Danh sách tích phân với hàm lượng giác

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 ${\displaystyle \cos(x)}$ là đạo hàm của hàm số ${\displaystyle \sin(x)}$, ta có

${\displaystyle \int a\cos nx\,dx={\frac {a}{n}}\sin nx+C}$

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

${\displaystyle \int \sin ax\,dx=-{\frac {1}{a}}\cos ax+C}$

${\displaystyle \int \sin ^{2}{ax}\,dx={\frac {x}{2}}-{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}-{\frac {1}{2a}}\sin ax\cos ax+C}$

${\displaystyle \int \sin ^{3}{ax}\,dx={\frac {\cos 3ax}{12a}}-{\frac {3\cos ax}{4a}}+C}$

${\displaystyle \int x\sin ^{2}{ax}\,dx={\frac {x^{2}}{4}}-{\frac {x}{4a}}\sin 2ax-{\frac {1}{8a^{2}}}\cos 2ax+C}$

${\displaystyle \int x^{2}\sin ^{2}{ax}\,dx={\frac {x^{3}}{6}}-\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax-{\frac {x}{4a^{2}}}\cos 2ax+C}$

${\displaystyle \int x\sin ax\,dx={\frac {\sin ax}{a^{2}}}-{\frac {x\cos ax}{a}}+C}$

${\displaystyle \int (\sin b_{1}x)(\sin b_{2}x)\,dx={\frac {\sin((b_{2}-b_{1})x)}{2(b_{2}-b_{1})}}-{\frac {\sin((b_{1}+b_{2})x)}{2(b_{1}+b_{2})}}+C\qquad {\mbox{(}}|b_{1}|\neq |b_{2}|{\mbox{)}}}$

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

${\displaystyle \int {\frac {dx}{\sin ax}}=-{\frac {1}{a}}\ln {\left|\csc {ax}+\cot {ax}\right|}+C}$

${\displaystyle \int {\frac {dx}{\sin ^{n}ax}}={\frac {\cos ax}{a(1-n)\sin ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\sin ^{n-2}ax}}\qquad {\mbox{(}}n>1{\mbox{)}}}$

${\displaystyle {\begin{aligned}\int x^{n}\sin ax\,dx&=-{\frac {x^{n}}{a}}\cos ax+{\frac {n}{a}}\int x^{n-1}\cos ax\,dx\\&=\sum _{k=0}^{2k\leq n}(-1)^{k+1}{\frac {x^{n-2k}}{a^{1+2k}}}{\frac {n!}{(n-2k)!}}\cos ax+\sum _{k=0}^{2k+1\leq n}(-1)^{k}{\frac {x^{n-1-2k}}{a^{2+2k}}}{\frac {n!}{(n-2k-1)!}}\sin ax\\&=-\sum _{k=0}^{n}{\frac {x^{n-k}}{a^{1+k}}}{\frac {n!}{(n-k)!}}\cos \left(ax+k{\frac {\pi }{2}}\right)\qquad {\mbox{(}}n>0{\mbox{)}}\end{aligned}}}$

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

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

${\displaystyle \int {\frac {dx}{1\pm \sin ax}}={\frac {1}{a}}\tan \left({\frac {ax}{2}}\mp {\frac {\pi }{4}}\right)+C}$

${\displaystyle \int {\frac {x\,dx}{1+\sin ax}}={\frac {x}{a}}\tan \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)+{\frac {2}{a^{2}}}\ln \left|\cos \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)\right|+C}$

${\displaystyle \int {\frac {x\,dx}{1-\sin ax}}={\frac {x}{a}}\cot \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)+{\frac {2}{a^{2}}}\ln \left|\sin \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)\right|+C}$

${\displaystyle \int {\frac {\sin ax\,dx}{1\pm \sin ax}}=\pm x+{\frac {1}{a}}\tan \left({\frac {\pi }{4}}\mp {\frac {ax}{2}}\right)+C}$

Tích phân chỉ chứa hàm cos

${\displaystyle \int \cos ax\,dx={\frac {1}{a}}\sin ax+C}$

${\displaystyle \int \cos ^{2}{ax}\,dx={\frac {x}{2}}+{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}+{\frac {1}{2a}}\sin ax\cos ax+C}$

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

${\displaystyle \int x\cos ax\,dx={\frac {\cos ax}{a^{2}}}+{\frac {x\sin ax}{a}}+C}$

${\displaystyle \int x^{2}\cos ^{2}{ax}\,dx={\frac {x^{3}}{6}}+\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax+{\frac {x}{4a^{2}}}\cos 2ax+C}$

${\displaystyle {\begin{aligned}\int x^{n}\cos ax\,dx&={\frac {x^{n}\sin ax}{a}}-{\frac {n}{a}}\int x^{n-1}\sin ax\,dx\\&=\sum _{k=0}^{2k+1\leq n}(-1)^{k}{\frac {x^{n-2k-1}}{a^{2+2k}}}{\frac {n!}{(n-2k-1)!}}\cos ax+\sum _{k=0}^{2k\leq n}(-1)^{k}{\frac {x^{n-2k}}{a^{1+2k}}}{\frac {n!}{(n-2k)!}}\sin ax\\&=\sum _{k=0}^{n}(-1)^{\lfloor k/2\rfloor }{\frac {x^{n-k}}{a^{1+k}}}{\frac {n!}{(n-k)!}}\cos \left(ax-{\frac {(-1)^{k}+1}{2}}{\frac {\pi }{2}}\right)\\&=\sum _{k=0}^{n}{\frac {x^{n-k}}{a^{1+k}}}{\frac {n!}{(n-k)!}}\sin \left(ax+k{\frac {\pi }{2}}\right)\qquad {\mbox{(}}n>0{\mbox{)}}\end{aligned}}}$

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

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

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

${\displaystyle \int {\frac {dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(}}n>1{\mbox{)}}}$

${\displaystyle \int {\frac {dx}{1+\cos ax}}={\frac {1}{a}}\tan {\frac {ax}{2}}+C}$

${\displaystyle \int {\frac {dx}{1-\cos ax}}=-{\frac {1}{a}}\cot {\frac {ax}{2}}+C}$

${\displaystyle \int {\frac {x\,dx}{1+\cos ax}}={\frac {x}{a}}\tan {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left|\cos {\frac {ax}{2}}\right|+C}$

${\displaystyle \int {\frac {x\,dx}{1-\cos ax}}=-{\frac {x}{a}}\cot {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left|\sin {\frac {ax}{2}}\right|+C}$

${\displaystyle \int {\frac {\cos ax\,dx}{1+\cos ax}}=x-{\frac {1}{a}}\tan {\frac {ax}{2}}+C}$

${\displaystyle \int {\frac {\cos ax\,dx}{1-\cos ax}}=-x-{\frac {1}{a}}\cot {\frac {ax}{2}}+C}$

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

Tích phân chỉ chứa hàm tan

${\displaystyle \int \tan ax\,dx=-{\frac {1}{a}}\ln |\cos ax|+C={\frac {1}{a}}\ln |\sec ax|+C\,\!}$

${\displaystyle \int \tan ^{2}{x}\,dx=\tan {x}-x+C}$

${\displaystyle \int \tan ^{n}ax\,dx={\frac {1}{a(n-1)}}\tan ^{n-1}ax-\int \tan ^{n-2}ax\,dx\qquad (n\neq 1)\,\!}$

${\displaystyle \int {\frac {dx}{q\tan ax+p}}={\frac {1}{p^{2}+q^{2}}}(px+{\frac {q}{a}}\ln |q\sin ax+p\cos ax|)+C\qquad (p^{2}+q^{2}\neq 0)\,\!}$

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

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

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

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

Tích phân chỉ chứa hàm secant

Xem thêm thông tin: Tích phân của hàm secant

${\displaystyle \int \sec {ax}\,dx={\frac {1}{a}}\ln {\left|\sec {ax}+\tan {ax}\right|}+C}$

${\displaystyle \int \sec ^{2}{x}\,dx=\tan {x}+C}$

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

${\displaystyle \int \sec ^{n}{x}\,dx={\frac {\sec ^{n-2}{x}\tan {x}}{n-1}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{x}\,dx}$^[1]

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

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

Tích phân chỉ chứa hàm cosecant

${\displaystyle {\begin{aligned}\int \csc {ax}\,dx&=-{\frac {1}{a}}\ln {\left|\csc {ax}+\cot {ax}\right|}+C\\&={\frac {1}{a}}\ln {\left|\csc {ax}-\cot {ax}\right|}+C\\&={\frac {1}{a}}\ln {\left|\tan {\left({\frac {ax}{2}}\right)}\right|}+C\end{aligned}}}$

${\displaystyle \int \csc ^{2}{x}\,dx=-\cot {x}+C}$

${\displaystyle {\begin{aligned}\int \csc ^{3}{x}\,dx&=-{\frac {1}{2}}\csc x\cot x-{\frac {1}{2}}\ln |\csc x+\cot x|+C\\&=-{\frac {1}{2}}\csc x\cot x+{\frac {1}{2}}\ln |\csc x-\cot x|+C\end{aligned}}}$

${\displaystyle \int \csc ^{n}{ax}\,dx=-{\frac {\csc ^{n-2}{ax}\cot {ax}}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \csc ^{n-2}{ax}\,dx\qquad {\mbox{ (}}n\neq 1{\mbox{)}}}$

${\displaystyle \int {\frac {dx}{\csc {x}+1}}=x-{\frac {2}{\cot {\frac {x}{2}}+1}}+C}$

${\displaystyle \int {\frac {dx}{\csc {x}-1}}=-x+{\frac {2}{\cot {\frac {x}{2}}-1}}+C}$

Tích phân chỉ chứa hàm cotang

${\displaystyle \int \cot ax\,dx={\frac {1}{a}}\ln |\sin ax|+C}$

${\displaystyle \int \cot ^{2}{x}\,dx=-\cot {x}-x+C}$

${\displaystyle \int \cot ^{n}ax\,dx=-{\frac {1}{a(n-1)}}\cot ^{n-1}ax-\int \cot ^{n-2}ax\,dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$

${\displaystyle \int {\frac {dx}{1+\cot ax}}=\int {\frac {\tan ax\,dx}{\tan ax+1}}={\frac {x}{2}}-{\frac {1}{2a}}\ln |\sin ax+\cos ax|+C}$

${\displaystyle \int {\frac {dx}{1-\cot ax}}=\int {\frac {\tan ax\,dx}{\tan ax-1}}={\frac {x}{2}}+{\frac {1}{2a}}\ln |\sin ax-\cos ax|+C}$

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.

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

${\displaystyle \int {\frac {dx}{(\cos ax\pm \sin ax)^{2}}}={\frac {1}{2a}}\tan \left(ax\mp {\frac {\pi }{4}}\right)+C}$

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

${\displaystyle \int {\frac {\cos ax\,dx}{\cos ax\pm \sin ax}}={\frac {x}{2}}\pm {\frac {1}{2a}}\ln \left|\sin ax\pm \cos ax\right|+C}$

${\displaystyle \int {\frac {\sin ax\,dx}{\cos ax\pm \sin ax}}=\pm {\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax\pm \cos ax\right|+C}$

${\displaystyle \int {\frac {\cos ax\,dx}{(\sin ax)(1+\cos ax)}}=-{\frac {1}{4a}}\tan ^{2}{\frac {ax}{2}}+{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C}$

${\displaystyle \int {\frac {\cos ax\,dx}{(\sin ax)(1-\cos ax)}}=-{\frac {1}{4a}}\cot ^{2}{\frac {ax}{2}}-{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C}$

${\displaystyle \int {\frac {\sin ax\,dx}{(\cos ax)(1+\sin ax)}}={\frac {1}{4a}}\cot ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)+{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C}$

${\displaystyle \int {\frac {\sin ax\,dx}{(\cos ax)(1-\sin ax)}}={\frac {1}{4a}}\tan ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)-{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C}$

${\displaystyle \int (\sin ax)(\cos ax)\,dx={\frac {1}{2a}}\sin ^{2}ax+C}$

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

${\displaystyle \int (\sin ^{n}ax)(\cos ax)\,dx={\frac {1}{a(n+1)}}\sin ^{n+1}ax+C\qquad {\mbox{(}}n\neq -1{\mbox{)}}}$

${\displaystyle \int (\sin ax)(\cos ^{n}ax)\,dx=-{\frac {1}{a(n+1)}}\cos ^{n+1}ax+C\qquad {\mbox{(}}n\neq -1{\mbox{)}}}$

${\displaystyle {\begin{aligned}\int (\sin ^{n}ax)(\cos ^{m}ax)\,dx&=-{\frac {(\sin ^{n-1}ax)(\cos ^{m+1}ax)}{a(n+m)}}+{\frac {n-1}{n+m}}\int (\sin ^{n-2}ax)(\cos ^{m}ax)\,dx\qquad {\mbox{(}}m,n>0{\mbox{)}}\\&={\frac {(\sin ^{n+1}ax)(\cos ^{m-1}ax)}{a(n+m)}}+{\frac {m-1}{n+m}}\int (\sin ^{n}ax)(\cos ^{m-2}ax)\,dx\qquad {\mbox{(}}m,n>0{\mbox{)}}\end{aligned}}}$

${\displaystyle \int {\frac {dx}{(\sin ax)(\cos ax)}}={\frac {1}{a}}\ln \left|\tan ax\right|+C}$

${\displaystyle \int {\frac {dx}{(\sin ax)(\cos ^{n}ax)}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+\int {\frac {dx}{(\sin ax)(\cos ^{n-2}ax)}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$

${\displaystyle \int {\frac {dx}{(\sin ^{n}ax)(\cos ax)}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+\int {\frac {dx}{(\sin ^{n-2}ax)(\cos ax)}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$

${\displaystyle \int {\frac {\sin ax\,dx}{\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+C\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$

${\displaystyle \int {\frac {\sin ^{2}ax\,dx}{\cos ax}}=-{\frac {1}{a}}\sin ax+{\frac {1}{a}}\ln \left|\tan \left({\frac {\pi }{4}}+{\frac {ax}{2}}\right)\right|+C}$

${\displaystyle \int {\frac {\sin ^{2}ax\,dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}-{\frac {1}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$

${\displaystyle \int {\frac {\sin ^{n}ax\,dx}{\cos ax}}=-{\frac {\sin ^{n-1}ax}{a(n-1)}}+\int {\frac {\sin ^{n-2}ax\,dx}{\cos ax}}\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$

${\displaystyle \int {\frac {\sin ^{n}ax\,dx}{\cos ^{m}ax}}={\begin{cases}{\dfrac {\sin ^{n+1}ax}{a(m-1)\cos ^{m-1}ax}}-{\dfrac {n-m+2}{m-1}}\displaystyle \int {\dfrac {\sin ^{n}ax\,dx}{\cos ^{m-2}ax}}&{\mbox{(}}m\neq 1{\mbox{)}}\\{\dfrac {\sin ^{n-1}ax}{a(m-1)\cos ^{m-1}ax}}-{\dfrac {n-1}{m-1}}\displaystyle \int {\dfrac {\sin ^{n-2}ax\,dx}{\cos ^{m-2}ax}}&{\mbox{(}}m\neq 1{\mbox{)}}\\-{\dfrac {\sin ^{n-1}ax}{a(n-m)\cos ^{m-1}ax}}+{\dfrac {n-1}{n-m}}\displaystyle \int {\dfrac {\sin ^{n-2}ax\,dx}{\cos ^{m}ax}}&{\mbox{(}}m\neq n{\mbox{)}}\end{cases}}}$

${\displaystyle \int {\frac {\cos ax\,dx}{\sin ^{n}ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+C\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$

${\displaystyle \int {\frac {\cos ^{2}ax\,dx}{\sin ax}}={\frac {1}{a}}\left(\cos ax+\ln \left|\tan {\frac {ax}{2}}\right|\right)+C}$

${\displaystyle \int {\frac {\cos ^{2}ax\,dx}{\sin ^{n}ax}}=-{\frac {1}{n-1}}\left({\frac {\cos ax}{a\sin ^{n-1}ax}}+\int {\frac {dx}{\sin ^{n-2}ax}}\right)\qquad {\mbox{(}}n\neq 1{\mbox{)}}}$

${\displaystyle \int {\frac {\cos ^{n}ax\,dx}{\sin ^{m}ax}}={\begin{cases}-{\dfrac {\cos ^{n+1}ax}{a(m-1)\sin ^{m-1}ax}}-{\dfrac {n-m+2}{m-1}}\displaystyle \int {\dfrac {\cos ^{n}ax\,dx}{\sin ^{m-2}ax}}&{\mbox{(}}m\neq 1{\mbox{)}}\\-{\dfrac {\cos ^{n-1}ax}{a(m-1)\sin ^{m-1}ax}}-{\dfrac {n-1}{m-1}}\displaystyle \int {\dfrac {\cos ^{n-2}ax\,dx}{\sin ^{m-2}ax}}&{\mbox{(}}m\neq 1{\mbox{)}}\\{\dfrac {\cos ^{n-1}ax}{a(n-m)\sin ^{m-1}ax}}+{\dfrac {n-1}{n-m}}\displaystyle \int {\dfrac {\cos ^{n-2}ax\,dx}{\sin ^{m}ax}}&{\mbox{(}}m\neq n{\mbox{)}}\end{cases}}}$

Tích phân chứa hàm sin và tang

${\displaystyle \int \sin ax\tan ax\,dx={\frac {1}{a}}(\ln |\sec ax+\tan ax|-\sin ax)+C\,\!}$

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

Tích phân chứa hàm cos và tang

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

Tích phân chứa hàm sin và cotang

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

Tích phân chứa hàm cos và cotang

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

Tích phân chứa hàm secant và tang

${\displaystyle \int (\sec x)(\tan x)\,dx=\sec x+C}$

Tích phân chứa hàm cosecant và cotang

${\displaystyle \int (\csc x)(\cot x)\,dx=-\csc x+C}$

Tích phân trên một phần tư đường tròn

${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{n}x\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{n}x\,dx={\begin{cases}{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdots {\frac {3}{4}}\cdot {\frac {1}{2}}\cdot {\frac {\pi }{2}},&n=2,4,6,8,\ldots \\{\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdots {\frac {4}{5}}\cdot {\frac {2}{3}},&n=3,5,7,9,\ldots \\1,&n=1\end{cases}}}$

Tích phân với giới hạn đối xứng

${\displaystyle \int _{-c}^{c}\sin {x}\,dx=0}$

${\displaystyle \int _{-c}^{c}\cos {x}\,dx=2\int _{0}^{c}\cos {x}\,dx=2\int _{-c}^{0}\cos {x}\,dx=2\sin {c}}$

${\displaystyle \int _{-c}^{c}\tan {x}\,dx=0}$

${\displaystyle \int _{-{\frac {a}{2}}}^{\frac {a}{2}}x^{2}\cos ^{2}{\frac {n\pi x}{a}}\,dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad }$ (n là số nguyên dương lẻ)

${\displaystyle \int _{\frac {-a}{2}}^{\frac {a}{2}}x^{2}\sin ^{2}{\frac {n\pi x}{a}}\,dx={\frac {a^{3}(n^{2}\pi ^{2}-6(-1)^{n})}{24n^{2}\pi ^{2}}}={\frac {a^{3}}{24}}(1-6{\frac {(-1)^{n}}{n^{2}\pi ^{2}}})\qquad }$ (n là số nguyên dương)

Tích phân trên toàn bộ đường tròn

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