Khác biệt giữa các bản “Danh sách tích phân với hàm lượng giác”

Đã lùi lại sửa đổi 22467828 của 42.112.234.251 (Thảo luận)
(Đã lùi lại sửa đổi 22467828 của 42.112.234.251 (Thảo luận))
: <math>\int x\sin^2 {ax}\;dx = \frac{x^2}{4} - \frac{x}{4a} \sin 2ax - \frac{1}{8a^2} \cos 2ax +C\!</math>
: <math>\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\!</math>
: <math>\int\sin b_1x\sin b_2x\;dx = \frac{\sin((b_1-b_2)x)}{2(b_1-b_2)}-\frac{\sin((b_1+b_2)x)}{2(b_1+b_2)}+C \qquad\mbox{(for }|b_1|\neq|b_2|\mbox{)}\,\!</math>
: <math>\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{(for }n>2\mbox{)}\,\!</math>
: <math>\int\frac{dx}{\sin ax} = \frac{1}{a}\ln \left|\tan\frac{ax}{2}\right|+C</math>
: <math>\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{(for }n>1\mbox{)}\,\!</math>
: <math>\int x\sin ax\;dx = \frac{\sin ax}{a^2}-\frac{x\cos ax}{a}+C\,\!</math>
: <math>\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 \qquad\mbox{(for }n>0\mbox{)}\,\!</math>
: <math>\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)}{24n^2\pi^2} \qquad\mbox{(for }n=2,4,6...\mbox{)}\,\!</math>
: <math>\int\frac{\sin ax}{x} dx = \sum_{n=0}^\infty (-1)^n\frac{(ax)^{2n+1}}{(2n+1)\cdot (2n+1)!} +C\,\!</math>
: <math>\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\,\!</math>
: <math>\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</math>
 
== Tích phân chỉ chứa hàm [[hàmHàm lượng giác|cos]] ==
 
: <math>\int\cos ax\;dx = \frac{1}{a}\sin ax+C\,\!</math>
 
: <math>\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\!</math>
 
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: <math>\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{(for }n>0\mbox{)}\,\!</math>
 
: <math>\int x\cos ax\;dx = \frac{\cos ax}{a^2} + \frac{x\sin ax}{a}+C\,\!</math>
: <math>\int\frac{\cos ax}{x} dx = \ln|ax|+\sum_{k=1}^\infty (-1)^k\frac{(ax)^{2k}}{2k\cdot(2k)!}+C\,\!</math>
 
: <math>\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{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\int\frac{dx}{\cos ax} = \frac{1}{a}\ln\left|\tan\left(\frac{ax}{2}+\frac{\pi}{4}\right)\right|+C</math>
 
: <math>\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{(for }n>1\mbox{)}\,\!</math>
 
: <math>\int\frac{dx}{1+\cos ax} = \frac{1}{a}\tan\frac{ax}{2}+C\,\!</math>
: <math>\int\frac{\cos ax\;dx}{1-\cos ax} = -x-\frac{1}{a}\cot\frac{ax}{2}+C\,\!</math>
 
: <math>\int\cos a_1x\cos a_2x\;dx = \frac{\sin(a_1-a_2)x}{2(a_1-a_2)}+\frac{\sin(a_1+a_2)x}{2(a_1+a_2)}+C \qquad\mbox{(for }|a_1|\neq|a_2|\mbox{)}\,\!</math>
 
== Tích phân chỉ chứa hàm [[hàm lượng giác|tan]] ==
: <math>\int\tan ax\;dx = -\frac{1}{a}\ln|\cos ax|+C = \frac{1}{a}\ln|\sec ax|+C\,\!</math>
 
: <math>\int\tan^n ax\;dx = \frac{1}{a(n-1)}\tan^{n-1} ax-\int\tan^{n-2} ax\;dx \qquad\mbox{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\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{(for }p^2 + q^2\neq 0\mbox{)}\,\!</math>
 
: <math>\int\frac{dx}{\tan ax + 1} = \frac{x}{2} + \frac{1}{2a}\ln|\sin ax + \cos ax|+C\,\!</math>
:<math>\int \sec^2{x} \, dx = \tan{x}+C</math>
 
:<math>\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{ (for }n \ne 1\mbox{)}\,\!</math>
 
:<math>\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</math><ref>Stewart, James. Calculus: Early Transcendentals, 6th Edition. Thomson: 2008</ref>
 
:<math>\int \frac{dx}{\sec{x} - 1} = - x - \cot{\frac{x}{2}}+C</math>
<!--
 
<!-- In the 17th century, the integral of the secant function was the subject of a well-known conjecture posed in the 1640s by Henry Bond. The problem was solved by [[Isaac Barrow]].<ref>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</ref> It was originally for the purposes of [[cartography]] that this was needed. -->
 
== Tích phân chỉ chứa hàm [[hàm lượng giác|cosecant]] ==
:<math>\int \csc^2{x} \, dx = -\cot{x}+C</math>
 
:<math>\int \csc^n{ax} \, dx = -\frac{\csc^{n-1}{ax} \csc{ax}}{a(n-1)} \,+\, \frac{n-2}{n-1}\int \csc^{n-2}{ax} \, dx \qquad \mbox{ (for }n \ne 1\mbox{)}\,</math>
 
:<math>\int \frac{dx}{\csc{x} + 1} = x - \frac{2\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}+\sin{\frac{x}{2}}}+C</math>
 
:<math>\int\cot ax\;dx = \frac{1}{a}\ln|\sin ax|+C\,\!</math>
: <math>\int\cot^n ax\;dx = -\frac{1}{a(n-1)}\cot^{n-1} ax - \int\cot^{n-2} ax\;dx \qquad\mbox{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\int\cot^n ax\;dx = -\frac{1}{a(n-1)}\cot^{n-1} ax - \int\cot^{n-2} ax\;dx \qquad\mbox{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\int\frac{dx}{1 + \cot ax} = \int\frac{\tan ax\;dx}{\tan ax+1}\,\!</math>
 
: <math>\int\frac{dx}{1 - \cot ax} = \int\frac{\tan ax\;dx}{\tan ax-1}\,\!</math>
 
: <math>\int\sin ax\cos ax\;dx = -\frac{1}{2a}\cos^2 ax +C\,\!</math>
 
: <math>\int\sin a_1x\cos a_2x\;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{(for }|a_1|\neq|a_2|\mbox{)}\,\!</math>
 
: <math>\int\sin^n ax\cos ax\;dx = \frac{1}{a(n+1)}\sin^{n+1} ax +C\qquad\mbox{(for }n\neq -1\mbox{)}\,\!</math>
 
: <math>\int\sin ax\cos^n ax\;dx = -\frac{1}{a(n+1)}\cos^{n+1} ax +C\qquad\mbox{(for }n\neq -1\mbox{)}\,\!</math>
 
: <math>\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{(for }m,n>0\mbox{)}\,\!</math>
: và: <math>\int\sin^n ax\cos^m ax\;dx = \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{(for }m,n>0\mbox{)}\,\!</math>
 
: và: <math>\int\sin^n ax\cos^m ax\;dx = \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{(for }m,n>0\mbox{)}\,\!</math>
 
: <math>\int\frac{dx}{\sin ax\cos ax} = \frac{1}{a}\ln\left|\tan ax\right|+C</math>
 
: <math>\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{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\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{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\int\frac{\sin ax\;dx}{\cos^n ax} = \frac{1}{a(n-1)\cos^{n-1} ax} +C\qquad\mbox{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\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</math>
 
: <math>\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{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\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{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\int\frac{\sin^n ax\;dx}{\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\;dx}{\cos^{m-2} ax} \qquad\mbox{(for }m\neq 1\mbox{)}\,\!</math>
: và: <math>\int\frac{\sin^n ax\;dx}{\cos^m ax} = -\frac{\sin^{n-1} ax}{a(n-m-1)\cos^{m-1} ax}-+\frac{n-1}{mn-1m}\int\frac{\sin^{n-2} ax\;dx}{\cos^{m-2} ax} \qquad\mbox{(for }m\neq 1\mbox{n)}\,\!</math>
: và: <math>\int\frac{\cossin^n ax\;dx}{\sincos^m ax} = -\frac{\cossin^{n-1} ax}{a(m-1)\sincos^{m-1} ax} - \frac{n-1}{m-1}\int\frac{\cossin^{n-2} ax\;dx}{\sincos^{m-2} ax} \qquad\mbox{(for }m\neq 1\mbox{)}\,\!</math>
 
: và: <math>\int\frac{\sin^ncos ax\;dx}{\cossin^mn ax} = -\frac{\sin^{n-1} ax}{a(n-m1)\cossin^{m-1} ax}+\frac{n-1}{n-m}\int\frac{\sin^{n-2} ax\;dx}{\cos^m ax} +C\qquad\mbox{(for }mn\neq n\mbox{1)}\,\!</math>
 
: và: <math>\int\frac{\sin^n ax\;dx}{\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\;dx}{\cos^{m-2} ax} \qquad\mbox{(for }m\neq 1\mbox{)}\,\!</math>
 
: <math>\int\frac{\cos ax\;dx}{\sin^n ax} = -\frac{1}{a(n-1)\sin^{n-1} ax} +C\qquad\mbox{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\int\frac{\cos^2 ax\;dx}{\sin ax} = \frac{1}{a}\left(\cos ax+\ln\left|\tan\frac{ax}{2}\right|\right) +C</math>
 
: <math>\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{(for }n\neq 1\mbox{)}</math>
 
: <math>\int\frac{\cos^n ax\;dx}{\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\;dx}{\sin^{m-2} ax} \qquad\mbox{(for }m\neq 1\mbox{)}\,\!</math>
: và: <math>\int\frac{\cos^n ax\;dx}{\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\;dx}{\sin^m ax} \qquad(m\neq n)\,\!</math>
 
: và: <math>\int\frac{\cos^n ax\;dx}{\sin^m ax} = -\frac{\cos^{n-1} ax}{a(n-m-1)\sin^{m-1} ax} +- \frac{n-1}{nm-m1}\int\frac{\cos^{n-2} ax\;dx}{\sin^{m-2} ax} \qquad\mbox{(for }m\neq n\mbox{1)}\,\!</math>
 
: và: <math>\int\frac{\cos^n ax\;dx}{\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\;dx}{\sin^{m-2} ax} \qquad\mbox{(for }m\neq 1\mbox{)}\,\!</math>
 
== Tích phân chứa hàm [[sin]] và [[hàm lượng giác|tang]] ==
 
: <math>\int \sin ax \tan ax\;dx = \frac{1}{a}(\ln|\sec ax + \tan ax| - \sin ax)+C\,\!</math>
: <math>\int\frac{\costan^n ax\;dx}{\sin^n2 ax} = -\frac{1}{a(n-1)}\sintan^{n-1} (ax}) +C\qquad\mbox{(for }n\neq 1\mbox{)}\,\!</math>
 
: <math>\int\frac{\tan^n ax\;dx}{\sin^2 ax} = \frac{1}{a(n-1)}\tan^{n-1} (ax) +C\qquad\mbox{(for }n\neq 1\mbox{)}\,\!</math>
 
== Tích phân chứa hàm [[hàm lượng giác|cos]] và [[hàm lượng giác|tang]] ==
 
: <math>\int\frac{\tan^n ax\;dx}{\cos^2 ax} = \frac{1}{a(n+1)}\tan^{n+1} ax +C\qquad\mbox{(for }n\neq -1\mbox{)}\,\!</math>
 
== Tích phân chứa hàm [[sin]] và [[hàm lượng giác|cotang]] ==
 
: <math>\int\frac{\cot^n ax\;dx}{\sin^2 ax} = -\frac{1}{a(n+1)}\cot^{n+1} ax +C\qquad\mbox{(for }n\neq -1\mbox{)}\,\!</math>
 
== Tích phân chứa hàm [[hàm lượng giác|cos]] và [[hàm lượng giác|cotang]] ==
 
: <math>\int\frac{\cot^n ax\;dx}{\cos^2 ax} = \frac{1}{a(1-n)}\tan^{1-n} ax +C\qquad\mbox{(for }n\neq 1\mbox{)}\,\!</math>
 
== Tích phân với giới hạn đối xứng ==
: <math>\int_{{-c}}^{{c}}\cos {x}\;dx = 2\int_{{0}}^{{c}}\cos {x}\;dx = 2\int_{{-c}}^{{0}}\cos {x}\;dx = 2\sin {c} \!</math>
: <math>\int_{{-c}}^{{c}}\tan {x}\;dx = 0 \!</math>
: <math>\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\mbox{(for }n=1,3,5...\mbox{)}\,\!</math>
 
==Tham khảo==