三角函数积分表此条目没有列出任何参考或来源。 (2017年12月26日)维基百科所有的内容都应该可供查证。请协助补充可靠来源以改善这篇条目。无法查证的内容可能会因为异议提出而移除。以下是部分三角函数的积分表（省略积分常数）： 目录 1 积分只有sin的函数 2 积分只有cos的函数 3 积分只有tan的函数 4 积分只有sec的函数 5 积分只有csc的函数 6 积分只有cot的函数 7 积分只有sin和cos的函数 8 积分只有sin和tan的函数 9 积分只有cos和tan的函数 10 积分只有sin和cot的函数 11 积分只有cos和cot的函数 12 积分只有tan和cot的函数 积分只有sin的函数 ∫ 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 余矢(Coversine)函数(参阅正矢(versine)函数)） ∫ 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的函数 ∫ 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的函数 ∫ 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的函数 ∫ 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的函数 ∫ 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的函数 ∫ 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的函数 ∫ 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的函数 ∫ 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的函数 ∫ 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的函数 ∫ 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的函数 ∫ 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的函数 ∫ 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{)}}\,\!}