รายการปริพันธ์ของฟังก์ชันตรีโกณมิติ

ต่อไปนี้เป็นรายการปริพันธ์ (ฟังก์ชันปฏิยานุพันธ์) ของฟังก์ชันตรีโกณมิติ โดยทั่วไป ถ้าฟังก์ชัน ${\displaystyle \sin(x)}$ เป็นฟังก์ชันตรีโกณมิติใด ๆ และ ${\displaystyle \cos(x)}$ เป็นอนุพันธ์ของมัน แล้ว

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

ในทุกสูตร กำหนดให้ ค่าคงตัว a เป็นจำนวนที่ไม่ใช่ศูนย์ และ C เป็นค่าคงตัวของการหาปริพันธ์

ปริพัทธ์ที่เกี่ยวข้องเฉพาะไซน์

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ปริพัทธ์ที่เกี่ยวข้องเฉพาะโคไซน์

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ปริพัทธ์ที่เกี่ยวข้องเฉพาะแทนเจนต์

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

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

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

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

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

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

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

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

ปริพัทธ์ที่เกี่ยวข้องเฉพาะเซแคนต์

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

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

${\displaystyle \int \sec ^{3}{x}\,dx={\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln |\sec x+\tan x|+C.}$

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

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

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

ปริพัทธ์ที่เกี่ยวข้องเฉพาะโคเซแคนต์

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

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

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

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

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

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

ปริพัทธ์ที่เกี่ยวข้องเฉพาะโคแทนเจนต์

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

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

${\displaystyle \int {\frac {dx}{1+\cot ax}}=\int {\frac {\tan ax\,dx}{\tan ax+1}}}$

${\displaystyle \int {\frac {dx}{1-\cot ax}}=\int {\frac {\tan ax\,dx}{\tan ax-1}}}$

ปริพัทธ์ที่เกี่ยวข้องทั้งไซน์และโคไซน์

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ปริพัทธ์ที่เกี่ยวข้องทั้งไซน์และแทนเจนต์

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

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

ปริพัทธ์ที่เกี่ยวข้องทั้งโคไซน์และแทนเจนต์

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

ปริพัทธ์ที่เกี่ยวข้องทั้งไซน์และโคแทนเจนต์

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

ปริพัทธ์ที่เกี่ยวข้องทั้งโคไซน์และโคแทนเจนต์

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

ปริพัทธ์ที่เกี่ยวข้องทั้งเซแคนต์และแทนเจนต์

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

ปริพัทธ์ที่เกี่ยวข้องทั้งโคเซแคนต์และโคแทนเจนต์

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