三角関数の原始関数の一覧
∫ 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 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!
∫ 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}
∫ sec n  x d x = sec n − 2  x tan  x n − 1 + n − 2 n − 1 ∫ sec n − 2  x d x {\displaystyle \int \sec ^{n}{x}\,\mathrm {d} x={\frac {\sec ^{n-2}{x}\tan {x}}{n-1}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{x}\,\mathrm {d} x}
∫ 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}
∫ d x sec  x − 1 = − x − cot  x 2 + C {\displaystyle \int {\frac {\mathrm {d} x}{\sec {x}-1}}=-x-\cot {\frac {x}{2}}+C}
∫ tan n  a x d x cos 2  a x = 1 a ( n + 1 ) tan n + 1  a x + C ( 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{(}}n\neq -1{\mbox{)}}\,\!}
∫ 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 ( 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{(}}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\,\!}
∫ 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\,\!}