三角函数积分表

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三角学
历史（英语：History of trigonometry） 三角函数 (反三角函数) 广义三角函数（英语：Generalized trigonometry）
参考
恒等式 精确值 三角表（英语：Trigonometric tables） 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
查 论 编

以下是部份三角函數的積分表（省略积分常数）：

${\displaystyle \int \sin cx\;dx=-{\frac {1}{c}}\cos cx\,\!}$

${\displaystyle \int \sin ^{n}cx\;dx=-{\frac {1}{nc}}\sin ^{n-1}cx\cos cx+{\frac {n-1}{n}}\int \sin ^{n-2}cx\;dx\qquad (}$其中${\displaystyle n>0\,\!)}$

${\displaystyle \int {\sqrt {1-\sin {x}}}\,dx=\int {\sqrt {\operatorname {cvs} {x}}}\,dx=2{\frac {\cos {\frac {x}{2}}+\sin {\frac {x}{2}}}{\cos {\frac {x}{2}}-\sin {\frac {x}{2}}}}{\sqrt {\operatorname {cvs} {x}}}(=2{\sqrt {1+\sin x}})}$（其中 ${\displaystyle \operatorname {cvs} {x}}$ 是 餘矢(Coversine)函數(參閱正矢(versine)函數)）

${\displaystyle \int x\sin cx\;dx={\frac {\sin cx}{c^{2}}}-{\frac {x\cos cx}{c}}\,\!}$

${\displaystyle \int x^{n}\sin cx\;dx=-{\frac {x^{n}}{c}}\cos cx+{\frac {n}{c}}\int x^{n-1}\cos cx\;dx\qquad (}$其中${\displaystyle n>0\,\!)}$

${\displaystyle \int _{-{\frac {a}{2}}}^{\frac {a}{2}}x^{2}\sin ^{2}{\frac {n\pi x}{a}}\;dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad (}$其中${\displaystyle n=2,4,6...\,\!)}$

${\displaystyle \int {\frac {\sin cx}{x}}dx=\sum _{i=0}^{\infty }(-1)^{i}{\frac {(cx)^{2i+1}}{(2i+1)\cdot (2i+1)!}}\,\!}$

${\displaystyle \int {\frac {\sin cx}{x^{n}}}dx=-{\frac {\sin cx}{(n-1)x^{n-1}}}+{\frac {c}{n-1}}\int {\frac {\cos cx}{x^{n-1}}}dx\,\!}$

${\displaystyle \int {\frac {dx}{\sin cx}}={\frac {1}{c}}\ln \left|\tan {\frac {cx}{2}}\right|}$

${\displaystyle \int {\frac {dx}{\sin ^{n}cx}}={\frac {\cos cx}{c(1-n)\sin ^{n-1}cx}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\sin ^{n-2}cx}}\qquad (}$其中${\displaystyle n>1\,\!)}$

${\displaystyle \int {\frac {dx}{1\pm \sin cx}}={\frac {1}{c}}\tan \left({\frac {cx}{2}}\mp {\frac {\pi }{4}}\right)}$

${\displaystyle \int {\frac {x\;dx}{1+\sin cx}}={\frac {x}{c}}\tan \left({\frac {cx}{2}}-{\frac {\pi }{4}}\right)+{\frac {2}{c^{2}}}\ln \left|\cos \left({\frac {cx}{2}}-{\frac {\pi }{4}}\right)\right|}$

${\displaystyle \int {\frac {x\;dx}{1-\sin cx}}={\frac {x}{c}}\cot \left({\frac {\pi }{4}}-{\frac {cx}{2}}\right)+{\frac {2}{c^{2}}}\ln \left|\sin \left({\frac {\pi }{4}}-{\frac {cx}{2}}\right)\right|}$

${\displaystyle \int {\frac {\sin cx\;dx}{1\pm \sin cx}}=\pm x+{\frac {1}{c}}\tan \left({\frac {\pi }{4}}\mp {\frac {cx}{2}}\right)}$

${\displaystyle \int \sin c_{1}x\sin c_{2}x\;dx={\frac {\sin(c_{1}-c_{2})x}{2(c_{1}-c_{2})}}-{\frac {\sin(c_{1}+c_{2})x}{2(c_{1}+c_{2})}}\qquad (}$其中${\displaystyle |c_{1}|\neq |c_{2}|\,\!)}$

${\displaystyle \int \cos cx\;dx={\frac {1}{c}}\sin cx\,\!}$

${\displaystyle \int \cos ^{n}cx\;dx={\frac {1}{nc}}\cos ^{n-1}cx\sin cx+{\frac {n-1}{n}}\int \cos ^{n-2}cx\;dx\qquad {\mbox{(}}n>0{\mbox{)}}\,\!}$

${\displaystyle \int x\cos cx\;dx={\frac {\cos cx}{c^{2}}}+{\frac {x\sin cx}{c}}\,\!}$

${\displaystyle \int x^{n}\cos cx\;dx={\frac {x^{n}\sin cx}{c}}-{\frac {n}{c}}\int x^{n-1}\sin cx\;dx\,\!}$

${\displaystyle \int _{-{\frac {a}{2}}}^{\frac {a}{2}}x^{2}\cos ^{2}{\frac {n\pi x}{a}}\;dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad {\mbox{(}}n=1,3,5...{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\cos cx}{x}}dx=\ln |cx|+\sum _{i=1}^{\infty }(-1)^{i}{\frac {(cx)^{2i}}{2i\cdot (2i)!}}\,\!}$

${\displaystyle \int {\frac {\cos cx}{x^{n}}}dx=-{\frac {\cos cx}{(n-1)x^{n-1}}}-{\frac {c}{n-1}}\int {\frac {\sin cx}{x^{n-1}}}dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {dx}{\cos cx}}={\frac {1}{c}}\ln \left|\tan \left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)\right|}$

${\displaystyle \int {\frac {dx}{\cos ^{n}cx}}={\frac {\sin cx}{c(n-1)cos^{n-1}cx}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cos ^{n-2}cx}}\qquad {\mbox{(}}n>1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {dx}{1+\cos cx}}={\frac {1}{c}}\tan {\frac {cx}{2}}\,\!}$

${\displaystyle \int {\frac {dx}{1-\cos cx}}=-{\frac {1}{c}}\cot {\frac {cx}{2}}\,\!}$

${\displaystyle \int {\frac {x\;dx}{1+\cos cx}}={\frac {x}{c}}\tan {\frac {cx}{2}}+{\frac {2}{c^{2}}}\ln \left|\cos {\frac {cx}{2}}\right|}$

${\displaystyle \int {\frac {x\;dx}{1-\cos cx}}=-{\frac {x}{c}}\cot {\frac {cx}{2}}+{\frac {2}{c^{2}}}\ln \left|\sin {\frac {cx}{2}}\right|}$

${\displaystyle \int {\frac {\cos cx\;dx}{1+\cos cx}}=x-{\frac {1}{c}}\tan {\frac {cx}{2}}\,\!}$

${\displaystyle \int {\frac {\cos cx\;dx}{1-\cos cx}}=-x-{\frac {1}{c}}\cot {\frac {cx}{2}}\,\!}$

${\displaystyle \int \cos c_{1}x\cos c_{2}x\;dx={\frac {\sin(c_{1}-c_{2})x}{2(c_{1}-c_{2})}}+{\frac {\sin(c_{1}+c_{2})x}{2(c_{1}+c_{2})}}\qquad {\mbox{(}}|c_{1}|\neq |c_{2}|{\mbox{)}}\,\!}$

${\displaystyle \int \tan cx\;dx=-{\frac {1}{c}}\ln |\cos cx|\ ={\frac {1}{c}}\ln |\sec cx|\,\!}$

${\displaystyle \int \tan ^{n}cx\;dx={\frac {1}{c(n-1)}}\tan ^{n-1}cx-\int \tan ^{n-2}cx\;dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {dx}{\tan cx+1}}={\frac {x}{2}}+{\frac {1}{2c}}\ln |\sin cx+\cos cx|\,\!}$

${\displaystyle \int {\frac {dx}{\tan cx-1}}=-{\frac {x}{2}}+{\frac {1}{2c}}\ln |\sin cx-\cos cx|\,\!}$

${\displaystyle \int {\frac {\tan cx\;dx}{\tan cx+1}}={\frac {x}{2}}-{\frac {1}{2c}}\ln |\sin cx+\cos cx|\,\!}$

${\displaystyle \int {\frac {\tan cx\;dx}{\tan cx-1}}={\frac {x}{2}}+{\frac {1}{2c}}\ln |\sin cx-\cos cx|\,\!}$

${\displaystyle \int \sec {cx}\,dx={\frac {1}{c}}\ln {\left|\sec {cx}+\tan {cx}\right|}}$

${\displaystyle \int \sec ^{2}x{\mbox{d}}x=\tan x+C}$

${\displaystyle \int \sec ^{n}{cx}\,dx={\frac {\sec ^{n-2}{cx}\tan {cx}}{c(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{cx}\,dx\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {dx}{\sec {x}+1}}=x-\tan {\frac {x}{2}}}$

${\displaystyle \int \csc {cx}\,dx={\frac {1}{c}}\ln {\left|\tan({\frac {cx}{2}})\right|}}$

${\displaystyle \int \csc ^{2}x{\mbox{d}}x=-\cot x+C}$

${\displaystyle \int \csc ^{n}{cx}\,dx=-{\frac {\csc ^{n-2}{cx}\cot {cx}}{c(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \csc ^{n-2}{cx}\,dx\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int \cot cx\;dx={\frac {1}{c}}\ln |\sin cx|\,\!}$

${\displaystyle \int \cot ^{n}cx\;dx=-{\frac {1}{c(n-1)}}\cot ^{n-1}cx-\int \cot ^{n-2}cx\;dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {dx}{1+\cot cx}}=\int {\frac {\tan cx\;dx}{\tan cx+1}}\,\!}$

${\displaystyle \int {\frac {dx}{1-\cot cx}}=\int {\frac {\tan cx\;dx}{\tan cx-1}}\,\!}$

${\displaystyle \int {\frac {dx}{\cos cx\pm \sin cx}}={\frac {1}{c{\sqrt {2}}}}\ln \left|\tan \left({\frac {cx}{2}}\pm {\frac {\pi }{8}}\right)\right|}$

${\displaystyle \int {\frac {dx}{(\cos cx\pm \sin cx)^{2}}}={\frac {1}{2c}}\tan \left(cx\mp {\frac {\pi }{4}}\right)}$

${\displaystyle \int {\frac {dx}{(\cos x+\sin x)^{n}}}={\frac {1}{n-1}}\left[{\frac {\sin x-\cos x}{(\cos x+\sin x)^{n-1}}}-2(n-2)\int {\frac {dx}{(\cos x+\sin x)^{n-2}}}\right]}$

${\displaystyle \int {\frac {\cos cx\;dx}{\cos cx+\sin cx}}={\frac {x}{2}}+{\frac {1}{2c}}\ln \left|\sin cx+\cos cx\right|}$

${\displaystyle \int {\frac {\cos cx\;dx}{\cos cx-\sin cx}}={\frac {x}{2}}-{\frac {1}{2c}}\ln \left|\sin cx-\cos cx\right|}$

${\displaystyle \int {\frac {\sin cx\;dx}{\cos cx+\sin cx}}={\frac {x}{2}}-{\frac {1}{2c}}\ln \left|\sin cx+\cos cx\right|}$

${\displaystyle \int {\frac {\sin cx\;dx}{\cos cx-\sin cx}}=-{\frac {x}{2}}-{\frac {1}{2c}}\ln \left|\sin cx-\cos cx\right|}$

${\displaystyle \int {\frac {\cos cx\;dx}{\sin cx(1+\cos cx)}}=-{\frac {1}{4c}}\tan ^{2}{\frac {cx}{2}}+{\frac {1}{2c}}\ln \left|\tan {\frac {cx}{2}}\right|}$

${\displaystyle \int {\frac {\cos cx\;dx}{\sin cx(1+-\cos cx)}}=-{\frac {1}{4c}}\cot ^{2}{\frac {cx}{2}}-{\frac {1}{2c}}\ln \left|\tan {\frac {cx}{2}}\right|}$

${\displaystyle \int {\frac {\sin cx\;dx}{\cos cx(1+\sin cx)}}={\frac {1}{4c}}\cot ^{2}\left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)+{\frac {1}{2c}}\ln \left|\tan \left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)\right|}$

${\displaystyle \int {\frac {\sin cx\;dx}{\cos cx(1-\sin cx)}}={\frac {1}{4c}}\tan ^{2}\left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)-{\frac {1}{2c}}\ln \left|\tan \left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)\right|}$

${\displaystyle \int \sin cx\cos cx\;dx={\frac {1}{2c}}\sin ^{2}cx\,\!}$

${\displaystyle \int \sin c_{1}x\cos c_{2}x\;dx=-{\frac {\cos(c_{1}+c_{2})x}{2(c_{1}+c_{2})}}-{\frac {\cos(c_{1}-c_{2})x}{2(c_{1}-c_{2})}}\qquad {\mbox{(for }}|c_{1}|\neq |c_{2}|{\mbox{)}}\,\!}$

${\displaystyle \int \sin ^{n}cx\cos cx\;dx={\frac {1}{c(n+1)}}\sin ^{n+1}cx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int \sin cx\cos ^{n}cx\;dx=-{\frac {1}{c(n+1)}}\cos ^{n+1}cx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int \sin ^{n}cx\cos ^{m}cx\;dx=-{\frac {\sin ^{n-1}cx\cos ^{m+1}cx}{c(n+m)}}+{\frac {n-1}{n+m}}\int \sin ^{n-2}cx\cos ^{m}cx\;dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!}$

also: ${\displaystyle \int \sin ^{n}cx\cos ^{m}cx\;dx={\frac {\sin ^{n+1}cx\cos ^{m-1}cx}{c(n+m)}}+{\frac {m-1}{n+m}}\int \sin ^{n}cx\cos ^{m-2}cx\;dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {dx}{\sin cx\cos cx}}={\frac {1}{c}}\ln \left|\tan cx\right|}$

${\displaystyle \int {\frac {dx}{\sin cx\cos ^{n}cx}}={\frac {1}{c(n-1)\cos ^{n-1}cx}}+\int {\frac {dx}{\sin cx\cos ^{n-2}cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {dx}{\sin ^{n}cx\cos cx}}=-{\frac {1}{c(n-1)\sin ^{n-1}cx}}+\int {\frac {dx}{\sin ^{n-2}cx\cos cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\sin cx\;dx}{\cos ^{n}cx}}={\frac {1}{c(n-1)\cos ^{n-1}cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\sin ^{2}cx\;dx}{\cos cx}}=-{\frac {1}{c}}\sin cx+{\frac {1}{c}}\ln \left|\tan \left({\frac {\pi }{4}}+{\frac {cx}{2}}\right)\right|}$

${\displaystyle \int {\frac {\sin ^{2}cx\;dx}{\cos ^{n}cx}}={\frac {\sin cx}{c(n-1)\cos ^{n-1}cx}}-{\frac {1}{n-1}}\int {\frac {dx}{\cos ^{n-2}cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos cx}}=-{\frac {\sin ^{n-1}cx}{c(n-1)}}+\int {\frac {\sin ^{n-2}cx\;dx}{\cos cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos ^{m}cx}}={\frac {\sin ^{n+1}cx}{c(m-1)\cos ^{m-1}cx}}-{\frac {n-m+2}{m-1}}\int {\frac {\sin ^{n}cx\;dx}{\cos ^{m-2}cx}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}$

also: ${\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos ^{m}cx}}=-{\frac {\sin ^{n-1}cx}{c(n-m)\cos ^{m-1}cx}}+{\frac {n-1}{n-m}}\int {\frac {\sin ^{n-2}cx\;dx}{\cos ^{m}cx}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!}$

also: ${\displaystyle \int {\frac {\sin ^{n}cx\;dx}{\cos ^{m}cx}}={\frac {\sin ^{n-1}cx}{c(m-1)\cos ^{m-1}cx}}-{\frac {n-1}{m-1}}\int {\frac {\sin ^{n-2}cx\;dx}{\cos ^{m-2}cx}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\cos cx\;dx}{\sin ^{n}cx}}=-{\frac {1}{c(n-1)\sin ^{n-1}cx}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\cos ^{2}cx\;dx}{\sin cx}}={\frac {1}{c}}\left(\cos cx+\ln \left|\tan {\frac {cx}{2}}\right|\right)}$

${\displaystyle \int {\frac {\cos ^{2}cx\;dx}{\sin ^{n}cx}}=-{\frac {1}{n-1}}\left({\frac {\cos cx}{c\sin ^{n-1}cx)}}+\int {\frac {dx}{\sin ^{n-2}cx}}\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}$

${\displaystyle \int {\frac {\cos ^{n}cx\;dx}{\sin ^{m}cx}}=-{\frac {\cos ^{n+1}cx}{c(m-1)\sin ^{m-1}cx}}-{\frac {n-m+2}{m-1}}\int {\frac {cos^{n}cx\;dx}{\sin ^{m-2}cx}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}$

also: ${\displaystyle \int {\frac {\cos ^{n}cx\;dx}{\sin ^{m}cx}}={\frac {\cos ^{n-1}cx}{c(n-m)\sin ^{m-1}cx}}+{\frac {n-1}{n-m}}\int {\frac {cos^{n-2}cx\;dx}{\sin ^{m}cx}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!}$

also: ${\displaystyle \int {\frac {\cos ^{n}cx\;dx}{\sin ^{m}cx}}=-{\frac {\cos ^{n-1}cx}{c(m-1)\sin ^{m-1}cx}}-{\frac {n-1}{m-1}}\int {\frac {cos^{n-2}cx\;dx}{\sin ^{m-2}cx}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int \sin cx\tan cx\;dx={\frac {1}{c}}(\ln |\sec cx+\tan cx|-\sin cx)\,\!}$

${\displaystyle \int {\frac {\tan ^{n}cx\;dx}{\sin ^{2}cx}}={\frac {1}{c(n-1)}}\tan ^{n-1}(cx)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\tan ^{n}cx\;dx}{\cos ^{2}cx}}={\frac {1}{c(n+1)}}\tan ^{n+1}cx\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\cot ^{n}cx\;dx}{\sin ^{2}cx}}={\frac {-1}{c(n+1)}}\cot ^{n+1}cx\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\cot ^{n}cx\;dx}{\cos ^{2}cx}}={\frac {1}{c(1-n)}}\tan ^{1-n}cx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$

${\displaystyle \int {\frac {\tan ^{m}(cx)}{\cot ^{n}(cx)}}\;dx={\frac {1}{c(m+n-1)}}\tan ^{m+n-1}(cx)-\int {\frac {\tan ^{m-2}(cx)}{\cot ^{n}(cx)}}\;dx\qquad {\mbox{(for }}m+n\neq 1{\mbox{)}}\,\!}$