Tích phân hàm số toán lủy thừa e

$\int e^{cx}dx=\frac{1}{c}{e}^{cx}$

$\int a^{cx}dx=\frac{1}{c\ln a}{a}^{cx}\text{( }a>0,a\ne 1\text{)}$

$\int x{e}^{cx}dx=\frac{e^{cx}}{c^2}(cx-1)$

$\int x^2{e}^{cx}dx=e^{cx}(\frac{x^2}{c}-\frac{2x}{c^2}+\frac{2}{c^3})$

$\int x^n{e}^{cx}dx=\frac{1}{c}{x}^n{e}^{cx}-\frac{n}{c}\int x^{n-1}{e}^{cx}dx$

$\int \frac{e^{cx}dx}{x}=\ln |x|+\sum _{i=1}^{\mathrm{\infty }}\frac{(cx{)}^i}{i\cdot i!}$

$\int \frac{e^{cx}dx}{x^n}=\frac{1}{n-1}(-\frac{e^{cx}}{x^{n-1}}+c\int \frac{e^{cx}}{x^{n-1}}dx)\text{( }n\ne 1\text{)}$

$\int e^{cx}\ln xdx=\frac{1}{c}{e}^{cx}\ln |x|-\mathrm{Ei}(cx)$

$\int e^{cx}\sin bxdx=\frac{e^{cx}}{c^2+b^2}(c\sin bx-b\cos bx)$

$\int e^{cx}\cos bxdx=\frac{e^{cx}}{c^2+b^2}(c\cos bx+b\sin bx)$

$\int e^{cx}{\sin }^{n}xdx=\frac{e^{cx}{\sin }^{n-1}x}{c^2+n^2}(c\sin x-n\cos x)+\frac{n(n-1)}{c^2+n^2}\int e^{cx}{\sin }^{n-2}xdx$

$\int e^{cx}{\cos }^{n}xdx=\frac{e^{cx}{\cos }^{n-1}x}{c^2+n^2}(c\cos x+n\sin x)+\frac{n(n-1)}{c^2+n^2}\int e^{cx}{\cos }^{n-2}xdx$

$\int x{e}^{c{x}^2}dx=\frac{1}{2c}{e}^{c{x}^2}$

$\int \frac{1}{\sigma \sqrt{2\pi }}{e}^{-(x-\mu {)}^2/2{\sigma }^2}dx=\frac{1}{2\sigma }(1+\text{erf}\frac{x-\mu }{\sigma \sqrt{2}})$

$\int e^{x^2}dx=e^{x^2}\left(\sum _{j=0}^{n-1}{c}_{2j}\frac{1}{x^{2j+1}}\right)+(2n-1)c_{2n-2}\int \frac{e^{x^2}}{x^{2n}}dx(n>0),$

với $c_{2j}=\frac{1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1}}=\frac{2j!}{j!2^{2j+1}}.$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}dx=\sqrt{\frac{\pi }{a}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{2n}{e}^{-x^2/a^2}dx=\sqrt{\pi }\frac{(2n)!}{n!}{\left(\frac{a}{2}\right)}^{2n+1}$

$\int \sinh cxdx=\frac{1}{c}\cosh cx$

$\int \cosh cxdx=\frac{1}{c}\sinh cx$

$\int {\sinh }^{2}cxdx=\frac{1}{4c}\sinh 2cx-\frac{x}{2}$

$\int {\cosh }^{2}cxdx=\frac{1}{4c}\sinh 2cx+\frac{x}{2}$

$\int {\sinh }^{n}cxdx=\frac{1}{cn}{\sinh }^{n-1}cx\cosh cx-\frac{n-1}{n}\int {\sinh }^{n-2}cxdx\text{( }n>0\text{)}$

hay: $\int {\sinh }^{n}cxdx=\frac{1}{c(n+1)}{\sinh }^{n+1}cx\cosh cx-\frac{n+2}{n+1}\int {\sinh }^{n+2}cxdx\text{( }n<0\text{, }n\ne -1\text{)}$

$\int {\cosh }^{n}cxdx=\frac{1}{cn}\sinh cx{\cosh }^{n-1}cx+\frac{n-1}{n}\int {\cosh }^{n-2}cxdx\text{( }n>0\text{)}$

hay: $\int {\cosh }^{n}cxdx=-\frac{1}{c(n+1)}\sinh cx{\cosh }^{n+1}cx-\frac{n+2}{n+1}\int {\cosh }^{n+2}cxdx\text{( }n<0\text{, }n\ne -1\text{)}$

$\int \frac{dx}{\sinh cx}=\frac{1}{c}\ln |\tanh \frac{cx}{2}|$

hay: $\int \frac{dx}{\sinh cx}=\frac{1}{c}\ln \left|\frac{\cosh cx-1}{\sinh cx}\right|$

hay: $\int \frac{dx}{\sinh cx}=\frac{1}{c}\ln \left|\frac{\sinh cx}{\cosh cx+1}\right|$

hay: $\int \frac{dx}{\sinh cx}=\frac{1}{c}\ln \left|\frac{\cosh cx-1}{\cosh cx+1}\right|$

$\int \frac{dx}{\cosh cx}=\frac{2}{c}\arctan e^{cx}$

$\int \frac{dx}{{\sinh }^{n}cx}=\frac{\cosh cx}{c(n-1){\sinh }^{n-1}cx}-\frac{n-2}{n-1}\int \frac{dx}{{\sinh }^{n-2}cx}\text{( }n\ne 1\text{)}$

$\int \frac{dx}{{\cosh }^{n}cx}=\frac{\sinh cx}{c(n-1){\cosh }^{n-1}cx}+\frac{n-2}{n-1}\int \frac{dx}{{\cosh }^{n-2}cx}\text{( }n\ne 1\text{)}$

$\int \frac{{\cosh }^{n}cx}{{\sinh }^{m}cx}dx=\frac{{\cosh }^{n-1}cx}{c(n-m){\sinh }^{m-1}cx}+\frac{n-1}{n-m}\int \frac{{\cosh }^{n-2}cx}{{\sinh }^{m}cx}dx\text{( }m\ne n\text{)}$

hay: $\int \frac{{\cosh }^{n}cx}{{\sinh }^{m}cx}dx=-\frac{{\cosh }^{n+1}cx}{c(m-1){\sinh }^{m-1}cx}+\frac{n-m+2}{m-1}\int \frac{{\cosh }^{n}cx}{{\sinh }^{m-2}cx}dx\text{( }m\ne 1\text{)}$

hay: $\int \frac{{\cosh }^{n}cx}{{\sinh }^{m}cx}dx=-\frac{{\cosh }^{n-1}cx}{c(m-1){\sinh }^{m-1}cx}+\frac{n-1}{m-1}\int \frac{{\cosh }^{n-2}cx}{{\sinh }^{m-2}cx}dx\text{( }m\ne 1\text{)}$

$\int \frac{{\sinh }^{m}cx}{{\cosh }^{n}cx}dx=\frac{{\sinh }^{m-1}cx}{c(m-n){\cosh }^{n-1}cx}+\frac{m-1}{m-n}\int \frac{{\sinh }^{m-2}cx}{{\cosh }^{n}cx}dx\text{( }m\ne n\text{)}$

hay: $\int \frac{{\sinh }^{m}cx}{{\cosh }^{n}cx}dx=\frac{{\sinh }^{m+1}cx}{c(n-1){\cosh }^{n-1}cx}+\frac{m-n+2}{n-1}\int \frac{{\sinh }^{m}cx}{{\cosh }^{n-2}cx}dx\text{( }n\ne 1\text{)}$

hay: $\int \frac{{\sinh }^{m}cx}{{\cosh }^{n}cx}dx=-\frac{{\sinh }^{m-1}cx}{c(n-1){\cosh }^{n-1}cx}+\frac{m-1}{n-1}\int \frac{{\sinh }^{m-2}cx}{{\cosh }^{n-2}cx}dx\text{( }n\ne 1\text{)}$

$\int x\sinh cxdx=\frac{1}{c}x\cosh cx-\frac{1}{c^2}\sinh cx$

$\int x\cosh cxdx=\frac{1}{c}x\sinh cx-\frac{1}{c^2}\cosh cx$

$\int \tanh cxdx=\frac{1}{c}\ln |\cosh cx|$

$\int \coth cxdx=\frac{1}{c}\ln |\sinh cx|$

$\int {\tanh }^{n}cxdx=-\frac{1}{c(n-1)}{\tanh }^{n-1}cx+\int {\tanh }^{n-2}cxdx\text{( }n\ne 1\text{)}$

$\int {\coth }^{n}cxdx=-\frac{1}{c(n-1)}{\coth }^{n-1}cx+\int {\coth }^{n-2}cxdx\text{( }n\ne 1\text{)}$

$\int \sinh bx\sinh cxdx=\frac{1}{b^2-c^2}(b\sinh cx\cosh bx-c\cosh cx\sinh bx)\text{( }{b}^2\ne c^2\text{)}$

$\int \cosh bx\cosh cxdx=\frac{1}{b^2-c^2}(b\sinh bx\cosh cx-c\sinh cx\cosh bx)\text{( }{b}^2\ne c^2\text{)}$

$\int \cosh bx\sinh cxdx=\frac{1}{b^2-c^2}(b\sinh bx\sinh cx-c\cosh bx\cosh cx)\text{( }{b}^2\ne c^2\text{)}$

$\int \sinh (ax+b)\sin (cx+d)dx=\frac{a}{a^2+c^2}\cosh (ax+b)\sin (cx+d)-\frac{c}{a^2+c^2}\sinh (ax+b)\cos (cx+d)$

$\int \sinh (ax+b)\cos (cx+d)dx=\frac{a}{a^2+c^2}\cosh (ax+b)\cos (cx+d)+\frac{c}{a^2+c^2}\sinh (ax+b)\sin (cx+d)$

$\int \cosh (ax+b)\sin (cx+d)dx=\frac{a}{a^2+c^2}\sinh (ax+b)\sin (cx+d)-\frac{c}{a^2+c^2}\cosh (ax+b)\cos (cx+d)$

$\int \cosh (ax+b)\cos (cx+d)dx=\frac{a}{a^2+c^2}\sinh (ax+b)\cos (cx+d)+\frac{c}{a^2+c^2}\cosh (ax+b)\sin (cx+d)$

Assume $(x^2>a^2)$, for $(x^2<a^2)$, see next section:

$\int xsdx=\frac{1}{3}{s}^3$

$\int \frac{sdx}{x}=s-a\arccos \left|\frac{a}{x}\right|$

$\int \frac{dx}{s}=\int \frac{dx}{\sqrt{x^2-a^2}}=\ln \left|\frac{x+s}{a}\right|$

Note that $\ln \left|\frac{x+s}{a}\right|=\mathrm{s}\mathrm{g}\mathrm{n}(x)\mathrm{arcosh}\left|\frac{x}{a}\right|=\frac{1}{2}\ln \left(\frac{x+s}{x-s}\right)$, where the positive value of $\mathrm{arcosh}\left|\frac{x}{a}\right|$ is to be taken.

$\int \frac{xdx}{s}=s$

$\int \frac{xdx}{s^3}=-\frac{1}{s}$

$\int \frac{xdx}{s^5}=-\frac{1}{3s^3}$

$\int \frac{xdx}{s^7}=-\frac{1}{5s^5}$

$\int \frac{xdx}{s^{2n+1}}=-\frac{1}{(2n-1)s^{2n-1}}$

$\int \frac{x^{2m}dx}{s^{2n+1}}=-\frac{1}{2n-1}\frac{x^{2m-1}}{s^{2n-1}}+\frac{2m-1}{2n-1}\int \frac{x^{2m-2}dx}{s^{2n-1}}$

$\int \frac{x^{2}dx}{s}=\frac{xs}{2}+\frac{a^2}{2}\ln \left|\frac{x+s}{a}\right|$

$\int \frac{x^{2}dx}{s^3}=-\frac{x}{s}+\ln \left|\frac{x+s}{a}\right|$

$\int \frac{x^{4}dx}{s}=\frac{x^{3}s}{4}+\frac{3}{8}{a}^{2}xs+\frac{3}{8}{a}^4\ln \left|\frac{x+s}{a}\right|$

$\int \frac{x^{4}dx}{s^3}=\frac{xs}{2}-\frac{a^{2}x}{s}+\frac{3}{2}{a}^2\ln \left|\frac{x+s}{a}\right|$

$\int \frac{x^{4}dx}{s^5}=-\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}+\ln \left|\frac{x+s}{a}\right|$

$\int \frac{x^{2m}dx}{s^{2n+1}}=(-1{)}^{n-m}\frac{1}{a^{2(n-m)}}\sum _{i=0}^{n-m-1}\frac{1}{2(m+i)+1}(\genfrac{}{}{0ex}{}{n-m-1}{i})\frac{x^{2(m+i)+1}}{s^{2(m+i)+1}}\text{(}n>m\ge 0\text{)}$

$\int \frac{dx}{s^3}=-\frac{1}{a^2}\frac{x}{s}$

$\int \frac{dx}{s^5}=\frac{1}{a^4}[\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}]$

$\int \frac{dx}{s^7}=-\frac{1}{a^6}[\frac{x}{s}-\frac{2}{3}\frac{x^3}{s^3}+\frac{1}{5}\frac{x^5}{s^5}]$

$\int \frac{dx}{s^9}=\frac{1}{a^8}[\frac{x}{s}-\frac{3}{3}\frac{x^3}{s^3}+\frac{3}{5}\frac{x^5}{s^5}-\frac{1}{7}\frac{x^7}{s^7}]$

$\int \frac{x^{2}dx}{s^5}=-\frac{1}{a^2}\frac{x^3}{3s^3}$

$\int \frac{x^{2}dx}{s^7}=\frac{1}{a^4}[\frac{1}{3}\frac{x^3}{s^3}-\frac{1}{5}\frac{x^5}{s^5}]$

$\int \frac{x^{2}dx}{s^9}=-\frac{1}{a^6}[\frac{1}{3}\frac{x^3}{s^3}-\frac{2}{5}\frac{x^5}{s^5}+\frac{1}{7}\frac{x^7}{s^7}]$

$\int udx=\frac{1}{2}(xu+a^2\arcsin \frac{x}{a})\text{(}|x|\le |a|\text{)}$

$\int xudx=-\frac{1}{3}{u}^3\text{(}|x|\le |a|\text{)}$

$\int x^{2}udx=-\frac{x}{4}{u}^3+\frac{a^2}{8}(xu+a^2\arcsin \frac{x}{a})\text{(}|x|\le |a|\text{)}$

$\int \frac{udx}{x}=u-a\ln \left|\frac{a+u}{x}\right|\text{(}|x|\le |a|\text{)}$

$\int \frac{dx}{u}=\arcsin \frac{x}{a}\text{(}|x|\le |a|\text{)}$

$\int \frac{x^{2}dx}{u}=\frac{1}{2}(-xu+a^2\arcsin \frac{x}{a})\text{(}|x|\le |a|\text{)}$

$\int udx=\frac{1}{2}(xu-\mathrm{sgn}x\mathrm{arcosh}\left|\frac{x}{a}\right|)\text{(for }|x|\ge |a|\text{)}$

$\int \frac{x}{u}dx=-u\text{(}|x|\le |a|\text{)}$

Tích phân hàm hợp (Integrals involving) $R=\sqrt{a{x}^2+bx+c}$

Assume (ax² + bx + c) cannot be reduced to the following expression (px + q)² for some p and q.

$\int \frac{dx}{R}=\frac{1}{\sqrt{a}}\ln |2\sqrt{a}R+2ax+b|\text{(for }a>0\text{)}$

$\int \frac{dx}{R}=\frac{1}{\sqrt{a}}\mathrm{arsinh}\frac{2ax+b}{\sqrt{4ac-b^2}}\text{(for }a>0\text{, }4ac-b^2>0\text{)}$

$\int \frac{dx}{R}=\frac{1}{\sqrt{a}}\ln |2ax+b|\text{(for }a>0\text{, }4ac-b^2=0\text{)}$

$\int \frac{dx}{R}=-\frac{1}{\sqrt{-a}}\arcsin \frac{2ax+b}{\sqrt{b^2-4ac}}\text{(for }a<0\text{, }4ac-b^2<0\text{, }|2ax+b|<\sqrt{b^2-4ac}\text{)}$

$\int \frac{dx}{R^3}=\frac{4ax+2b}{(4ac-b^2)R}$

$\int \frac{dx}{R^5}=\frac{4ax+2b}{3(4ac-b^2)R}(\frac{1}{R^2}+\frac{8a}{4ac-b^2})$

$\int \frac{dx}{R^{2n+1}}=\frac{2}{(2n-1)(4ac-b^2)}(\frac{2ax+b}{R^{2n-1}}+4a(n-1)\int \frac{dx}{R^{2n-1}})$

$\int \frac{x}{R}dx=\frac{R}{a}-\frac{b}{2a}\int \frac{dx}{R}$

$\int \frac{x}{R^3}dx=-\frac{2bx+4c}{(4ac-b^2)R}$

$\int \frac{x}{R^{2n+1}}dx=-\frac{1}{(2n-1)a{R}^{2n-1}}-\frac{b}{2a}\int \frac{dx}{R^{2n+1}}$

$\int \frac{dx}{xR}=-\frac{1}{\sqrt{c}}\ln \left(\frac{2\sqrt{c}R+bx+2c}{x}\right)$

$\int \frac{dx}{xR}=-\frac{1}{\sqrt{c}}\mathrm{arsinh}\left(\frac{bx+2c}{|x|\sqrt{4ac-b^2}}\right)$

$\int Sdx=\frac{2S^3}{3a}$

$\int \frac{dx}{S}=\frac{2S}{a}$

$\int \frac{dx}{xS}=\{\begin{array}{cc}-\frac{2}{\sqrt{b}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{h}\left(\frac{S}{\sqrt{b}}\right)& \text{(for }b>0,ax>0\text{)}\\ -\frac{2}{\sqrt{b}}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\left(\frac{S}{\sqrt{b}}\right)& \text{(for }b>0,ax<0\text{)}\\ \frac{2}{\sqrt{-b}}\arctan \left(\frac{S}{\sqrt{-b}}\right)& \text{(for }b<0\text{)}\end{array}$

$\int \frac{S}{x}dx=\{\begin{array}{cc}2(S-\sqrt{b}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{h}\left(\frac{S}{\sqrt{b}}\right))& \text{(for }b>0,ax>0\text{)}\\ 2(S-\sqrt{b}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{h}\left(\frac{S}{\sqrt{b}}\right))& \text{(for }b>0,ax<0\text{)}\\ 2(S-\sqrt{-b}\arctan \left(\frac{S}{\sqrt{-b}}\right))& \text{(for }b<0\text{)}\end{array}$

$\int \frac{x^n}{S}dx=\frac{2}{a(2n+1)}(x^{n}S-bn\int \frac{x^{n-1}}{S}dx)$

$\int x^{n}Sdx=\frac{2}{a(2n+3)}(x^n{S}^3-nb\int x^{n-1}Sdx)$

$\int \frac{1}{x^{n}S}dx=-\frac{1}{b(n-1)}(\frac{S}{x^{n-1}}+(n-\frac{3}{2})a\int \frac{dx}{x^{n-1}S})$

$\int \sin axdx=-\frac{1}{a}\cos ax+C$

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

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

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

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

$\int \sin b_{1}x\sin b_{2}xdx=\frac{\sin ((b_1-b_2)x)}{2(b_1-b_2)}-\frac{\sin ((b_1+b_2)x)}{2(b_1+b_2)}+C\text{(for }|b_1|\ne |b_2|\text{)}$

$\int {\sin }^{n}axdx=-\frac{{\sin }^{n-1}ax\cos ax}{na}+\frac{n-1}{n}\int {\sin }^{n-2}axdx\text{(for }n>2\text{)}$

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

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

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

$\int x^n\sin axdx=-\frac{x^n}{a}\cos ax+\frac{n}{a}\int x^{n-1}\cos axdx=\sum _{k=0}^{2k\le n}(-1{)}^{k+1}\frac{x^{n-2k}}{a^{1+2k}}\frac{n!}{(n-2k)!}\cos ax+\sum _{k=0}^{2k+1\le n}(-1{)}^k\frac{x^{n-1-2k}}{a^{2+2k}}\frac{n!}{(n-2k-1)!}\sin ax\text{(for }n>0\text{)}$

$\underset{\frac{-a}{2}}{\overset{\frac{a}{2}}{\int }}{x}^2{\sin }^2\frac{n\pi x}{a}dx=\frac{a^3(n^2{\pi }^2-6)}{24n^2{\pi }^2}\text{(for }n=2,4,6...\text{)}$

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

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

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

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

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

$\int \frac{\sin axdx}{1\pm \sin ax}=\pm x+\frac{1}{a}\tan (\frac{\pi }{4}\mp \frac{ax}{2})+C$

$\int \cos axdx=\frac{1}{a}\sin ax+C$

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

$\int {\cos }^{n}axdx=\frac{{\cos }^{n-1}ax\sin ax}{na}+\frac{n-1}{n}\int {\cos }^{n-2}axdx\text{(for }n>0\text{)}$

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

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

$\int x^n\cos axdx=\frac{x^n\sin ax}{a}-\frac{n}{a}\int x^{n-1}\sin axdx=\sum _{k=0}^{2k+1\le n}(-1{)}^k\frac{x^{n-2k-1}}{a^{2+2k}}\frac{n!}{(n-2k-1)!}\cos ax+\sum _{k=0}^{2k\le n}(-1{)}^k\frac{x^{n-2k}}{a^{1+2k}}\frac{n!}{(n-2k)!}\sin ax$

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

$\int \frac{\cos ax}{x^n}dx=-\frac{\cos ax}{(n-1)x^{n-1}}-\frac{a}{n-1}\int \frac{\sin ax}{x^{n-1}}dx\text{(for }n\ne 1\text{)}$

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

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

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

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

$\int \frac{xdx}{1+\cos ax}=\frac{x}{a}\tan \frac{ax}{2}+\frac{2}{a^2}\ln |\cos \frac{ax}{2}|+C$

$\int \frac{xdx}{1-\cos ax}=-\frac{x}{a}\cot \frac{ax}{2}+\frac{2}{a^2}\ln |\sin \frac{ax}{2}|+C$

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

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

$\int \cos a_{1}x\cos a_{2}xdx=\frac{\sin (a_1-a_2)x}{2(a_1-a_2)}+\frac{\sin (a_1+a_2)x}{2(a_1+a_2)}+C\text{(for }|a_1|\ne |a_2|\text{)}$

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

$\int {\tan }^{n}axdx=\frac{1}{a(n-1)}{\tan }^{n-1}ax-\int {\tan }^{n-2}axdx\text{(for }n\ne 1\text{)}$

$\int \frac{dx}{q\tan ax+p}=\frac{1}{p^2+q^2}(px+\frac{q}{a}\ln |q\sin ax+p\cos ax|)+C\text{(for }{p}^2+q^2\ne 0\text{)}$

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

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

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

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

$\int \sec axdx=\frac{1}{a}\ln |\sec ax+\tan ax|+C$

$\int {\sec }^{2}xdx=\tan x+C$

$\int {\sec }^{n}axdx=\frac{{\sec }^{n-2}ax\tan ax}{a(n-1)}+\frac{n-2}{n-1}\int {\sec }^{n-2}axdx\text{ (for }n\ne 1\text{)}$

$\int {\sec }^{n}xdx=\frac{{\sec }^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int {\sec }^{n-2}xdx$

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

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