Danh sách tích phân với hàm lượng giác

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Đây là danh sách tích phân (nguyên hàm) của các hàm lượng giác. Đối với tích phân của chứa hàm lượng giác và hàm mũ, xem Danh sách tích phân với hàm mũ. Đối với danh sách đầy đủ các tích phân, xem Danh sách tích phân. Đối với danh sách các tích phân đặc biệt của các hàm lượng giác, xem Tích phân lượng giác.

Nhìn chung, với cos ⁡ ( x ) {\displaystyle \cos(x)} là đạo hàm của hàm số sin ⁡ ( x ) {\displaystyle \sin(x)} , ta có

∫ a cos ⁡ n x d x = a n sin ⁡ n x + C {\displaystyle \int a\cos nx\,dx={\frac {a}{n}}\sin nx+C}

Trong mọi công thức dưới đây, a là một hằng số khác không và C ký hiệu cho hằng số tích phân.

Tích phân chỉ chứa hàm sin

∫ sin ⁡ a x d x = − 1 a cos ⁡ a x + C {\displaystyle \int \sin ax\,dx=-{\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}\,dx={\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}\,dx={\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}\,dx={\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}\,dx={\frac {x^{3}}{6}}-\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax-{\frac {x}{4a^{2}}}\cos 2ax+C}
∫ x sin ⁡ a x d x = sin ⁡ a x a 2 − x cos ⁡ a x a + C {\displaystyle \int x\sin ax\,dx={\frac {\sin ax}{a^{2}}}-{\frac {x\cos ax}{a}}+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 ( | b 1 | ≠ | b 2 | ) {\displaystyle \int (\sin b_{1}x)(\sin b_{2}x)\,dx={\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{(}}|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 ( n > 0 ) {\displaystyle \int \sin ^{n}{ax}\,dx=-{\frac {\sin ^{n-1}ax\cos ax}{na}}+{\frac {n-1}{n}}\int \sin ^{n-2}ax\,dx\qquad {\mbox{(}}n>0{\mbox{)}}}
∫ d x sin ⁡ a x = − 1 a ln ⁡ | csc ⁡ a x + cot ⁡ a x | + C {\displaystyle \int {\frac {dx}{\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 ( n > 1 ) {\displaystyle \int {\frac {dx}{\sin ^{n}ax}}={\frac {\cos ax}{a(1-n)\sin ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\sin ^{n-2}ax}}\qquad {\mbox{(}}n>1{\mbox{)}}}
∫ 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 = − ∑ k = 0 n x n − k a 1 + k n ! ( n − k ) ! cos ⁡ ( a x + k π 2 ) ( n > 0 ) {\displaystyle {\begin{aligned}\int x^{n}\sin ax\,dx&=-{\frac {x^{n}}{a}}\cos ax+{\frac {n}{a}}\int x^{n-1}\cos ax\,dx\\&=\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\\&=-\sum _{k=0}^{n}{\frac {x^{n-k}}{a^{1+k}}}{\frac {n!}{(n-k)!}}\cos \left(ax+k{\frac {\pi }{2}}\right)\qquad {\mbox{(}}n>0{\mbox{)}}\end{aligned}}}
∫ 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}}\,dx=\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}}}\,dx=-{\frac {\sin ax}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}\int {\frac {\cos ax}{x^{n-1}}}\,dx}
∫ d x 1 ± sin ⁡ a x = 1 a tan ⁡ ( a x 2 ∓ π 4 ) + C {\displaystyle \int {\frac {dx}{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\,dx}{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\,dx}{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\,dx}{1\pm \sin ax}}=\pm x+{\frac {1}{a}}\tan \left({\frac {\pi }{4}}\mp {\frac {ax}{2}}\right)+C}

Tích phân chỉ chứa hàm cos

∫ cos ⁡ a x d x = 1 a sin ⁡ a x + C {\displaystyle \int \cos ax\,dx={\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}\,dx={\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\,dx={\frac {\cos ^{n-1}ax\sin ax}{na}}+{\frac {n-1}{n}}\int \cos ^{n-2}ax\,dx\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\,dx={\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}\,dx={\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 = ∑ k = 0 n ( − 1 ) ⌊ k / 2 ⌋ x n − k a 1 + k n ! ( n − k ) ! cos ⁡ ( a x − ( − 1 ) k + 1 2 π 2 ) = ∑ k = 0 n x n − k a 1 + k n ! ( n − k ) ! sin ⁡ ( a x + k π 2 ) ( n > 0 ) {\displaystyle {\begin{aligned}\int x^{n}\cos ax\,dx&={\frac {x^{n}\sin ax}{a}}-{\frac {n}{a}}\int x^{n-1}\sin ax\,dx\\&=\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\\&=\sum _{k=0}^{n}(-1)^{\lfloor k/2\rfloor }{\frac {x^{n-k}}{a^{1+k}}}{\frac {n!}{(n-k)!}}\cos \left(ax-{\frac {(-1)^{k}+1}{2}}{\frac {\pi }{2}}\right)\\&=\sum _{k=0}^{n}{\frac {x^{n-k}}{a^{1+k}}}{\frac {n!}{(n-k)!}}\sin \left(ax+k{\frac {\pi }{2}}\right)\qquad {\mbox{(}}n>0{\mbox{)}}\end{aligned}}}
∫ 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}}\,dx=\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 ( n ≠ 1 ) {\displaystyle \int {\frac {\cos ax}{x^{n}}}\,dx=-{\frac {\cos ax}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\int {\frac {\sin ax}{x^{n-1}}}\,dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}}
∫ d x cos ⁡ a x = 1 a ln ⁡ | tan ⁡ ( a x 2 + π 4 ) | + C {\displaystyle \int {\frac {dx}{\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 ( n > 1 ) {\displaystyle \int {\frac {dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(}}n>1{\mbox{)}}}
∫ d x 1 + cos ⁡ a x = 1 a tan ⁡ a x 2 + C {\displaystyle \int {\frac {dx}{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 {dx}{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\,dx}{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\,dx}{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\,dx}{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\,dx}{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 ( | a 1 | ≠ | a 2 | ) {\displaystyle \int (\cos a_{1}x)(\cos a_{2}x)\,dx={\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{(}}|a_{1}|\neq |a_{2}|{\mbox{)}}}

Tích phân chỉ chứa hàm tan

∫ tan ⁡ a x d x = − 1 a ln ⁡ | cos ⁡ a x | + C = 1 a ln ⁡ | sec ⁡ a x | + C {\displaystyle \int \tan ax\,dx=-{\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}\,dx=\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 ( n ≠ 1 ) {\displaystyle \int \tan ^{n}ax\,dx={\frac {1}{a(n-1)}}\tan ^{n-1}ax-\int \tan ^{n-2}ax\,dx\qquad (n\neq 1)\,\!}
∫ 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 ( p 2 + q 2 ≠ 0 ) {\displaystyle \int {\frac {dx}{q\tan ax+p}}={\frac {1}{p^{2}+q^{2}}}(px+{\frac {q}{a}}\ln |q\sin ax+p\cos ax|)+C\qquad (p^{2}+q^{2}\neq 0)\,\!}
∫ d x tan ⁡ a x + 1 = x 2 + 1 2 a ln ⁡ | sin ⁡ a x + cos ⁡ a x | + C {\displaystyle \int {\frac {dx}{\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 {dx}{\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\,dx}{\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\,dx}{\tan ax-1}}={\frac {x}{2}}+{\frac {1}{2a}}\ln |\sin ax-\cos ax|+C\,\!}

Tích phân chỉ chứa hàm secant

∫ sec ⁡ a x d x = 1 a ln ⁡ | sec ⁡ a x + tan ⁡ a x | + C {\displaystyle \int \sec {ax}\,dx={\frac {1}{a}}\ln {\left|\sec {ax}+\tan {ax}\right|}+C}
∫ sec 2 ⁡ x d x = tan ⁡ x + C {\displaystyle \int \sec ^{2}{x}\,dx=\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 ( n ≠ 1 ) {\displaystyle \int \sec ^{n}{ax}\,dx={\frac {\sec ^{n-2}{ax}\tan {ax}}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{ax}\,dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}\,\!}
∫ 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}\,dx={\frac {\sec ^{n-2}{x}\tan {x}}{n-1}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{x}\,dx} [1]
∫ d x sec ⁡ x + 1 = x − tan ⁡ x 2 + C {\displaystyle \int {\frac {dx}{\sec {x}+1}}=x-\tan {\frac {x}{2}}+C}
∫ d x sec ⁡ x − 1 = − x − cot ⁡ x 2 + C {\displaystyle \int {\frac {dx}{\sec {x}-1}}=-x-\cot {\frac {x}{2}}+C}

Tích phân chỉ chứa hàm cosecant

∫ csc ⁡ a x d x = − 1 a ln ⁡ | csc ⁡ a x + cot ⁡ a x | + C = 1 a ln ⁡ | csc ⁡ a x − cot ⁡ a x | + C = 1 a ln ⁡ | tan ⁡ ( a x 2 ) | + C {\displaystyle {\begin{aligned}\int \csc {ax}\,dx&=-{\frac {1}{a}}\ln {\left|\csc {ax}+\cot {ax}\right|}+C\\&={\frac {1}{a}}\ln {\left|\csc {ax}-\cot {ax}\right|}+C\\&={\frac {1}{a}}\ln {\left|\tan {\left({\frac {ax}{2}}\right)}\right|}+C\end{aligned}}}
∫ csc 2 ⁡ x d x = − cot ⁡ x + C {\displaystyle \int \csc ^{2}{x}\,dx=-\cot {x}+C}
∫ csc 3 ⁡ x d x = − 1 2 csc ⁡ x cot ⁡ x − 1 2 ln ⁡ | csc ⁡ x + cot ⁡ x | + C = − 1 2 csc ⁡ x cot ⁡ x + 1 2 ln ⁡ | csc ⁡ x − cot ⁡ x | + C {\displaystyle {\begin{aligned}\int \csc ^{3}{x}\,dx&=-{\frac {1}{2}}\csc x\cot x-{\frac {1}{2}}\ln |\csc x+\cot x|+C\\&=-{\frac {1}{2}}\csc x\cot x+{\frac {1}{2}}\ln |\csc x-\cot x|+C\end{aligned}}}
∫ csc n ⁡ a x d x = − csc n − 2 ⁡ a x cot ⁡ a x a ( n − 1 ) + n − 2 n − 1 ∫ csc n − 2 ⁡ a x d x  ( n ≠ 1 ) {\displaystyle \int \csc ^{n}{ax}\,dx=-{\frac {\csc ^{n-2}{ax}\cot {ax}}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \csc ^{n-2}{ax}\,dx\qquad {\mbox{ (}}n\neq 1{\mbox{)}}}
∫ d x csc ⁡ x + 1 = x − 2 cot ⁡ x 2 + 1 + C {\displaystyle \int {\frac {dx}{\csc {x}+1}}=x-{\frac {2}{\cot {\frac {x}{2}}+1}}+C}
∫ d x csc ⁡ x − 1 = − x + 2 cot ⁡ x 2 − 1 + C {\displaystyle \int {\frac {dx}{\csc {x}-1}}=-x+{\frac {2}{\cot {\frac {x}{2}}-1}}+C}

Tích phân chỉ chứa hàm cotang

∫ cot ⁡ a x d x = 1 a ln ⁡ | sin ⁡ a x | + C {\displaystyle \int \cot ax\,dx={\frac {1}{a}}\ln |\sin ax|+C}
∫ cot 2 ⁡ x d x = − cot ⁡ x − x + C {\displaystyle \int \cot ^{2}{x}\,dx=-\cot {x}-x+C}
∫ cot n ⁡ a x d x = − 1 a ( n − 1 ) cot n − 1 ⁡ a x − ∫ cot n − 2 ⁡ a x d x ( n ≠ 1 ) {\displaystyle \int \cot ^{n}ax\,dx=-{\frac {1}{a(n-1)}}\cot ^{n-1}ax-\int \cot ^{n-2}ax\,dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}}
∫ d x 1 + cot ⁡ a x = ∫ 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 {dx}{1+\cot ax}}=\int {\frac {\tan ax\,dx}{\tan ax+1}}={\frac {x}{2}}-{\frac {1}{2a}}\ln |\sin ax+\cos ax|+C}
∫ d x 1 − cot ⁡ a x = ∫ 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 {dx}{1-\cot ax}}=\int {\frac {\tan ax\,dx}{\tan ax-1}}={\frac {x}{2}}+{\frac {1}{2a}}\ln |\sin ax-\cos ax|+C}

Tích phân chứa hàm sin và cos

Tích phân một hàm hữu tỉ (phân thức) của sin và cos có thể được tính bằng quy tắc Bioche.

∫ d x cos ⁡ a x ± sin ⁡ a x = 1 a 2 ln ⁡ | tan ⁡ ( a x 2 ± π 8 ) | + C {\displaystyle \int {\frac {dx}{\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 {dx}{(\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 {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 ⁡ 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\,dx}{\cos ax\pm \sin ax}}={\frac {x}{2}}\pm {\frac {1}{2a}}\ln \left|\sin ax\pm \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\,dx}{\cos ax\pm \sin ax}}=\pm {\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax\pm \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\,dx}{(\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\,dx}{(\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\,dx}{(\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\,dx}{(\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 sin 2 ⁡ a x + C {\displaystyle \int (\sin ax)(\cos ax)\,dx={\frac {1}{2a}}\sin ^{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 ( | a 1 | ≠ | a 2 | ) {\displaystyle \int (\sin a_{1}x)(\cos a_{2}x)\,dx=-{\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{(}}|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 ( n ≠ − 1 ) {\displaystyle \int (\sin ^{n}ax)(\cos ax)\,dx={\frac {1}{a(n+1)}}\sin ^{n+1}ax+C\qquad {\mbox{(}}n\neq -1{\mbox{)}}}
∫ ( sin ⁡ a x ) ( cos n ⁡ a x ) d x = − 1 a ( n + 1 ) cos n + 1 ⁡ a x + C ( n ≠ − 1 ) {\displaystyle \int (\sin ax)(\cos ^{n}ax)\,dx=-{\frac {1}{a(n+1)}}\cos ^{n+1}ax+C\qquad {\mbox{(}}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 ( m , n > 0 ) = ( 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 ( m , n > 0 ) {\displaystyle {\begin{aligned}\int (\sin ^{n}ax)(\cos ^{m}ax)\,dx&=-{\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)\,dx\qquad {\mbox{(}}m,n>0{\mbox{)}}\\&={\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)\,dx\qquad {\mbox{(}}m,n>0{\mbox{)}}\end{aligned}}}
∫ d x ( sin ⁡ a x ) ( cos ⁡ a x ) = 1 a ln ⁡ | tan ⁡ a x | + C {\displaystyle \int {\frac {dx}{(\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 ) ( n ≠ 1 ) {\displaystyle \int {\frac {dx}{(\sin ax)(\cos ^{n}ax)}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+\int {\frac {dx}{(\sin ax)(\cos ^{n-2}ax)}}\qquad {\mbox{(}}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 ) ( n ≠ 1 ) {\displaystyle \int {\frac {dx}{(\sin ^{n}ax)(\cos ax)}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+\int {\frac {dx}{(\sin ^{n-2}ax)(\cos ax)}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}}
∫ sin ⁡ a x d x cos n ⁡ a x = 1 a ( n − 1 ) cos n − 1 ⁡ a x + C ( n ≠ 1 ) {\displaystyle \int {\frac {\sin ax\,dx}{\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+C\qquad {\mbox{(}}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\,dx}{\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 ( n ≠ 1 ) {\displaystyle \int {\frac {\sin ^{2}ax\,dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}-{\frac {1}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(}}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 ( n ≠ 1 ) {\displaystyle \int {\frac {\sin ^{n}ax\,dx}{\cos ax}}=-{\frac {\sin ^{n-1}ax}{a(n-1)}}+\int {\frac {\sin ^{n-2}ax\,dx}{\cos ax}}\qquad {\mbox{(}}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 ( m ≠ 1 ) 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 ( m ≠ 1 ) − 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 ( m ≠ n ) {\displaystyle \int {\frac {\sin ^{n}ax\,dx}{\cos ^{m}ax}}={\begin{cases}{\dfrac {\sin ^{n+1}ax}{a(m-1)\cos ^{m-1}ax}}-{\dfrac {n-m+2}{m-1}}\displaystyle \int {\dfrac {\sin ^{n}ax\,dx}{\cos ^{m-2}ax}}&{\mbox{(}}m\neq 1{\mbox{)}}\\{\dfrac {\sin ^{n-1}ax}{a(m-1)\cos ^{m-1}ax}}-{\dfrac {n-1}{m-1}}\displaystyle \int {\dfrac {\sin ^{n-2}ax\,dx}{\cos ^{m-2}ax}}&{\mbox{(}}m\neq 1{\mbox{)}}\\-{\dfrac {\sin ^{n-1}ax}{a(n-m)\cos ^{m-1}ax}}+{\dfrac {n-1}{n-m}}\displaystyle \int {\dfrac {\sin ^{n-2}ax\,dx}{\cos ^{m}ax}}&{\mbox{(}}m\neq n{\mbox{)}}\end{cases}}}
∫ cos ⁡ a x d x sin n ⁡ a x = − 1 a ( n − 1 ) sin n − 1 ⁡ a x + C ( n ≠ 1 ) {\displaystyle \int {\frac {\cos ax\,dx}{\sin ^{n}ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+C\qquad {\mbox{(}}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\,dx}{\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 ) ( n ≠ 1 ) {\displaystyle \int {\frac {\cos ^{2}ax\,dx}{\sin ^{n}ax}}=-{\frac {1}{n-1}}\left({\frac {\cos ax}{a\sin ^{n-1}ax}}+\int {\frac {dx}{\sin ^{n-2}ax}}\right)\qquad {\mbox{(}}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 ( m ≠ 1 ) − 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 ( m ≠ 1 ) 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 ( m ≠ n ) {\displaystyle \int {\frac {\cos ^{n}ax\,dx}{\sin ^{m}ax}}={\begin{cases}-{\dfrac {\cos ^{n+1}ax}{a(m-1)\sin ^{m-1}ax}}-{\dfrac {n-m+2}{m-1}}\displaystyle \int {\dfrac {\cos ^{n}ax\,dx}{\sin ^{m-2}ax}}&{\mbox{(}}m\neq 1{\mbox{)}}\\-{\dfrac {\cos ^{n-1}ax}{a(m-1)\sin ^{m-1}ax}}-{\dfrac {n-1}{m-1}}\displaystyle \int {\dfrac {\cos ^{n-2}ax\,dx}{\sin ^{m-2}ax}}&{\mbox{(}}m\neq 1{\mbox{)}}\\{\dfrac {\cos ^{n-1}ax}{a(n-m)\sin ^{m-1}ax}}+{\dfrac {n-1}{n-m}}\displaystyle \int {\dfrac {\cos ^{n-2}ax\,dx}{\sin ^{m}ax}}&{\mbox{(}}m\neq n{\mbox{)}}\end{cases}}}

Tích phân chứa hàm sin và tang

∫ 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\,dx={\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 ( n ≠ 1 ) {\displaystyle \int {\frac {\tan ^{n}ax\,dx}{\sin ^{2}ax}}={\frac {1}{a(n-1)}}\tan ^{n-1}(ax)+C\qquad (n\neq 1)\,\!}

Tích phân chứa hàm cos và tang

∫ 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\,dx}{\cos ^{2}ax}}={\frac {1}{a(n+1)}}\tan ^{n+1}ax+C\qquad (n\neq -1)\,\!}

Tích phân chứa hàm sin và cotang

∫ cot n ⁡ a x d x sin 2 ⁡ a x = − 1 a ( n + 1 ) cot n + 1 ⁡ a x + C ( n ≠ − 1 ) {\displaystyle \int {\frac {\cot ^{n}ax\,dx}{\sin ^{2}ax}}=-{\frac {1}{a(n+1)}}\cot ^{n+1}ax+C\qquad (n\neq -1)\,\!}

Tích phân chứa hàm cos và cotang

∫ cot n ⁡ a x d x cos 2 ⁡ a x = 1 a ( 1 − n ) tan 1 − n ⁡ a x + C ( n ≠ 1 ) {\displaystyle \int {\frac {\cot ^{n}ax\,dx}{\cos ^{2}ax}}={\frac {1}{a(1-n)}}\tan ^{1-n}ax+C\qquad (n\neq 1)\,\!}

Tích phân chứa hàm secant và tang

∫ ( sec ⁡ x ) ( tan ⁡ x ) d x = sec ⁡ x + C {\displaystyle \int (\sec x)(\tan x)\,dx=\sec x+C}

Tích phân chứa hàm cosecant và cotang

∫ ( csc ⁡ x ) ( cot ⁡ x ) d x = − csc ⁡ x + C {\displaystyle \int (\csc x)(\cot x)\,dx=-\csc x+C}

Tích phân trên một phần tư đường tròn

∫ 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 , n = 2 , 4 , 6 , 8 , … n − 1 n ⋅ n − 3 n − 2 ⋯ 4 5 ⋅ 2 3 , n = 3 , 5 , 7 , 9 , … 1 , n = 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}},&n=2,4,6,8,\ldots \\{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdots {\frac {4}{5}}\cdot {\frac {2}{3}},&n=3,5,7,9,\ldots \\1,&n=1\end{cases}}}

Tích phân với giới hạn đối xứng

∫ − c c sin ⁡ x d x = 0 {\displaystyle \int _{-c}^{c}\sin {x}\,dx=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}\,dx=2\int _{0}^{c}\cos {x}\,dx=2\int _{-c}^{0}\cos {x}\,dx=2\sin {c}}
∫ − c c tan ⁡ x d x = 0 {\displaystyle \int _{-c}^{c}\tan {x}\,dx=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 {\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 } (n là số nguyên dương lẻ)
∫ − 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 ) {\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(-1)^{n})}{24n^{2}\pi ^{2}}}={\frac {a^{3}}{24}}(1-6{\frac {(-1)^{n}}{n^{2}\pi ^{2}}})\qquad } (n là số nguyên dương)

Tích phân trên toàn bộ đường tròn

∫ 0 2 π sin 2 m + 1 ⁡ x cos 2 n + 1 ⁡ x d x = 0 m , n ∈ Z {\displaystyle \int _{0}^{2\pi }\sin ^{2m+1}{x}\cos ^{2n+1}{x}\,dx=0\qquad m,n\in \mathbb {Z} }

Tham khảo

  1. ^ Stewart, James. Calculus: Early Transcendentals, 6th Edition. Thomson: 2008
  • Gradshteĭn, I. S. (2015). Table of Integrals, Series, and Products. Waltham, MA: Academic Press. ISBN 978-0-12-384933-5. OCLC 893676501.