ত্রিকোণমিতিক অপেক্ষকের সমাকলের তালিকা

ত্রিকোণমিতিক অপেক্ষকের সমাকলের তালিকা নিচে প্রদত্ত হল। ∫ sin ⁡ a x d x = − 1 a cos ⁡ a x + C {\displaystyle \int \sin ax\;\mathrm {d} x=-{\frac {1}{a}}\cos ax+C\,\!} ∫ sin 2 ⁡ a x d x = x 2 − 1 4 a sin ⁡ 2 a x + C = x 2 − 1 2 a sin ⁡ a x cos ⁡ a x + C {\displaystyle \int \sin ^{2}{ax}\;\mathrm {d} x={\frac {x}{2}}-{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}-{\frac {1}{2a}}\sin ax\cos ax+C\!} ∫ sin 3 ⁡ a x d x = cos ⁡ 3 a x 12 a − 3 cos ⁡ a x 4 a + C {\displaystyle \int \sin ^{3}{ax}\;\mathrm {d} x={\frac {\cos 3ax}{12a}}-{\frac {3\cos ax}{4a}}+C\!} ∫ x sin 2 ⁡ a x d x = x 2 4 − x 4 a sin ⁡ 2 a x − 1 8 a 2 cos ⁡ 2 a x + C {\displaystyle \int x\sin ^{2}{ax}\;\mathrm {d} x={\frac {x^{2}}{4}}-{\frac {x}{4a}}\sin 2ax-{\frac {1}{8a^{2}}}\cos 2ax+C\!} ∫ x 2 sin 2 ⁡ a x d x = x 3 6 − ( x 2 4 a − 1 8 a 3 ) sin ⁡ 2 a x − x 4 a 2 cos ⁡ 2 a x + C {\displaystyle \int x^{2}\sin ^{2}{ax}\;\mathrm {d} x={\frac {x^{3}}{6}}-\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax-{\frac {x}{4a^{2}}}\cos 2ax+C\!} ∫ sin ⁡ b 1 x sin ⁡ b 2 x d x = sin ⁡ ( ( b 2 − b 1 ) x ) 2 ( b 2 − b 1 ) − sin ⁡ ( ( b 1 + b 2 ) x ) 2 ( b 1 + b 2 ) + C (for  | b 1 | ≠ | b 2 | ) {\displaystyle \int \sin b_{1}x\sin b_{2}x\;\mathrm {d} x={\frac {\sin((b_{2}-b_{1})x)}{2(b_{2}-b_{1})}}-{\frac {\sin((b_{1}+b_{2})x)}{2(b_{1}+b_{2})}}+C\qquad {\mbox{(for }}|b_{1}|\neq |b_{2}|{\mbox{)}}\,\!} ∫ sin n ⁡ a x d x = − sin n − 1 ⁡ a x cos ⁡ a x n a + n − 1 n ∫ sin n − 2 ⁡ a x d x (for  n > 0 ) {\displaystyle \int \sin ^{n}{ax}\;\mathrm {d} x=-{\frac {\sin ^{n-1}ax\cos ax}{na}}+{\frac {n-1}{n}}\int \sin ^{n-2}ax\;\mathrm {d} x\qquad {\mbox{(for }}n>0{\mbox{)}}\,\!} ∫ d x sin ⁡ a x = − 1 a ln ⁡ | csc ⁡ a x + cot ⁡ a x | + C {\displaystyle \int {\frac {\mathrm {d} x}{\sin ax}}=-{\frac {1}{a}}\ln {\left|\csc {ax}+\cot {ax}\right|}+C} ∫ d x sin n ⁡ a x = cos ⁡ a x a ( 1 − n ) sin n − 1 ⁡ a x + n − 2 n − 1 ∫ d x sin n − 2 ⁡ a x (for  n > 1 ) {\displaystyle \int {\frac {\mathrm {d} x}{\sin ^{n}ax}}={\frac {\cos ax}{a(1-n)\sin ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {\mathrm {d} x}{\sin ^{n-2}ax}}\qquad {\mbox{(for }}n>1{\mbox{)}}\,\!} ∫ x sin ⁡ a x d x = sin ⁡ a x a 2 − x cos ⁡ a x a + C {\displaystyle \int x\sin ax\;\mathrm {d} x={\frac {\sin ax}{a^{2}}}-{\frac {x\cos ax}{a}}+C\,\!} ∫ x n sin ⁡ a x d x = − x n a cos ⁡ a x + n a ∫ x n − 1 cos ⁡ a x d x = ∑ k = 0 2 k ≤ n ( − 1 ) k + 1 x n − 2 k a 1 + 2 k n ! ( n − 2 k ) ! cos ⁡ a x + ∑ k = 0 2 k + 1 ≤ n ( − 1 ) k x n − 1 − 2 k a 2 + 2 k n ! ( n − 2 k − 1 ) ! sin ⁡ a x (for  n > 0 ) {\displaystyle \int x^{n}\sin ax\;\mathrm {d} x=-{\frac {x^{n}}{a}}\cos ax+{\frac {n}{a}}\int x^{n-1}\cos ax\;\mathrm {d} x=\sum _{k=0}^{2k\leq n}(-1)^{k+1}{\frac {x^{n-2k}}{a^{1+2k}}}{\frac {n!}{(n-2k)!}}\cos ax+\sum _{k=0}^{2k+1\leq n}(-1)^{k}{\frac {x^{n-1-2k}}{a^{2+2k}}}{\frac {n!}{(n-2k-1)!}}\sin ax\qquad {\mbox{(for }}n>0{\mbox{)}}\,\!} ∫ sin ⁡ a x x d x = ∑ n = 0 ∞ ( − 1 ) n ( a x ) 2 n + 1 ( 2 n + 1 ) ⋅ ( 2 n + 1 ) ! + C {\displaystyle \int {\frac {\sin ax}{x}}\mathrm {d} x=\sum _{n=0}^{\infty }(-1)^{n}{\frac {(ax)^{2n+1}}{(2n+1)\cdot (2n+1)!}}+C\,\!} ∫ sin ⁡ a x x n d x = − sin ⁡ a x ( n − 1 ) x n − 1 + a n − 1 ∫ cos ⁡ a x x n − 1 d x {\displaystyle \int {\frac {\sin ax}{x^{n}}}\mathrm {d} x=-{\frac {\sin ax}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}\int {\frac {\cos ax}{x^{n-1}}}\mathrm {d} x\,\!} ∫ d x 1 ± sin ⁡ a x = 1 a tan ⁡ ( a x 2 ∓ π 4 ) + C {\displaystyle \int {\frac {\mathrm {d} x}{1\pm \sin ax}}={\frac {1}{a}}\tan \left({\frac {ax}{2}}\mp {\frac {\pi }{4}}\right)+C} ∫ x d x 1 + sin ⁡ a x = x a tan ⁡ ( a x 2 − π 4 ) + 2 a 2 ln ⁡ | cos ⁡ ( a x 2 − π 4 ) | + C {\displaystyle \int {\frac {x\;\mathrm {d} x}{1+\sin ax}}={\frac {x}{a}}\tan \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)+{\frac {2}{a^{2}}}\ln \left|\cos \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)\right|+C} ∫ x d x 1 − sin ⁡ a x = x a cot ⁡ ( π 4 − a x 2 ) + 2 a 2 ln ⁡ | sin ⁡ ( π 4 − a x 2 ) | + C {\displaystyle \int {\frac {x\;\mathrm {d} x}{1-\sin ax}}={\frac {x}{a}}\cot \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)+{\frac {2}{a^{2}}}\ln \left|\sin \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)\right|+C} ∫ sin ⁡ a x d x 1 ± sin ⁡ a x = ± x + 1 a tan ⁡ ( π 4 ∓ a x 2 ) + C {\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{1\pm \sin ax}}=\pm x+{\frac {1}{a}}\tan \left({\frac {\pi }{4}}\mp {\frac {ax}{2}}\right)+C} ∫ cos ⁡ a x d x = 1 a sin ⁡ a x + C {\displaystyle \int \cos ax\;\mathrm {d} x={\frac {1}{a}}\sin ax+C\,\!} ∫ cos 2 ⁡ a x d x = x 2 + 1 4 a sin ⁡ 2 a x + C = x 2 + 1 2 a sin ⁡ a x cos ⁡ a x + C {\displaystyle \int \cos ^{2}{ax}\;\mathrm {d} x={\frac {x}{2}}+{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}+{\frac {1}{2a}}\sin ax\cos ax+C\!} ∫ cos n ⁡ a x d x = cos n − 1 ⁡ a x sin ⁡ a x n a + n − 1 n ∫ cos n − 2 ⁡ a x d x (for  n > 0 ) {\displaystyle \int \cos ^{n}ax\;\mathrm {d} x={\frac {\cos ^{n-1}ax\sin ax}{na}}+{\frac {n-1}{n}}\int \cos ^{n-2}ax\;\mathrm {d} x\qquad {\mbox{(for }}n>0{\mbox{)}}\,\!} ∫ x cos ⁡ a x d x = cos ⁡ a x a 2 + x sin ⁡ a x a + C {\displaystyle \int x\cos ax\;\mathrm {d} x={\frac {\cos ax}{a^{2}}}+{\frac {x\sin ax}{a}}+C\,\!} ∫ x 2 cos 2 ⁡ a x d x = x 3 6 + ( x 2 4 a − 1 8 a 3 ) sin ⁡ 2 a x + x 4 a 2 cos ⁡ 2 a x + C {\displaystyle \int x^{2}\cos ^{2}{ax}\;\mathrm {d} x={\frac {x^{3}}{6}}+\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax+{\frac {x}{4a^{2}}}\cos 2ax+C\!} ∫ x n cos ⁡ a x d x = x n sin ⁡ a x a − n a ∫ x n − 1 sin ⁡ a x d x = ∑ k = 0 2 k + 1 ≤ n ( − 1 ) k x n − 2 k − 1 a 2 + 2 k n ! ( n − 2 k − 1 ) ! cos ⁡ a x + ∑ k = 0 2 k ≤ n ( − 1 ) k x n − 2 k a 1 + 2 k n ! ( n − 2 k ) ! sin ⁡ a x {\displaystyle \int x^{n}\cos ax\;\mathrm {d} x={\frac {x^{n}\sin ax}{a}}-{\frac {n}{a}}\int x^{n-1}\sin ax\;\mathrm {d} x\,=\sum _{k=0}^{2k+1\leq n}(-1)^{k}{\frac {x^{n-2k-1}}{a^{2+2k}}}{\frac {n!}{(n-2k-1)!}}\cos ax+\sum _{k=0}^{2k\leq n}(-1)^{k}{\frac {x^{n-2k}}{a^{1+2k}}}{\frac {n!}{(n-2k)!}}\sin ax\!} ∫ cos ⁡ a x x d x = ln ⁡ | a x | + ∑ k = 1 ∞ ( − 1 ) k ( a x ) 2 k 2 k ⋅ ( 2 k ) ! + C {\displaystyle \int {\frac {\cos ax}{x}}\mathrm {d} x=\ln |ax|+\sum _{k=1}^{\infty }(-1)^{k}{\frac {(ax)^{2k}}{2k\cdot (2k)!}}+C\,\!} ∫ cos ⁡ a x x n d x = − cos ⁡ a x ( n − 1 ) x n − 1 − a n − 1 ∫ sin ⁡ a x x n − 1 d x (for  n ≠ 1 ) {\displaystyle \int {\frac {\cos ax}{x^{n}}}\mathrm {d} x=-{\frac {\cos ax}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\int {\frac {\sin ax}{x^{n-1}}}\mathrm {d} x\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ d x cos ⁡ a x = 1 a ln ⁡ | tan ⁡ ( a x 2 + π 4 ) | + C {\displaystyle \int {\frac {\mathrm {d} x}{\cos ax}}={\frac {1}{a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C} ∫ d x cos n ⁡ a x = sin ⁡ a x a ( n − 1 ) cos n − 1 ⁡ a x + n − 2 n − 1 ∫ d x cos n − 2 ⁡ a x (for  n > 1 ) {\displaystyle \int {\frac {\mathrm {d} x}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {\mathrm {d} x}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n>1{\mbox{)}}\,\!} ∫ d x 1 + cos ⁡ a x = 1 a tan ⁡ a x 2 + C {\displaystyle \int {\frac {\mathrm {d} x}{1+\cos ax}}={\frac {1}{a}}\tan {\frac {ax}{2}}+C\,\!} ∫ d x 1 − cos ⁡ a x = − 1 a cot ⁡ a x 2 + C {\displaystyle \int {\frac {\mathrm {d} x}{1-\cos ax}}=-{\frac {1}{a}}\cot {\frac {ax}{2}}+C} ∫ x d x 1 + cos ⁡ a x = x a tan ⁡ a x 2 + 2 a 2 ln ⁡ | cos ⁡ a x 2 | + C {\displaystyle \int {\frac {x\;\mathrm {d} x}{1+\cos ax}}={\frac {x}{a}}\tan {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left|\cos {\frac {ax}{2}}\right|+C} ∫ x d x 1 − cos ⁡ a x = − x a cot ⁡ a x 2 + 2 a 2 ln ⁡ | sin ⁡ a x 2 | + C {\displaystyle \int {\frac {x\;\mathrm {d} x}{1-\cos ax}}=-{\frac {x}{a}}\cot {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left|\sin {\frac {ax}{2}}\right|+C} ∫ cos ⁡ a x d x 1 + cos ⁡ a x = x − 1 a tan ⁡ a x 2 + C {\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{1+\cos ax}}=x-{\frac {1}{a}}\tan {\frac {ax}{2}}+C\,\!} ∫ cos ⁡ a x d x 1 − cos ⁡ a x = − x − 1 a cot ⁡ a x 2 + C {\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{1-\cos ax}}=-x-{\frac {1}{a}}\cot {\frac {ax}{2}}+C\,\!} ∫ cos ⁡ a 1 x cos ⁡ a 2 x d x = sin ⁡ ( ( a 2 − a 1 ) x ) 2 ( a 2 − a 1 ) + sin ⁡ ( ( a 2 + a 1 ) x ) 2 ( a 2 + a 1 ) + C (for  | a 1 | ≠ | a 2 | ) {\displaystyle \int \cos a_{1}x\cos a_{2}x\;\mathrm {d} x={\frac {\sin((a_{2}-a_{1})x)}{2(a_{2}-a_{1})}}+{\frac {\sin((a_{2}+a_{1})x)}{2(a_{2}+a_{1})}}+C\qquad {\mbox{(for }}|a_{1}|\neq |a_{2}|{\mbox{)}}\,\!} ∫ tan ⁡ a x d x = − 1 a ln ⁡ | cos ⁡ a x | + C = 1 a ln ⁡ | sec ⁡ a x | + C {\displaystyle \int \tan ax\;\mathrm {d} x=-{\frac {1}{a}}\ln |\cos ax|+C={\frac {1}{a}}\ln |\sec ax|+C\,\!} ∫ tan 2 ⁡ x d x = tan ⁡ x − x + C {\displaystyle \int \tan ^{2}{x}\,\mathrm {d} x=\tan {x}-x+C} ∫ tan n ⁡ a x d x = 1 a ( n − 1 ) tan n − 1 ⁡ a x − ∫ tan n − 2 ⁡ a x d x (for  n ≠ 1 ) {\displaystyle \int \tan ^{n}ax\;\mathrm {d} x={\frac {1}{a(n-1)}}\tan ^{n-1}ax-\int \tan ^{n-2}ax\;\mathrm {d} x\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ d x q tan ⁡ a x + p = 1 p 2 + q 2 ( p x + q a ln ⁡ | q sin ⁡ a x + p cos ⁡ a x | ) + C (for  p 2 + q 2 ≠ 0 ) {\displaystyle \int {\frac {\mathrm {d} x}{q\tan ax+p}}={\frac {1}{p^{2}+q^{2}}}(px+{\frac {q}{a}}\ln |q\sin ax+p\cos ax|)+C\qquad {\mbox{(for }}p^{2}+q^{2}\neq 0{\mbox{)}}\,\!} ∫ d x tan ⁡ a x + 1 = x 2 + 1 2 a ln ⁡ | sin ⁡ a x + cos ⁡ a x | + C {\displaystyle \int {\frac {\mathrm {d} x}{\tan ax+1}}={\frac {x}{2}}+{\frac {1}{2a}}\ln |\sin ax+\cos ax|+C\,\!} ∫ d x tan ⁡ a x − 1 = − x 2 + 1 2 a ln ⁡ | sin ⁡ a x − cos ⁡ a x | + C {\displaystyle \int {\frac {\mathrm {d} x}{\tan ax-1}}=-{\frac {x}{2}}+{\frac {1}{2a}}\ln |\sin ax-\cos ax|+C\,\!} ∫ tan ⁡ a x d x tan ⁡ a x + 1 = x 2 − 1 2 a ln ⁡ | sin ⁡ a x + cos ⁡ a x | + C {\displaystyle \int {\frac {\tan ax\;\mathrm {d} x}{\tan ax+1}}={\frac {x}{2}}-{\frac {1}{2a}}\ln |\sin ax+\cos ax|+C\,\!} ∫ tan ⁡ a x d x tan ⁡ a x − 1 = x 2 + 1 2 a ln ⁡ | sin ⁡ a x − cos ⁡ a x | + C {\displaystyle \int {\frac {\tan ax\;\mathrm {d} x}{\tan ax-1}}={\frac {x}{2}}+{\frac {1}{2a}}\ln |\sin ax-\cos ax|+C\,\!} দেখুন সেকান্ট অপেক্ষকের সমাকল ∫ sec ⁡ a x d x = 1 a ln ⁡ | sec ⁡ a x + tan ⁡ a x | + C {\displaystyle \int \sec {ax}\,\mathrm {d} x={\frac {1}{a}}\ln {\left|\sec {ax}+\tan {ax}\right|}+C} ∫ sec 2 ⁡ x d x = tan ⁡ x + C {\displaystyle \int \sec ^{2}{x}\,\mathrm {d} x=\tan {x}+C} ∫ sec 3 ⁡ x d x = 1 2 sec ⁡ x tan ⁡ x + 1 2 ln ⁡ | sec ⁡ x + tan ⁡ x | + C . {\displaystyle \int \sec ^{3}x\,dx={\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln |\sec x+\tan x|+C.} ∫ sec n ⁡ a x d x = sec n − 2 ⁡ a x tan ⁡ a x a ( n − 1 ) + n − 2 n − 1 ∫ sec n − 2 ⁡ a x d x  (for  n ≠ 1 ) {\displaystyle \int \sec ^{n}{ax}\,\mathrm {d} x={\frac {\sec ^{n-2}{ax}\tan {ax}}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{ax}\,\mathrm {d} x\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,\!} ∫ d x sec ⁡ x + 1 = x − tan ⁡ x 2 + C {\displaystyle \int {\frac {\mathrm {d} x}{\sec {x}+1}}=x-\tan {\frac {x}{2}}+C} ∫ csc ⁡ ( a x ) d x = − 1 a ln ⁡ | csc ⁡ a x + cot ⁡ a x | + C {\displaystyle \int \csc(ax)\mathrm {d} x=-{\frac {1}{a}}\ln {\left|\csc {ax}+\cot {ax}\right|}+C} ∫ csc 2 ⁡ x d x = − cot ⁡ x + C {\displaystyle \int \csc ^{2}{x}\,\mathrm {d} x=-\cot {x}+C} ∫ csc n ⁡ a x d x = − csc n − 1 ⁡ ( a x ) cos ⁡ ( a x ) a ( n − 1 ) + n − 2 n − 1 ∫ csc n − 2 ⁡ a x d x  (for  n ≠ 1 ) {\displaystyle \int \csc ^{n}{ax}\,\mathrm {d} x=-{\frac {\csc ^{n-1}\left(ax\right)\cos \left(ax\right)}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \csc ^{n-2}{ax}\,\mathrm {d} x\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,\!} ∫ d x csc ⁡ x + 1 = x − 2 sin ⁡ x 2 cos ⁡ x 2 + sin ⁡ x 2 + C {\displaystyle \int {\frac {\mathrm {d} x}{\csc {x}+1}}=x-{\frac {2\sin {\frac {x}{2}}}{\cos {\frac {x}{2}}+\sin {\frac {x}{2}}}}+C} ∫ d x csc ⁡ x − 1 = 2 sin ⁡ x 2 cos ⁡ x 2 − sin ⁡ x 2 − x + C {\displaystyle \int {\frac {\mathrm {d} x}{\csc {x}-1}}={\frac {2\sin {\frac {x}{2}}}{\cos {\frac {x}{2}}-\sin {\frac {x}{2}}}}-x+C} ∫ d x cos ⁡ a x ± sin ⁡ a x = 1 a 2 ln ⁡ | tan ⁡ ( a x 2 ± π 8 ) | + C {\displaystyle \int {\frac {\mathrm {d} x}{\cos ax\pm \sin ax}}={\frac {1}{a{\sqrt {2}}}}\ln \left|\tan \left({\frac {ax}{2}}\pm {\frac {\pi }{8}}\right)\right|+C} ∫ d x ( cos ⁡ a x ± sin ⁡ a x ) 2 = 1 2 a tan ⁡ ( a x ∓ π 4 ) + C {\displaystyle \int {\frac {\mathrm {d} x}{(\cos ax\pm \sin ax)^{2}}}={\frac {1}{2a}}\tan \left(ax\mp {\frac {\pi }{4}}\right)+C} ∫ 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 {\mathrm {d} x}{(\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 {\mathrm {d} x}{(\cos x+\sin x)^{n-2}}}\right)} ∫ cos ⁡ a x d x cos ⁡ a x + sin ⁡ a x = x 2 + 1 2 a ln ⁡ | sin ⁡ a x + cos ⁡ a x | + C {\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{\cos ax+\sin ax}}={\frac {x}{2}}+{\frac {1}{2a}}\ln \left|\sin ax+\cos ax\right|+C} ∫ cos ⁡ a x d x cos ⁡ a x − sin ⁡ a x = x 2 − 1 2 a ln ⁡ | sin ⁡ a x − cos ⁡ a x | + C {\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{\cos ax-\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax-\cos ax\right|+C} ∫ sin ⁡ a x d x cos ⁡ a x + sin ⁡ a x = x 2 − 1 2 a ln ⁡ | sin ⁡ a x + cos ⁡ a x | + C {\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{\cos ax+\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax+\cos ax\right|+C} ∫ sin ⁡ a x d x cos ⁡ a x − sin ⁡ a x = − x 2 − 1 2 a ln ⁡ | sin ⁡ a x − cos ⁡ a x | + C {\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{\cos ax-\sin ax}}=-{\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax-\cos ax\right|+C} ∫ cos ⁡ a x d x sin ⁡ a x ( 1 + cos ⁡ a x ) = − 1 4 a tan 2 ⁡ a x 2 + 1 2 a ln ⁡ | tan ⁡ a x 2 | + C {\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{\sin ax(1+\cos ax)}}=-{\frac {1}{4a}}\tan ^{2}{\frac {ax}{2}}+{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C} ∫ cos ⁡ a x d x sin ⁡ a x ( 1 − cos ⁡ a x ) = − 1 4 a cot 2 ⁡ a x 2 − 1 2 a ln ⁡ | tan ⁡ a x 2 | + C {\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{\sin ax(1-\cos ax)}}=-{\frac {1}{4a}}\cot ^{2}{\frac {ax}{2}}-{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C} ∫ sin ⁡ a x d x cos ⁡ a x ( 1 + sin ⁡ a x ) = 1 4 a cot 2 ⁡ ( a x 2 + π 4 ) + 1 2 a ln ⁡ | tan ⁡ ( a x 2 + π 4 ) | + C {\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{\cos ax(1+\sin ax)}}={\frac {1}{4a}}\cot ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)+{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C} ∫ sin ⁡ a x d x cos ⁡ a x ( 1 − sin ⁡ a x ) = 1 4 a tan 2 ⁡ ( a x 2 + π 4 ) − 1 2 a ln ⁡ | tan ⁡ ( a x 2 + π 4 ) | + C {\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{\cos ax(1-\sin ax)}}={\frac {1}{4a}}\tan ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)-{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C} ∫ sin ⁡ a x cos ⁡ a x d x = − 1 2 a cos 2 ⁡ a x + C {\displaystyle \int \sin ax\cos ax\;\mathrm {d} x=-{\frac {1}{2a}}\cos ^{2}ax+C\,\!} ∫ sin ⁡ a 1 x cos ⁡ a 2 x d x = − cos ⁡ ( ( a 1 − a 2 ) x ) 2 ( a 1 − a 2 ) − cos ⁡ ( ( a 1 + a 2 ) x ) 2 ( a 1 + a 2 ) + C (for  | a 1 | ≠ | a 2 | ) {\displaystyle \int \sin a_{1}x\cos a_{2}x\;\mathrm {d} x=-{\frac {\cos((a_{1}-a_{2})x)}{2(a_{1}-a_{2})}}-{\frac {\cos((a_{1}+a_{2})x)}{2(a_{1}+a_{2})}}+C\qquad {\mbox{(for }}|a_{1}|\neq |a_{2}|{\mbox{)}}\,\!} ∫ sin n ⁡ a x cos ⁡ a x d x = 1 a ( n + 1 ) sin n + 1 ⁡ a x + C (for  n ≠ − 1 ) {\displaystyle \int \sin ^{n}ax\cos ax\;\mathrm {d} x={\frac {1}{a(n+1)}}\sin ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!} ∫ sin ⁡ a x cos n ⁡ a x d x = − 1 a ( n + 1 ) cos n + 1 ⁡ a x + C (for  n ≠ − 1 ) {\displaystyle \int \sin ax\cos ^{n}ax\;\mathrm {d} x=-{\frac {1}{a(n+1)}}\cos ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!} ∫ sin n ⁡ a x cos m ⁡ a x d x = − sin n − 1 ⁡ a x cos m + 1 ⁡ a x a ( n + m ) + n − 1 n + m ∫ sin n − 2 ⁡ a x cos m ⁡ a x d x (for  m , n > 0 ) {\displaystyle \int \sin ^{n}ax\cos ^{m}ax\;\mathrm {d} x=-{\frac {\sin ^{n-1}ax\cos ^{m+1}ax}{a(n+m)}}+{\frac {n-1}{n+m}}\int \sin ^{n-2}ax\cos ^{m}ax\;\mathrm {d} x\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!} also: ∫ sin n ⁡ a x cos m ⁡ a x d x = sin n + 1 ⁡ a x cos m − 1 ⁡ a x a ( n + m ) + m − 1 n + m ∫ sin n ⁡ a x cos m − 2 ⁡ a x d x (for  m , n > 0 ) {\displaystyle \int \sin ^{n}ax\cos ^{m}ax\;\mathrm {d} x={\frac {\sin ^{n+1}ax\cos ^{m-1}ax}{a(n+m)}}+{\frac {m-1}{n+m}}\int \sin ^{n}ax\cos ^{m-2}ax\;\mathrm {d} x\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!} ∫ d x sin ⁡ a x cos ⁡ a x = 1 a ln ⁡ | tan ⁡ a x | + C {\displaystyle \int {\frac {\mathrm {d} x}{\sin ax\cos ax}}={\frac {1}{a}}\ln \left|\tan ax\right|+C} ∫ d x sin ⁡ a x cos n ⁡ a x = 1 a ( n − 1 ) cos n − 1 ⁡ a x + ∫ d x sin ⁡ a x cos n − 2 ⁡ a x (for  n ≠ 1 ) {\displaystyle \int {\frac {\mathrm {d} x}{\sin ax\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+\int {\frac {\mathrm {d} x}{\sin ax\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ d x sin n ⁡ a x cos ⁡ a x = − 1 a ( n − 1 ) sin n − 1 ⁡ a x + ∫ d x sin n − 2 ⁡ a x cos ⁡ a x (for  n ≠ 1 ) {\displaystyle \int {\frac {\mathrm {d} x}{\sin ^{n}ax\cos ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+\int {\frac {\mathrm {d} x}{\sin ^{n-2}ax\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ sin ⁡ a x d x cos n ⁡ a x = 1 a ( n − 1 ) cos n − 1 ⁡ a x + C (for  n ≠ 1 ) {\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ sin 2 ⁡ a x d x cos ⁡ a x = − 1 a sin ⁡ a x + 1 a ln ⁡ | tan ⁡ ( π 4 + a x 2 ) | + C {\displaystyle \int {\frac {\sin ^{2}ax\;\mathrm {d} x}{\cos ax}}=-{\frac {1}{a}}\sin ax+{\frac {1}{a}}\ln \left|\tan \left({\frac {\pi }{4}}+{\frac {ax}{2}}\right)\right|+C} ∫ sin 2 ⁡ a x d x cos n ⁡ a x = sin ⁡ a x a ( n − 1 ) cos n − 1 ⁡ a x − 1 n − 1 ∫ d x cos n − 2 ⁡ a x (for  n ≠ 1 ) {\displaystyle \int {\frac {\sin ^{2}ax\;\mathrm {d} x}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}-{\frac {1}{n-1}}\int {\frac {\mathrm {d} x}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ sin n ⁡ a x d x cos ⁡ a x = − sin n − 1 ⁡ a x a ( n − 1 ) + ∫ sin n − 2 ⁡ a x d x cos ⁡ a x (for  n ≠ 1 ) {\displaystyle \int {\frac {\sin ^{n}ax\;\mathrm {d} x}{\cos ax}}=-{\frac {\sin ^{n-1}ax}{a(n-1)}}+\int {\frac {\sin ^{n-2}ax\;\mathrm {d} x}{\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ sin n ⁡ a x d x cos m ⁡ a x = sin n + 1 ⁡ a x a ( m − 1 ) cos m − 1 ⁡ a x − n − m + 2 m − 1 ∫ sin n ⁡ a x d x cos m − 2 ⁡ a x (for  m ≠ 1 ) {\displaystyle \int {\frac {\sin ^{n}ax\;\mathrm {d} x}{\cos ^{m}ax}}={\frac {\sin ^{n+1}ax}{a(m-1)\cos ^{m-1}ax}}-{\frac {n-m+2}{m-1}}\int {\frac {\sin ^{n}ax\;\mathrm {d} x}{\cos ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!} also: ∫ sin n ⁡ a x d x cos m ⁡ a x = − sin n − 1 ⁡ a x a ( n − m ) cos m − 1 ⁡ a x + n − 1 n − m ∫ sin n − 2 ⁡ a x d x cos m ⁡ a x (for  m ≠ n ) {\displaystyle \int {\frac {\sin ^{n}ax\;\mathrm {d} x}{\cos ^{m}ax}}=-{\frac {\sin ^{n-1}ax}{a(n-m)\cos ^{m-1}ax}}+{\frac {n-1}{n-m}}\int {\frac {\sin ^{n-2}ax\;\mathrm {d} x}{\cos ^{m}ax}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!} also: ∫ sin n ⁡ a x d x cos m ⁡ a x = sin n − 1 ⁡ a x a ( m − 1 ) cos m − 1 ⁡ a x − n − 1 m − 1 ∫ sin n − 2 ⁡ a x d x cos m − 2 ⁡ a x (for  m ≠ 1 ) {\displaystyle \int {\frac {\sin ^{n}ax\;\mathrm {d} x}{\cos ^{m}ax}}={\frac {\sin ^{n-1}ax}{a(m-1)\cos ^{m-1}ax}}-{\frac {n-1}{m-1}}\int {\frac {\sin ^{n-2}ax\;\mathrm {d} x}{\cos ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!} ∫ cos ⁡ a x d x sin n ⁡ a x = − 1 a ( n − 1 ) sin n − 1 ⁡ a x + C (for  n ≠ 1 ) {\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{\sin ^{n}ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ cos 2 ⁡ a x d x sin ⁡ a x = 1 a ( cos ⁡ a x + ln ⁡ | tan ⁡ a x 2 | ) + C {\displaystyle \int {\frac {\cos ^{2}ax\;\mathrm {d} x}{\sin ax}}={\frac {1}{a}}\left(\cos ax+\ln \left|\tan {\frac {ax}{2}}\right|\right)+C} ∫ cos 2 ⁡ a x d x sin n ⁡ a x = − 1 n − 1 ( cos ⁡ a x a sin n − 1 ⁡ a x ) + ∫ d x sin n − 2 ⁡ a x ) (for  n ≠ 1 ) {\displaystyle \int {\frac {\cos ^{2}ax\;\mathrm {d} x}{\sin ^{n}ax}}=-{\frac {1}{n-1}}\left({\frac {\cos ax}{a\sin ^{n-1}ax)}}+\int {\frac {\mathrm {d} x}{\sin ^{n-2}ax}}\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}} ∫ cos n ⁡ a x d x sin m ⁡ a x = − cos n + 1 ⁡ a x a ( m − 1 ) sin m − 1 ⁡ a x − n − m + 2 m − 1 ∫ cos n ⁡ a x d x sin m − 2 ⁡ a x (for  m ≠ 1 ) {\displaystyle \int {\frac {\cos ^{n}ax\;\mathrm {d} x}{\sin ^{m}ax}}=-{\frac {\cos ^{n+1}ax}{a(m-1)\sin ^{m-1}ax}}-{\frac {n-m+2}{m-1}}\int {\frac {\cos ^{n}ax\;\mathrm {d} x}{\sin ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!} also: ∫ cos n ⁡ a x d x sin m ⁡ a x = cos n − 1 ⁡ a x a ( n − m ) sin m − 1 ⁡ a x + n − 1 n − m ∫ cos n − 2 ⁡ a x d x sin m ⁡ a x (for  m ≠ n ) {\displaystyle \int {\frac {\cos ^{n}ax\;\mathrm {d} x}{\sin ^{m}ax}}={\frac {\cos ^{n-1}ax}{a(n-m)\sin ^{m-1}ax}}+{\frac {n-1}{n-m}}\int {\frac {\cos ^{n-2}ax\;\mathrm {d} x}{\sin ^{m}ax}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!} also: ∫ cos n ⁡ a x d x sin m ⁡ a x = − cos n − 1 ⁡ a x a ( m − 1 ) sin m − 1 ⁡ a x − n − 1 m − 1 ∫ cos n − 2 ⁡ a x d x sin m − 2 ⁡ a x (for  m ≠ 1 ) {\displaystyle \int {\frac {\cos ^{n}ax\;\mathrm {d} x}{\sin ^{m}ax}}=-{\frac {\cos ^{n-1}ax}{a(m-1)\sin ^{m-1}ax}}-{\frac {n-1}{m-1}}\int {\frac {\cos ^{n-2}ax\;\mathrm {d} x}{\sin ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!} ∫ sin ⁡ a x tan ⁡ a x d x = 1 a ( ln ⁡ | sec ⁡ a x + tan ⁡ a x | − sin ⁡ a x ) + C {\displaystyle \int \sin ax\tan ax\;\mathrm {d} x={\frac {1}{a}}(\ln |\sec ax+\tan ax|-\sin ax)+C\,\!} ∫ tan n ⁡ a x d x sin 2 ⁡ a x = 1 a ( n − 1 ) tan n − 1 ⁡ ( a x ) + C (for  n ≠ 1 ) {\displaystyle \int {\frac {\tan ^{n}ax\;\mathrm {d} x}{\sin ^{2}ax}}={\frac {1}{a(n-1)}}\tan ^{n-1}(ax)+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ tan n ⁡ a x d x cos 2 ⁡ a x = 1 a ( n + 1 ) tan n + 1 ⁡ a x + C (for  n ≠ − 1 ) {\displaystyle \int {\frac {\tan ^{n}ax\;\mathrm {d} x}{\cos ^{2}ax}}={\frac {1}{a(n+1)}}\tan ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!} ∫ cot n ⁡ a x d x sin 2 ⁡ a x = − 1 a ( n + 1 ) cot n + 1 ⁡ a x + C (for  n ≠ − 1 ) {\displaystyle \int {\frac {\cot ^{n}ax\;\mathrm {d} x}{\sin ^{2}ax}}=-{\frac {1}{a(n+1)}}\cot ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!} ∫ cot n ⁡ a x d x cos 2 ⁡ a x = 1 a ( 1 − n ) tan 1 − n ⁡ a x + C (for  n ≠ 1 ) {\displaystyle \int {\frac {\cot ^{n}ax\;\mathrm {d} x}{\cos ^{2}ax}}={\frac {1}{a(1-n)}}\tan ^{1-n}ax+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ sec ⁡ x tan ⁡ x d x = sec ⁡ x + C {\displaystyle \int \sec x\tan x\;\mathrm {d} x=\sec x+C} ∫ 0 π 2 sin n ⁡ x d x = ∫ 0 π 2 cos n ⁡ x d x = { n − 1 n ⋅ n − 3 n − 2 ⋯ 3 4 ⋅ 1 2 ⋅ π 2 , if  n  is even n − 1 n ⋅ n − 3 n − 2 ⋯ 4 5 ⋅ 2 3 if  n  is odd and more than 1 {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}x\,dx={\begin{cases}{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdots {\frac {3}{4}}\cdot {\frac {1}{2}}\cdot {\frac {\pi }{2}},&{\text{if }}n{\text{ is even}}\\{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdots {\frac {4}{5}}\cdot {\frac {2}{3}}&{\text{if }}n{\text{ is odd and more than 1}}\end{cases}}} ∫ − c c sin ⁡ x d x = 0 {\displaystyle \int _{-c}^{c}\sin {x}\;\mathrm {d} x=0\!} ∫ − c c cos ⁡ x d x = 2 ∫ 0 c cos ⁡ x d x = 2 ∫ − c 0 cos ⁡ x d x = 2 sin ⁡ c {\displaystyle \int _{-c}^{c}\cos {x}\;\mathrm {d} x=2\int _{0}^{c}\cos {x}\;\mathrm {d} x=2\int _{-c}^{0}\cos {x}\;\mathrm {d} x=2\sin {c}\!} ∫ − c c tan ⁡ x d x = 0 {\displaystyle \int _{-c}^{c}\tan {x}\;\mathrm {d} x=0\!} ∫ − 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 (for  n = 1 , 3 , 5... ) {\displaystyle \int _{-{\frac {a}{2}}}^{\frac {a}{2}}x^{2}\cos ^{2}{\frac {n\pi x}{a}}\;\mathrm {d} x={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad {\mbox{(for }}n=1,3,5...{\mbox{)}}\,\!} ∫ − a 2 a 2 x 2 sin 2 ⁡ n π x a d x = a 3 ( n 2 π 2 − 6 ( − 1 ) n ) 24 n 2 π 2 = a 3 24 ( 1 − 6 ( − 1 ) n n 2 π 2 ) (for  n = 1 , 2 , 3 , . . . ) {\displaystyle \int _{\frac {-a}{2}}^{\frac {a}{2}}x^{2}\sin ^{2}{\frac {n\pi x}{a}}\;\mathrm {d} x={\frac {a^{3}(n^{2}\pi ^{2}-6(-1)^{n})}{24n^{2}\pi ^{2}}}={\frac {a^{3}}{24}}(1-6{\frac {(-1)^{n}}{n^{2}\pi ^{2}}})\qquad {\mbox{(for }}n=1,2,3,...{\mbox{)}}\,\!} ∫ 0 2 π sin 2 m + 1 ⁡ x cos 2 n + 1 ⁡ x d x = 0 { n , m } ∈ Z {\displaystyle \int _{0}^{2\pi }\sin ^{2m+1}{x}\cos ^{2n+1}{x}\;\mathrm {d} x=0\!\qquad \{n,m\}\in \mathbb {Z} } V. Frederick Rickey and Philip M. Tuchinsky, "An Application of Geography to Mathematics: History of the Integral of the Secant", Mathematics Magazine, volume 53, number 3, May 2980, pages 162–166