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

Bước tới điều hướng Bước tới tìm kiếm
clean up, replaced: → (2), → (7) using AWB
(Đã lùi lại sửa đổi 22467828 của (Thảo luận))
n (clean up, replaced: → (2), → (7) using AWB)
: <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(n>1)\,\!</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(n>0)\,\!</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(n=2,4,6...)\,\!</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 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\!</math>
: <math>\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 \!</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 \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 (n\ne 1)\,\!</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 - \tan{\frac{x}{2}}+C</math>
:<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\sin ax\cos^n ax\;dx = -\frac{1}{a(n+1)}\cos^{n+1} ax +C\qquad(n\neq -1)\,\!</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(m,n>0)\,\!</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(m,n>0)\,\!</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(n\neq -1)\,\!</math>
== Tích phân chứa hàm [[hàm lượng giác|cos]] và [[hàm lượng giác|cotang]] ==
: <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(n=1,3,5...)\,\!</math>
==Tham khảo==

Trình đơn chuyển hướng