sqrt {1-\sin {x}}}\,dx=\int {\sqrt {\operatorname {cvs} {x}}}\,dx=2{\frac {\cos {\frac {x}{2}}+\sin {\frac {x}{2}}}{\cos {\frac {x}{2}}-\sin {\frac {x}{2}}}}{\sqrt {\operatorname {cvs} {x}}}(=2{\sqrt {1+\sin x}})}
（其中
cvs
x
{\displaystyle \operatorname {cvs} {x}}
是
餘矢(Coversine)函數
(參閱
正矢(versine)函數
)）
∫
x
sin
c
x
d
x
=
sin
c
x
c
2
−
x
cos
c
x
c
{\displaystyle \int x\sin cx\;dx={\frac {\sin cx}{c^{2}}}-{\frac {x\cos cx}{c}}\,\!}
∫
x
n
sin
c
x
d
x
=
−
x
n
c
cos
c
x
+
n
c
∫
x
n
−
1
cos
c
x
d
x
(
{\displaystyle \int x^{n}\sin cx\;dx=-{\frac {x^{n}}{c}}\cos cx+{\frac {n}{c}}\int x^{n-1}\cos cx\;dx\qquad (}
其中
n
>
0
)
{\displaystyle n>0\,\!)}
∫
−
a
2
a
2
x
2
sin
2
n
π
x
a
d
x
=
a
3
(
n
2
π
2
−
6
)
24
n
2
π
2
(
{\displaystyle \int _{-{\frac {a}{2}}}^{\frac {a}{2}}x^{2}\sin ^{2}{\frac {n\pi x}{a}}\;dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad (}
其中
n
=
2
,
4
,
6...
)
{\displaystyle n=2,4,6...\,\!)}
∫
sin
c
x
x
d
x
=
∑
i
=
0
∞
(
−
1
)
i
(
c
x
)
2
i
+
1
(
2
i
+
1
)
⋅
(
2
i
+
1
)
!
{\displaystyle \int {\frac {\sin cx}{x}}dx=\sum _{i=0}^{\infty }(-1)^{i}{\frac {(cx)^{2i+1}}{(2i+1)\cdot (2i+1)!}}\,\!}
∫
sin
c
x
x
n
d
x
=
−
sin
c
x
(
n
−
1
)
x
n
−
1
+
c
n
−
1
∫
cos
c
x
x
n
−
1
d
x
{\displaystyle \int {\frac {\sin cx}{x^{n}}}dx=-{\frac {\sin cx}{(n-1)x^{n-1}}}+{\frac {c}{n-1}}\int {\frac {\cos cx}{x^{n-1}}}dx\,\!}
∫
d
x
sin
c
x
=
1
c
ln
|
tan
c
x
2
|
{\displaystyle \int {\frac {dx}{\sin cx}}={\frac {1}{c}}\ln \left|\tan {\frac {cx}{2}}\right|}
∫
d
x
sin
n
c
x
=
cos
c
x
c
(
1
−
n
)
sin
n
−
1
c
x
+
n
−
2
n
−
1
∫
d
x
sin
n
−
2
c
x
(
{\displaystyle \int {\frac {dx}{\sin ^{n}cx}}={\frac {\cos cx}{c(1-n)\sin ^{n-1}cx}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\sin ^{n-2}cx}}\qquad (}
其中
n
>
1
)
{\displaystyle n>1\,\!)}
∫
d
x
1
±
sin
c
x
=
1
c
tan
(
c
x
2
∓
π
4
)
{\displaystyle \int {\frac {dx}{1\pm \sin cx}}={\frac {1}{c}}\tan \left({\frac {cx}{2}}\mp {\frac {\pi }{4}}\right)}
∫
x
d
x
1
+
sin
c
x
=
x
c
tan
(
c
x
2
−
π
4
)
+
2
c
2
ln
|
cos
(
c
x
2
−
π
4
)
|
{\displaystyle \int {\frac {x\;dx}{1+\sin cx}}={\frac {x}{c}}\tan \left({\frac {cx}{2}}-{\frac {\pi }{4}}\right)+{\frac {2}{c^{2}}}\ln \left|\cos \left({\frac {cx}{2}}-{\frac {\pi }{4}}\right)\right|}
∫
x
d
x
1
−
sin
c
x
=
x
c
cot
(
π
4
−
c
x
2
)
+
2
c
2
ln
|
sin
(
π
4
−
c
x
2
)
|
{\displaystyle \int {\frac {x\;dx}{1-\sin cx}}={\frac {x}{c}}\cot \left({\frac {\pi }{4}}-{\frac {cx}{2}}\right)+{\frac {2}{c^{2}}}\ln \left|\sin \left({\frac {\pi }{4}}-{\frac {cx}{2}}\right)\right|}
∫
sin
c
x
d
x
1
±
sin
c
x
=
±
x
+
1
c
tan
(
π
4
∓
c
x
2
)
{\displaystyle \int {\frac {\sin cx\;dx}{1\pm \sin cx}}=\pm x+{\frac {1}{c}}\tan \left({\frac {\pi }{4}}\mp {\frac {cx}{2}}\right)}
∫
sin
c
1
x
sin
c
2
x
d
x
=
sin
(
c
1
−
c
2
)
x
2
(
c
1
−
c
2
)
−
sin
(
c
1
+
c
2
)
x
2
(
c
1
+
c
2
)
(
{\displaystyle \int \sin c_{1}x\sin c_{2}x\;dx={\frac {\sin(c_{1}-c_{2})x}{2(c_{1}-c_{2})}}-{\frac {\sin(c_{1}+c_{2})x}{2(c_{1}+c_{2})}}\qquad (}
其中
|
c
1
|
≠
|
c
2
|
)
{\displaystyle |c_{1}|\neq |c_{2}|\,\!)}
积分只有
cos
的函數
∫
cos
c
x
d
x
=
1
c
sin
c
x
{\displaystyle \int \cos cx\;dx={\frac {1}{c}}\sin cx\,\!}
∫
cos
n
c
x
d
x
=
1
n
c
cos
n
−
1
c
x
sin
c
x
+
n
−
1
n
∫
cos
n
−
2
c
x
d
x
(
n
>
0
)
{\displaystyle \int \cos ^{n}cx\;dx={\frac {1}{nc}}\cos ^{n-1}cx\sin cx+{\frac {n-1}{n}}\int \cos ^{n-2}cx\;dx\qquad {\mbox{(}}n>0{\mbox{)}}\,\!}
∫
x
cos
c
x
d
x
=
cos
c
x
c
2
+
x
sin
c
x
c
{\displaystyle \int x\cos cx\;dx={\frac {\cos cx}{c^{2}}}+{\frac {x\sin cx}{c}}\,\!}
∫
x
n
cos
c
x
d
x
=
x
n
sin
c
x
c
−
n
c
∫
x
n
−
1
sin
c
x
d
x
{\displaystyle \int x^{n}\cos cx\;dx={\frac {x^{n}\sin cx}{c}}-{\frac {n}{c}}\int x^{n-1}\sin cx\;dx\,\!}
∫
−
a
2
a
2
x
2
cos
2
n
π
x
a
d
x
=
a
3
(
n
2
π
2
−
6
)
24
n
2
π
2
(
n
=
1
,
3
,
5...
)
{\displaystyle \int _{-{\frac {a}{2}}}^{\frac {a}{2}}x^{2}\cos ^{2}{\frac {n\pi x}{a}}\;dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad {\mbox{(}}n=1,3,5...{\mbox{)}}\,\!}
∫
cos
c
x
x
d
x
=
ln
|
c
x
|
+
∑
i
=
1
∞
(
−
1
)
i
(
c
x
)
2
i
2
i
⋅
(
2
i
)
!
{\displaystyle \int {\frac {\cos cx}{x}}dx=\ln |cx|+\sum _{i=1}^{\infty }(-1)^{i}{\frac {(cx)^{2i}}{2i\cdot (2i)!}}\,\!}
∫
cos
c
x
x
n
d
x
=
−
cos
c
x
(
n
−
1
)
x
n
−
1
−
c
n
−
1
∫
sin
c
x
x
n
−
1
d
x
(
n
≠
1
)
{\displaystyle \int {\frac {\cos cx}{x^{n}}}dx=-{\frac {\cos cx}{(n-1)x^{n-1}}}-{\frac {c}{n-1}}\int {\frac {\sin cx}{x^{n-1}}}dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}\,\!}
∫
d
x
cos
c
x
=
1
c
ln
|
tan
(
c
x
2
+
π
4
)
|
{\displaystyle \int {\frac {dx}{\cos cx}}={\frac {1}{c}}\ln \left|\tan \left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)\right|}
∫
d
x
cos
n
c
x
=
sin
c
x
c
(
n
−
1
)
c
o
s
n
−
1
c
x
+
n
−
2
n
−
1
∫
d
x
cos
n
−
2
c
x
(
n
>
1
)
{\displaystyle \int {\frac {dx}{\cos ^{n}cx}}={\frac {\sin cx}{c(n-1)cos^{n-1}cx}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cos ^{n-2}cx}}\qquad {\mbox{(}}n>1{\mbox{)}}\,\!}
∫
d
x
1
+
cos
c
x
=
1
c
tan
c
x
2
{\displaystyle \int {\frac {dx}{1+\cos cx}}={\frac {1}{c}}\tan {\frac {cx}{2}}\,\!}
∫
d
x
1
−
cos
c
x
=
−
1
c
cot
c
x
2
{\displaystyle \int {\frac {dx}{1-\cos cx}}=-{\frac {1}{c}}\cot {\frac {cx}{2}}\,\!}
∫
x
d
x
1
+
cos
c
x
=
x
c
tan
c
x
2
+
2
c
2
ln
|
cos
c
x
2
|
{\displaystyle \int {\frac {x\;dx}{1+\cos cx}}={\frac {x}{c}}\tan {\frac {cx}{2}}+{\frac {2}{c^{2}}}\ln \left|\cos {\frac {cx}{2}}\right|}
∫
x
d
x
1
−
cos
c
x
=
−
x
c
cot
c
x
2
+
2
c
2
ln
|
sin
c
x
2
|
{\displaystyle \int {\frac {x\;dx}{1-\cos cx}}=-{\frac {x}{c}}\cot {\frac {cx}{2}}+{\frac {2}{c^{2}}}\ln \left|\sin {\frac {cx}{2}}\right|}
∫
cos
c
x
d
x
1
+
cos
c
x
=
x
−
1
c
tan
c
x
2
{\displaystyle \int {\frac {\cos cx\;dx}{1+\cos cx}}=x-{\frac {1}{c}}\tan {\frac {cx}{2}}\,\!}
∫
cos
c
x
d
x
1
−
cos
c
x
=
−
x
−
1
c
cot
c
x
2
{\displaystyle \int {\frac {\cos cx\;dx}{1-\cos cx}}=-x-{\frac {1}{c}}\cot {\frac {cx}{2}}\,\!}
∫
cos
c
1
x
cos
c
2
x
d
x
=
sin
(
c
1
−
c
2
)
x
2
(
c
1
−
c
2
)
+
sin
(
c
1
+
c
2
)
x
2
(
c
1
+
c
2
)
(
|
c
1
|
≠
|
c
2
|
)
{\displaystyle \int \cos c_{1}x\cos c_{2}x\;dx={\frac {\sin(c_{1}-c_{2})x}{2(c_{1}-c_{2})}}+{\frac {\sin(c_{1}+c_{2})x}{2(c_{1}+c_{2})}}\qquad {\mbox{(}}|c_{1}|\neq |c_{2}|{\mbox{)}}\,\!}
积分只有
tan
的函數
∫
tan
c
x
d
x
=
−
1
c
ln
|
cos
c
x
|
=
1
c
ln
|
sec
c
x
|
{\displaystyle \int \tan cx\;dx=-{\frac {1}{c}}\ln |\cos cx|\ ={\frac {1}{c}}\ln |\sec cx|\,\!}
∫
tan
n
c
x
d
x
=
1
c
(
n
−
1
)
tan
n
−
1
c
x
−
∫
tan
n
−
2
c
x
d
x
(for
n
≠
1
)
{\displaystyle \int \tan ^{n}cx\;dx={\frac {1}{c(n-1)}}\tan ^{n-1}cx-\int \tan ^{n-2}cx\;dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
tan
c
x
+
1
=
x
2
+
1
2
c
ln
|
sin
c
x
+
cos
c
x
|
{\displaystyle \int {\frac {dx}{\tan cx+1}}={\frac {x}{2}}+{\frac {1}{2c}}\ln |\sin cx+\cos cx|\,\!}
∫
d
x
tan
c
x
−
1
=
−
x
2
+
1
2
c
ln
|
sin
c
x
−
cos
c
x
|
{\displaystyle \int {\frac {dx}{\tan cx-1}}=-{\frac {x}{2}}+{\frac {1}{2c}}\ln |\sin cx-\cos cx|\,\!}
∫
tan
c
x
d
x
tan
c
x
+
1
=
x
2
−
1
2
c
ln
|
sin
c
x
+
cos
c
x
|
{\displaystyle \int {\frac {\tan cx\;dx}{\tan cx+1}}={\frac {x}{2}}-{\frac {1}{2c}}\ln |\sin cx+\cos cx|\,\!}
∫
tan
c
x
d
x
tan
c
x
−
1
=
x
2
+
1
2
c
ln
|
sin
c
x
−
cos
c
x
|
{\displaystyle \int {\frac {\tan cx\;dx}{\tan cx-1}}={\frac {x}{2}}+{\frac {1}{2c}}\ln |\sin cx-\cos cx|\,\!}
积分只有
sec
的函數
∫
sec
c
x
d
x
=
1
c
ln
|
sec
c
x
+
tan
c
x
|
{\displaystyle \int \sec {cx}\,dx={\frac {1}{c}}\ln {\left|\sec {cx}+\tan {cx}\right|}}
∫
sec
2
x
d
x
=
tan
x
+
C
{\displaystyle \int \sec ^{2}x{\mbox{d}}x=\tan x+C}
∫
sec
n
c
x
d
x
=
sec
n
−
2
c
x
tan
c
x
c
(
n
−
1
)
+
n
−
2
n
−
1
∫
sec
n
−
2
c
x
d
x
(for
n
≠
1
)
{\displaystyle \int \sec ^{n}{cx}\,dx={\frac {\sec ^{n-2}{cx}\tan {cx}}{c(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{cx}\,dx\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
sec
x
+
1
=
x
−
tan
x
2
{\displaystyle \int {\frac {dx}{\sec {x}+1}}=x-\tan {\frac {x}{2}}}
积分只有
csc
的函數
∫
csc
c
x
d
x
=
1
c
ln
|
csc
c
x
−
cot
c
x
|
{\displaystyle \int \csc {cx}\,dx={\frac {1}{c}}\ln {\left|\csc {cx}-\cot {cx}\right|}}
∫
csc
2
x
d
x
=
−
cot
x
+
C
{\displaystyle \int \csc ^{2}x{\mbox{d}}x=-\cot x+C}
∫
csc
n
c
x
d
x
=
−
csc
n
−
2
c
x
cot
c
x
c
(
n
−
1
)
+
n
−
2
n
−
1
∫
csc
n
−
2
c
x
d
x
(for
n
≠
1
)
{\displaystyle \int \csc ^{n}{cx}\,dx=-{\frac {\csc ^{n-2}{cx}\cot {cx}}{c(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \csc ^{n-2}{cx}\,dx\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,\!}
积分只有
cot
的函數
∫
cot
c
x
d
x
=
1
c
ln
|
sin
c
x
|
{\displaystyle \int \cot cx\;dx={\frac {1}{c}}\ln |\sin cx|\,\!}
∫
cot
n
c
x
d
x
=
−
1
c
(
n
−
1
)
cot
n
−
1
c
x
−
∫
cot
n
−
2
c
x
d
x
(for
n
≠
1
)
{\displaystyle \int \cot ^{n}cx\;dx=-{\frac {1}{c(n-1)}}\cot ^{n-1}cx-\int \cot ^{n-2}cx\;dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
1
+
cot
c
x
=
∫
tan
c
x
d
x
tan
c
x
+
1
{\displaystyle \int {\frac {dx}{1+\cot cx}}=\int {\frac {\tan cx\;dx}{\tan cx+1}}\,\!}
∫
d
x
1
−
cot
c
x
=
∫
tan
c
x
d
x
tan
c
x
−
1
{\displaystyle \int {\frac {dx}{1-\cot cx}}=\int {\frac {\tan cx\;dx}{\tan cx-1}}\,\!}
积分只有
sin
和
cos
的函數
∫
d
x
cos
c
x
±
sin
c
x
=
1
c
2
ln
|
tan
(
c
x
2
±
π
8
)
|
{\displaystyle \int {\frac {dx}{\cos cx\pm \sin cx}}={\frac {1}{c{\sqrt {2}}}}\ln \left|\tan \left({\frac {cx}{2}}\pm {\frac {\pi }{8}}\right)\right|}
∫
d
x
(
cos
c
x
±
sin
c
x
)
2
=
1
2
c
tan
(
c
x
∓
π
4
)
{\displaystyle \int {\frac {dx}{(\cos cx\pm \sin cx)^{2}}}={\frac {1}{2c}}\tan \left(cx\mp {\frac {\pi }{4}}\right)}
∫
d
x
(
cos
x
+
sin
x
)
n
=
1
n
−
1
[
sin
x
−
cos
x
(
cos
x
+
sin
x
)
n
−
1
−
2
(
n
−
2
)
∫
d
x
(
cos
x
+
sin
x
)
n
−
2
]
{\displaystyle \int {\frac {dx}{(\cos x+\sin x)^{n}}}={\frac {1}{n-1}}\left[{\frac {\sin x-\cos x}{(\cos x+\sin x)^{n-1}}}-2(n-2)\int {\frac {dx}{(\cos x+\sin x)^{n-2}}}\right]}
∫
cos
c
x
d
x
cos
c
x
+
sin
c
x
=
x
2
+
1
2
c
ln
|
sin
c
x
+
cos
c
x
|
{\displaystyle \int {\frac {\cos cx\;dx}{\cos cx+\sin cx}}={\frac {x}{2}}+{\frac {1}{2c}}\ln \left|\sin cx+\cos cx\right|}
∫
cos
c
x
d
x
cos
c
x
−
sin
c
x
=
x
2
−
1
2
c
ln
|
sin
c
x
−
cos
c
x
|
{\displaystyle \int {\frac {\cos cx\;dx}{\cos cx-\sin cx}}={\frac {x}{2}}-{\frac {1}{2c}}\ln \left|\sin cx-\cos cx\right|}
∫
sin
c
x
d
x
cos
c
x
+
sin
c
x
=
x
2
−
1
2
c
ln
|
sin
c
x
+
cos
c
x
|
{\displaystyle \int {\frac {\sin cx\;dx}{\cos cx+\sin cx}}={\frac {x}{2}}-{\frac {1}{2c}}\ln \left|\sin cx+\cos cx\right|}
∫
sin
c
x
d
x
cos
c
x
−
sin
c
x
=
−
x
2
−
1
2
c
ln
|
sin
c
x
−
cos
c
x
|
{\displaystyle \int {\frac {\sin cx\;dx}{\cos cx-\sin cx}}=-{\frac {x}{2}}-{\frac {1}{2c}}\ln \left|\sin cx-\cos cx\right|}
∫
cos
c
x
d
x
sin
c
x
(
1
+
cos
c
x
)
=
−
1
4
c
tan
2
c
x
2
+
1
2
c
ln
|
tan
c
x
2
|
{\displaystyle \int {\frac {\cos cx\;dx}{\sin cx(1+\cos cx)}}=-{\frac {1}{4c}}\tan ^{2}{\frac {cx}{2}}+{\frac {1}{2c}}\ln \left|\tan {\frac {cx}{2}}\right|}
∫
cos
c
x
d
x
sin
c
x
(
1
+
−
cos
c
x
)
=
−
1
4
c
cot
2
c
x
2
−
1
2
c
ln
|
tan
c
x
2
|
{\displaystyle \int {\frac {\cos cx\;dx}{\sin cx(1+-\cos cx)}}=-{\frac {1}{4c}}\cot ^{2}{\frac {cx}{2}}-{\frac {1}{2c}}\ln \left|\tan {\frac {cx}{2}}\right|}
∫
sin
c
x
d
x
cos
c
x
(
1
+
sin
c
x
)
=
1
4
c
cot
2
(
c
x
2
+
π
4
)
+
1
2
c
ln
|
tan
(
c
x
2
+
π
4
)
|
{\displaystyle \int {\frac {\sin cx\;dx}{\cos cx(1+\sin cx)}}={\frac {1}{4c}}\cot ^{2}\left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)+{\frac {1}{2c}}\ln \left|\tan \left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)\right|}
∫
sin
c
x
d
x
cos
c
x
(
1
−
sin
c
x
)
=
1
4
c
tan
2
(
c
x
2
+
π
4
)
−
1
2
c
ln
|
tan
(
c
x
2
+
π
4
)
|
{\displaystyle \int {\frac {\sin cx\;dx}{\cos cx(1-\sin cx)}}={\frac {1}{4c}}\tan ^{2}\left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)-{\frac {1}{2c}}\ln \left|\tan \left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)\right|}
∫
sin
c
x
cos
c
x
d
x
=
1
2
c
sin
2
c
x
{\displaystyle \int \sin cx\cos cx\;dx={\frac {1}{2c}}\sin ^{2}cx\,\!}
∫
sin
c
1
x
cos
c
2
x
d
x
=
−
cos
(
c
1
+
c
2
)
x
2
(
c
1
+
c
2
)
−
cos
(
c
1
−
c
2
)
x
2
(
c
1
−
c
2
)
(for
|
c
1
|
≠
|
c
2
|
)
{\displaystyle \int \sin c_{1}x\cos c_{2}x\;dx=-{\frac {\cos(c_{1}+c_{2})x}{2(c_{1}+c_{2})}}-{\frac {\cos(c_{1}-c_{2})x}{2(c_{1}-c_{2})}}\qquad {\mbox{(for }}|c_{1}|\neq |c_{2}|{\mbox{)}}\,\!}
∫
sin
n
c
x
cos
c
x
d
x
=
1
c
(
n
+
1
)
sin
n
+
1
c
x
(for
n
≠
1
)
{\displaystyle \int \sin ^{n}cx\cos cx\;dx={\frac {1}{c(n+1)}}\sin ^{n+1}cx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
sin
c
x
cos
n
c
x
d
x
=
−
1
c
(
n
+
1
)
cos
n
+
1
c
x
(for
n
≠
1
)
{\displaystyle \int \sin cx\cos ^{n}cx\;dx=-{\frac {1}{c(n+1)}}\cos ^{n+1}cx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
sin
n
c
x
cos
m
c
x
d
x
=
−
sin
n
−
1
c
x
cos
m
+
1
c
x
c
(
n
+
m
)
+
n
−
1
n
+
m
∫
sin
n
−
2
c
x
cos
m
c
x
d
x
(for
m
,
n
>
0
)
{\displaystyle \int \sin ^{n}cx\cos ^{m}cx\;dx=-{\frac {\sin ^{n-1}cx\cos ^{m+1}cx}{c(n+m)}}+{\frac {n-1}{n+m}}\int \sin ^{n-2}cx\cos ^{m}cx\;dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!}
also:
∫
sin
n
c
x
cos
m
c
x
d
x
=
sin
n
+
1
c
x
cos
m
−
1
c
x
c
(
n
+
m
)
+
m
−
1
n
+
m
∫
sin
n
c
x
cos
m
−
2
c
x
d
x
(for
m
,
n
>
0
)
{\displaystyle \int \sin ^{n}cx\cos ^{m}cx\;dx={\frac {\sin ^{n+1}cx\cos ^{m-1}cx}{c(n+m)}}+{\frac {m-1}{n+m}}\int \sin ^{n}cx\cos ^{m-2}cx\;dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!}
∫
d
x
sin
c
x
cos
c
x
=
1
c
ln
|
tan
c
x
|
{\displaystyle \int {\frac {dx}{\sin cx\cos cx}}={\frac {1}{c}}\ln \left|\tan cx\right|}
∫
d
x
sin
c
x
cos
n
c
x
=
1
c
(
n
−
1
)
cos
n
−
1
c
x
+
∫
d
x
sin
c
x
cos
n
−
2
c
x
(for
n
≠
1
)
{\displaystyle \int {\frac {dx}{\sin cx\cos ^{n}cx}}={\frac {1}{c(n-1)\cos ^{n-1}cx}}+\int {\frac {dx}{\sin cx\cos ^{n-2}cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
d
x
sin
n
c
x
cos
c
x
=
−
1
c
(
n
−
1
)
sin
n
−
1
c
x
+
∫
d
x
sin
n
−
2
c
x
cos
c
x
(for
n
≠
1
)
{\displaystyle \int {\frac {dx}{\sin ^{n}cx\cos cx}}=-{\frac {1}{c(n-1)\sin ^{n-1}cx}}+\int {\frac {dx}{\sin ^{n-2}cx\cos cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
sin
c
x
d
x
cos
n
c
x
=
1
c
(
n
−
1
)
cos
n
−
1
c
x
(for
n
≠
1
)
{\displaystyle \int {\frac {\sin cx\;dx}{\cos ^{n}cx}}={\frac {1}{c(n-1)\cos ^{n-1}cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
sin
2
c
x
d
x
cos
c
x
=
−
1
c
sin
c
x
+
1
c
ln
|
tan
(
π
4
+
c
x
2
)
|
{\displaystyle \int {\frac {\sin ^{2}cx\;dx}{\cos cx}}=-{\frac {1}{c}}\sin cx+{\frac {1}{c}}\ln \left|\tan \left({\frac {\pi }{4}}+{\frac {cx}{2}}\right)\right|}
∫
sin
2
c
x
d
x
cos
n
c
x
=
sin
c
x
c
(
n
−
1
)
cos
n
−
1
c
x
−
1
n
−
1
∫
d
x
cos
n
−
2
c
x
(for
n
≠
1
)
{\displaystyle \int {\frac {\sin ^{2}cx\;dx}{\cos ^{n}cx}}={\frac {\sin cx}{c(n-1)\cos ^{n-1}cx}}-{\frac {1}{n-1}}\int {\frac {dx}{\cos ^{n-2}cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
sin
n
c
x
d
x
cos
c
x
=
−
sin
n
−
1
c
x
c
(
n
−
1
)
+
∫
sin
n
−
2
c
x
d
x
cos
c
x
(for
n
≠
1
)
{\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos cx}}=-{\frac {\sin ^{n-1}cx}{c(n-1)}}+\int {\frac {\sin ^{n-2}cx\;dx}{\cos cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
sin
n
c
x
d
x
cos
m
c
x
=
sin
n
+
1
c
x
c
(
m
−
1
)
cos
m
−
1
c
x
−
n
−
m
+
2
m
−
1
∫
sin
n
c
x
d
x
cos
m
−
2
c
x
(for
m
≠
1
)
{\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos ^{m}cx}}={\frac {\sin ^{n+1}cx}{c(m-1)\cos ^{m-1}cx}}-{\frac {n-m+2}{m-1}}\int {\frac {\sin ^{n}cx\;dx}{\cos ^{m-2}cx}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
also:
∫
sin
n
c
x
d
x
cos
m
c
x
=
−
sin
n
−
1
c
x
c
(
n
−
m
)
cos
m
−
1
c
x
+
n
−
1
n
−
m
∫
sin
n
−
2
c
x
d
x
cos
m
c
x
(for
m
≠
n
)
{\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos ^{m}cx}}=-{\frac {\sin ^{n-1}cx}{c(n-m)\cos ^{m-1}cx}}+{\frac {n-1}{n-m}}\int {\frac {\sin ^{n-2}cx\;dx}{\cos ^{m}cx}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!}
also:
∫
sin
n
c
x
d
x
cos
m
c
x
=
sin
n
−
1
c
x
c
(
m
−
1
)
cos
m
−
1
c
x
−
n
−
1
m
−
1
∫
sin
n
−
2
c
x
d
x
cos
m
−
2
c
x
(for
m
≠
1
)
{\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos ^{m}cx}}={\frac {\sin ^{n-1}cx}{c(m-1)\cos ^{m-1}cx}}-{\frac {n-1}{m-1}}\int {\frac {\sin ^{n-2}cx\;dx}{\cos ^{m-2}cx}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
∫
cos
c
x
d
x
sin
n
c
x
=
−
1
c
(
n
−
1
)
sin
n
−
1
c
x
(for
n
≠
1
)
{\displaystyle \int {\frac {\cos cx\;dx}{\sin ^{n}cx}}=-{\frac {1}{c(n-1)\sin ^{n-1}cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
∫
cos
2
c
x
d
x
sin
c
x
=
1
c
(
cos
c
x
+
ln
|
tan
c
x
2
|
)
{\displaystyle \int {\frac {\cos ^{2}cx\;dx}{\sin cx}}={\frac {1}{c}}\left(\cos cx+\ln \left|\tan {\frac {cx}{2}}\right|\right)}
∫
cos
2
c
x
d
x
sin
n
c
x
=
−
1
n
−
1
(
cos
c
x
c
sin
n
−
1
c
x
)
+
∫
d
x
sin
n
−
2
c
x
)
(for
n
≠
1
)
{\displaystyle \int {\frac {\cos ^{2}cx\;dx}{\sin ^{n}cx}}=-{\frac {1}{n-1}}\left({\frac {\cos cx}{c\sin ^{n-1}cx)}}+\int {\frac {dx}{\sin ^{n-2}cx}}\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
∫
cos
n
c
x
d
x
sin
m
c
x
=
−
cos
n
+
1
c
x
c
(
m
−
1
)
sin
m
−
1
c
x
−
n
−
m
−
2
m
−
1
∫
c
o
s
n
c
x
d
x
sin
m
−
2
c
x
(for
m
≠
1
)
{\displaystyle \int {\frac {\cos ^{n}cx\;dx}{\sin ^{m}cx}}=-{\frac {\cos ^{n+1}cx}{c(m-1)\sin ^{m-1}cx}}-{\frac {n-m-2}{m-1}}\int {\frac {cos^{n}cx\;dx}{\sin ^{m-2}cx}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
also:
∫
cos
n
c
x
d
x
sin
m
c
x
=
cos
n
−
1
c
x
c
(
n
−
m
)
sin
m
−
1
c
x
+
n
−
1
n
−
m
∫
c
o
s
n
−
2
c
x
d
x
sin
m
c
x
(for
m
≠
n
)
{\displaystyle \int {\frac {\cos ^{n}cx\;dx}{\sin ^{m}cx}}={\frac {\cos ^{n-1}cx}{c(n-m)\sin ^{m-1}cx}}+{\frac {n-1}{n-m}}\int {\frac {cos^{n-2}cx\;dx}{\sin ^{m}cx}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!}
also:
∫
cos
n
c
x
d
x
sin
m
c
x
=
−
cos
n
−
1
c
x
c
(
m
−
1
)
sin
m
−
1
c
x
−
n
−
1
m
−
1
∫
c
o
s
n
−
2
c
x
d
x
sin
m
−
2
c
x
(for
m
≠
1
)
{\displaystyle \int {\frac {\cos ^{n}cx\;dx}{\sin ^{m}cx}}=-{\frac {\cos ^{n-1}cx}{c(m-1)\sin ^{m-1}cx}}-{\frac {n-1}{m-1}}\int {\frac {cos^{n-2}cx\;dx}{\sin ^{m-2}cx}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
积分只有
sin
和
tan
的函數
∫
sin
c
x
tan
c
x
d
x
=
1
c
(
ln
|
sec
c
x
+
tan
c
x
|
−
sin
c
x
)
{\displaystyle \int \sin cx\tan cx\;dx={\frac {1}{c}}(\ln |\sec cx+\tan cx|-\sin cx)\,\!}
∫
tan
n
c
x
d
x
sin
2
c
x
=
1
c
(
n
−
1
)
tan
n
−
1
(
c
x
)
(for
n
≠
1
)
{\displaystyle \int {\frac {\tan ^{n}cx\;dx}{\sin ^{2}cx}}={\frac {1}{c(n-1)}}\tan ^{n-1}(cx)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
积分只有
cos
和
tan
的函數
∫
tan
n
c
x
d
x
cos
2
c
x
=
1
c
(
n
+
1
)
tan
n
+
1
c
x
(for
n
≠
−
1
)
{\displaystyle \int {\frac {\tan ^{n}cx\;dx}{\cos ^{2}cx}}={\frac {1}{c(n+1)}}\tan ^{n+1}cx\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
积分只有
sin
和
cot
的函數
∫
cot
n
c
x
d
x
sin
2
c
x
=
−
1
c
(
n
+
1
)
cot
n
+
1
c
x
(for
n
≠
−
1
)
{\displaystyle \int {\frac {\cot ^{n}cx\;dx}{\sin ^{2}cx}}={\frac {-1}{c(n+1)}}\cot ^{n+1}cx\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
积分只有
cos
和
cot
的函數
∫
cot
n
c
x
d
x
cos
2
c
x
=
1
c
(
1
−
n
)
tan
1
−
n
c
x
(for
n
≠
1
)
{\displaystyle \int {\frac {\cot ^{n}cx\;dx}{\cos ^{2}cx}}={\frac {1}{c(1-n)}}\tan ^{1-n}cx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
积分只有
tan
和
cot
的函數
∫
tan
m
(
c
x
)
cot
n
(
c
x
)
d
x
=
1
c
(
m
+
n
−
1
)
tan
m
+
n
−
1
(
c
x
)
−
∫
tan
m
−
2
(
c
x
)
cot
n
(
c
x
)
d
x
(for
m
+
n
≠
1
)
{\displaystyle \int {\frac {\tan ^{m}(cx)}{\cot ^{n}(cx)}}\;dx={\frac {1}{c(m+n-1)}}\tan ^{m+n-1}(cx)-\int {\frac {\tan ^{m-2}(cx)}{\cot ^{n}(cx)}}\;dx\qquad {\mbox{(for }}m+n\neq 1{\mbox{)}}\,\!}
This article is issued from
Wikipedia
. The text is licensed under
Creative Commons - Attribution - Sharealike
. Additional terms may apply for the media files.
![{\displaystyle \int {\frac {dx}{(\cos x+\sin x)^{n}}}={\frac {1}{n-1}}\left[{\frac {\sin x-\cos x}{(\cos x+\sin x)^{n-1}}}-2(n-2)\int {\frac {dx}{(\cos x+\sin x)^{n-2}}}\right]}](../I/28c566bec8e0cc2b64413440fac71b6a79ea0143.svg)
