# Tích phân hàm số toán lủy thừa e

${\displaystyle \int e^{cx}\;dx={\frac {1}{c}}e^{cx}}$
${\displaystyle \int a^{cx}\;dx={\frac {1}{c\ln a}}a^{cx}\qquad {\mbox{( }}a>0,{\mbox{ }}a\neq 1{\mbox{)}}}$
${\displaystyle \int xe^{cx}\;dx={\frac {e^{cx}}{c^{2}}}(cx-1)}$
${\displaystyle \int x^{2}e^{cx}\;dx=e^{cx}\left({\frac {x^{2}}{c}}-{\frac {2x}{c^{2}}}+{\frac {2}{c^{3}}}\right)}$
${\displaystyle \int x^{n}e^{cx}\;dx={\frac {1}{c}}x^{n}e^{cx}-{\frac {n}{c}}\int x^{n-1}e^{cx}dx}$
${\displaystyle \int {\frac {e^{cx}\;dx}{x}}=\ln |x|+\sum _{i=1}^{\infty }{\frac {(cx)^{i}}{i\cdot i!}}}$
${\displaystyle \int {\frac {e^{cx}\;dx}{x^{n}}}={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,dx\right)\qquad {\mbox{( }}n\neq 1{\mbox{)}}}$
${\displaystyle \int e^{cx}\ln x\;dx={\frac {1}{c}}e^{cx}\ln |x|-\operatorname {Ei} \,(cx)}$
${\displaystyle \int e^{cx}\sin bx\;dx={\frac {e^{cx}}{c^{2}+b^{2}}}(c\sin bx-b\cos bx)}$
${\displaystyle \int e^{cx}\cos bx\;dx={\frac {e^{cx}}{c^{2}+b^{2}}}(c\cos bx+b\sin bx)}$
${\displaystyle \int e^{cx}\sin ^{n}x\;dx={\frac {e^{cx}\sin ^{n-1}x}{c^{2}+n^{2}}}(c\sin x-n\cos x)+{\frac {n(n-1)}{c^{2}+n^{2}}}\int e^{cx}\sin ^{n-2}x\;dx}$
${\displaystyle \int e^{cx}\cos ^{n}x\;dx={\frac {e^{cx}\cos ^{n-1}x}{c^{2}+n^{2}}}(c\cos x+n\sin x)+{\frac {n(n-1)}{c^{2}+n^{2}}}\int e^{cx}\cos ^{n-2}x\;dx}$
${\displaystyle \int xe^{cx^{2}}\;dx={\frac {1}{2c}}\;e^{cx^{2}}}$
${\displaystyle \int {1 \over \sigma {\sqrt {2\pi }}}\,e^{-{(x-\mu )^{2}/2\sigma ^{2}}}\;dx={\frac {1}{2\sigma }}(1+{\mbox{erf}}\,{\frac {x-\mu }{\sigma {\sqrt {2}}}})}$
${\displaystyle \int e^{x^{2}}\,dx=e^{x^{2}}\left(\sum _{j=0}^{n-1}c_{2j}\,{\frac {1}{x^{2j+1}}}\right)+(2n-1)c_{2n-2}\int {\frac {e^{x^{2}}}{x^{2n}}}\;dx\quad (n>0),}$
với ${\displaystyle c_{2j}={\frac {1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1}}}={\frac {2j\,!}{j!\,2^{2j+1}}}\ .}$
${\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}\,dx={\sqrt {\pi \over a}}}$
${\displaystyle \int _{0}^{\infty }x^{2n}e^{-{x^{2}}/{a^{2}}}\,dx={\sqrt {\pi }}{(2n)! \over {n!}}{\left({\frac {a}{2}}\right)}^{2n+1}}$

${\displaystyle \int \sinh cx\,dx={\frac {1}{c}}\cosh cx}$
${\displaystyle \int \cosh cx\,dx={\frac {1}{c}}\sinh cx}$
${\displaystyle \int \sinh ^{2}cx\,dx={\frac {1}{4c}}\sinh 2cx-{\frac {x}{2}}}$
${\displaystyle \int \cosh ^{2}cx\,dx={\frac {1}{4c}}\sinh 2cx+{\frac {x}{2}}}$
${\displaystyle \int \sinh ^{n}cx\,dx={\frac {1}{cn}}\sinh ^{n-1}cx\cosh cx-{\frac {n-1}{n}}\int \sinh ^{n-2}cx\,dx\qquad {\mbox{( }}n>0{\mbox{)}}}$
hay: ${\displaystyle \int \sinh ^{n}cx\,dx={\frac {1}{c(n+1)}}\sinh ^{n+1}cx\cosh cx-{\frac {n+2}{n+1}}\int \sinh ^{n+2}cx\,dx\qquad {\mbox{( }}n<0{\mbox{, }}n\neq -1{\mbox{)}}}$
${\displaystyle \int \cosh ^{n}cx\,dx={\frac {1}{cn}}\sinh cx\cosh ^{n-1}cx+{\frac {n-1}{n}}\int \cosh ^{n-2}cx\,dx\qquad {\mbox{( }}n>0{\mbox{)}}}$
hay: ${\displaystyle \int \cosh ^{n}cx\,dx=-{\frac {1}{c(n+1)}}\sinh cx\cosh ^{n+1}cx-{\frac {n+2}{n+1}}\int \cosh ^{n+2}cx\,dx\qquad {\mbox{( }}n<0{\mbox{, }}n\neq -1{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{\sinh cx}}={\frac {1}{c}}\ln \left|\tanh {\frac {cx}{2}}\right|}$
hay: ${\displaystyle \int {\frac {dx}{\sinh cx}}={\frac {1}{c}}\ln \left|{\frac {\cosh cx-1}{\sinh cx}}\right|}$
hay: ${\displaystyle \int {\frac {dx}{\sinh cx}}={\frac {1}{c}}\ln \left|{\frac {\sinh cx}{\cosh cx+1}}\right|}$
hay: ${\displaystyle \int {\frac {dx}{\sinh cx}}={\frac {1}{c}}\ln \left|{\frac {\cosh cx-1}{\cosh cx+1}}\right|}$
${\displaystyle \int {\frac {dx}{\cosh cx}}={\frac {2}{c}}\arctan e^{cx}}$
${\displaystyle \int {\frac {dx}{\sinh ^{n}cx}}={\frac {\cosh cx}{c(n-1)\sinh ^{n-1}cx}}-{\frac {n-2}{n-1}}\int {\frac {dx}{\sinh ^{n-2}cx}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{\cosh ^{n}cx}}={\frac {\sinh cx}{c(n-1)\cosh ^{n-1}cx}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cosh ^{n-2}cx}}\qquad {\mbox{( }}n\neq 1{\mbox{)}}}$
${\displaystyle \int {\frac {\cosh ^{n}cx}{\sinh ^{m}cx}}dx={\frac {\cosh ^{n-1}cx}{c(n-m)\sinh ^{m-1}cx}}+{\frac {n-1}{n-m}}\int {\frac {\cosh ^{n-2}cx}{\sinh ^{m}cx}}dx\qquad {\mbox{( }}m\neq n{\mbox{)}}}$
hay: ${\displaystyle \int {\frac {\cosh ^{n}cx}{\sinh ^{m}cx}}dx=-{\frac {\cosh ^{n+1}cx}{c(m-1)\sinh ^{m-1}cx}}+{\frac {n-m+2}{m-1}}\int {\frac {\cosh ^{n}cx}{\sinh ^{m-2}cx}}dx\qquad {\mbox{( }}m\neq 1{\mbox{)}}}$
hay: ${\displaystyle \int {\frac {\cosh ^{n}cx}{\sinh ^{m}cx}}dx=-{\frac {\cosh ^{n-1}cx}{c(m-1)\sinh ^{m-1}cx}}+{\frac {n-1}{m-1}}\int {\frac {\cosh ^{n-2}cx}{\sinh ^{m-2}cx}}dx\qquad {\mbox{( }}m\neq 1{\mbox{)}}}$
${\displaystyle \int {\frac {\sinh ^{m}cx}{\cosh ^{n}cx}}dx={\frac {\sinh ^{m-1}cx}{c(m-n)\cosh ^{n-1}cx}}+{\frac {m-1}{m-n}}\int {\frac {\sinh ^{m-2}cx}{\cosh ^{n}cx}}dx\qquad {\mbox{( }}m\neq n{\mbox{)}}}$
hay: ${\displaystyle \int {\frac {\sinh ^{m}cx}{\cosh ^{n}cx}}dx={\frac {\sinh ^{m+1}cx}{c(n-1)\cosh ^{n-1}cx}}+{\frac {m-n+2}{n-1}}\int {\frac {\sinh ^{m}cx}{\cosh ^{n-2}cx}}dx\qquad {\mbox{( }}n\neq 1{\mbox{)}}}$
hay: ${\displaystyle \int {\frac {\sinh ^{m}cx}{\cosh ^{n}cx}}dx=-{\frac {\sinh ^{m-1}cx}{c(n-1)\cosh ^{n-1}cx}}+{\frac {m-1}{n-1}}\int {\frac {\sinh ^{m-2}cx}{\cosh ^{n-2}cx}}dx\qquad {\mbox{( }}n\neq 1{\mbox{)}}}$
${\displaystyle \int x\sinh cx\,dx={\frac {1}{c}}x\cosh cx-{\frac {1}{c^{2}}}\sinh cx}$
${\displaystyle \int x\cosh cx\,dx={\frac {1}{c}}x\sinh cx-{\frac {1}{c^{2}}}\cosh cx}$
${\displaystyle \int \tanh cx\,dx={\frac {1}{c}}\ln |\cosh cx|}$
${\displaystyle \int \coth cx\,dx={\frac {1}{c}}\ln |\sinh cx|}$
${\displaystyle \int \tanh ^{n}cx\,dx=-{\frac {1}{c(n-1)}}\tanh ^{n-1}cx+\int \tanh ^{n-2}cx\,dx\qquad {\mbox{( }}n\neq 1{\mbox{)}}}$
${\displaystyle \int \coth ^{n}cx\,dx=-{\frac {1}{c(n-1)}}\coth ^{n-1}cx+\int \coth ^{n-2}cx\,dx\qquad {\mbox{( }}n\neq 1{\mbox{)}}}$
${\displaystyle \int \sinh bx\sinh cx\,dx={\frac {1}{b^{2}-c^{2}}}(b\sinh cx\cosh bx-c\cosh cx\sinh bx)\qquad {\mbox{( }}b^{2}\neq c^{2}{\mbox{)}}}$
${\displaystyle \int \cosh bx\cosh cx\,dx={\frac {1}{b^{2}-c^{2}}}(b\sinh bx\cosh cx-c\sinh cx\cosh bx)\qquad {\mbox{( }}b^{2}\neq c^{2}{\mbox{)}}}$
${\displaystyle \int \cosh bx\sinh cx\,dx={\frac {1}{b^{2}-c^{2}}}(b\sinh bx\sinh cx-c\cosh bx\cosh cx)\qquad {\mbox{( }}b^{2}\neq c^{2}{\mbox{)}}}$
${\displaystyle \int \sinh(ax+b)\sin(cx+d)\,dx={\frac {a}{a^{2}+c^{2}}}\cosh(ax+b)\sin(cx+d)-{\frac {c}{a^{2}+c^{2}}}\sinh(ax+b)\cos(cx+d)}$
${\displaystyle \int \sinh(ax+b)\cos(cx+d)\,dx={\frac {a}{a^{2}+c^{2}}}\cosh(ax+b)\cos(cx+d)+{\frac {c}{a^{2}+c^{2}}}\sinh(ax+b)\sin(cx+d)}$
${\displaystyle \int \cosh(ax+b)\sin(cx+d)\,dx={\frac {a}{a^{2}+c^{2}}}\sinh(ax+b)\sin(cx+d)-{\frac {c}{a^{2}+c^{2}}}\cosh(ax+b)\cos(cx+d)}$
${\displaystyle \int \cosh(ax+b)\cos(cx+d)\,dx={\frac {a}{a^{2}+c^{2}}}\sinh(ax+b)\cos(cx+d)+{\frac {c}{a^{2}+c^{2}}}\cosh(ax+b)\sin(cx+d)}$

Assume ${\displaystyle (x^{2}>a^{2})}$, for ${\displaystyle (x^{2}, see next section:

${\displaystyle \int xs\;dx={\frac {1}{3}}s^{3}}$
${\displaystyle \int {\frac {s\;dx}{x}}=s-a\arccos \left|{\frac {a}{x}}\right|}$
${\displaystyle \int {\frac {dx}{s}}=\int {\frac {dx}{\sqrt {x^{2}-a^{2}}}}=\ln \left|{\frac {x+s}{a}}\right|}$

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

${\displaystyle \int {\frac {x\;dx}{s}}=s}$
${\displaystyle \int {\frac {x\;dx}{s^{3}}}=-{\frac {1}{s}}}$
${\displaystyle \int {\frac {x\;dx}{s^{5}}}=-{\frac {1}{3s^{3}}}}$
${\displaystyle \int {\frac {x\;dx}{s^{7}}}=-{\frac {1}{5s^{5}}}}$
${\displaystyle \int {\frac {x\;dx}{s^{2n+1}}}=-{\frac {1}{(2n-1)s^{2n-1}}}}$
${\displaystyle \int {\frac {x^{2m}\;dx}{s^{2n+1}}}=-{\frac {1}{2n-1}}{\frac {x^{2m-1}}{s^{2n-1}}}+{\frac {2m-1}{2n-1}}\int {\frac {x^{2m-2}\;dx}{s^{2n-1}}}}$
${\displaystyle \int {\frac {x^{2}\;dx}{s}}={\frac {xs}{2}}+{\frac {a^{2}}{2}}\ln \left|{\frac {x+s}{a}}\right|}$
${\displaystyle \int {\frac {x^{2}\;dx}{s^{3}}}=-{\frac {x}{s}}+\ln \left|{\frac {x+s}{a}}\right|}$
${\displaystyle \int {\frac {x^{4}\;dx}{s}}={\frac {x^{3}s}{4}}+{\frac {3}{8}}a^{2}xs+{\frac {3}{8}}a^{4}\ln \left|{\frac {x+s}{a}}\right|}$
${\displaystyle \int {\frac {x^{4}\;dx}{s^{3}}}={\frac {xs}{2}}-{\frac {a^{2}x}{s}}+{\frac {3}{2}}a^{2}\ln \left|{\frac {x+s}{a}}\right|}$
${\displaystyle \int {\frac {x^{4}\;dx}{s^{5}}}=-{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}+\ln \left|{\frac {x+s}{a}}\right|}$
${\displaystyle \int {\frac {x^{2m}\;dx}{s^{2n+1}}}=(-1)^{n-m}{\frac {1}{a^{2(n-m)}}}\sum _{i=0}^{n-m-1}{\frac {1}{2(m+i)+1}}{n-m-1 \choose i}{\frac {x^{2(m+i)+1}}{s^{2(m+i)+1}}}\qquad {\mbox{(}}n>m\geq 0{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{s^{3}}}=-{\frac {1}{a^{2}}}{\frac {x}{s}}}$
${\displaystyle \int {\frac {dx}{s^{5}}}={\frac {1}{a^{4}}}\left[{\frac {x}{s}}-{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}\right]}$
${\displaystyle \int {\frac {dx}{s^{7}}}=-{\frac {1}{a^{6}}}\left[{\frac {x}{s}}-{\frac {2}{3}}{\frac {x^{3}}{s^{3}}}+{\frac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}$
${\displaystyle \int {\frac {dx}{s^{9}}}={\frac {1}{a^{8}}}\left[{\frac {x}{s}}-{\frac {3}{3}}{\frac {x^{3}}{s^{3}}}+{\frac {3}{5}}{\frac {x^{5}}{s^{5}}}-{\frac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}$
${\displaystyle \int {\frac {x^{2}\;dx}{s^{5}}}=-{\frac {1}{a^{2}}}{\frac {x^{3}}{3s^{3}}}}$
${\displaystyle \int {\frac {x^{2}\;dx}{s^{7}}}={\frac {1}{a^{4}}}\left[{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\frac {1}{5}}{\frac {x^{5}}{s^{5}}}\right]}$
${\displaystyle \int {\frac {x^{2}\;dx}{s^{9}}}=-{\frac {1}{a^{6}}}\left[{\frac {1}{3}}{\frac {x^{3}}{s^{3}}}-{\frac {2}{5}}{\frac {x^{5}}{s^{5}}}+{\frac {1}{7}}{\frac {x^{7}}{s^{7}}}\right]}$

${\displaystyle \int u\;dx={\frac {1}{2}}\left(xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$
${\displaystyle \int xu\;dx=-{\frac {1}{3}}u^{3}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$
${\displaystyle \int x^{2}u\;dx=-{\frac {x}{4}}u^{3}+{\frac {a^{2}}{8}}(xu+a^{2}\arcsin {\frac {x}{a}})\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$
${\displaystyle \int {\frac {u\;dx}{x}}=u-a\ln \left|{\frac {a+u}{x}}\right|\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{u}}=\arcsin {\frac {x}{a}}\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$
${\displaystyle \int {\frac {x^{2}\;dx}{u}}={\frac {1}{2}}\left(-xu+a^{2}\arcsin {\frac {x}{a}}\right)\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$
${\displaystyle \int u\;dx={\frac {1}{2}}\left(xu-\operatorname {sgn} x\,\operatorname {arcosh} \left|{\frac {x}{a}}\right|\right)\qquad {\mbox{(for }}|x|\geq |a|{\mbox{)}}}$
${\displaystyle \int {\frac {x}{u}}\;dx=-u\qquad {\mbox{(}}|x|\leq |a|{\mbox{)}}}$

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

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

${\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\ln \left|2{\sqrt {a}}R+2ax+b\right|\qquad {\mbox{(for }}a>0{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\,\operatorname {arsinh} {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}\qquad {\mbox{(for }}a>0{\mbox{, }}4ac-b^{2}>0{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{R}}={\frac {1}{\sqrt {a}}}\ln |2ax+b|\quad {\mbox{(for }}a>0{\mbox{, }}4ac-b^{2}=0{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{R}}=-{\frac {1}{\sqrt {-a}}}\arcsin {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}\qquad {\mbox{(for }}a<0{\mbox{, }}4ac-b^{2}<0{\mbox{, }}\left|2ax+b\right|<{\sqrt {b^{2}-4ac}}{\mbox{)}}}$
${\displaystyle \int {\frac {dx}{R^{3}}}={\frac {4ax+2b}{(4ac-b^{2})R}}}$
${\displaystyle \int {\frac {dx}{R^{5}}}={\frac {4ax+2b}{3(4ac-b^{2})R}}\left({\frac {1}{R^{2}}}+{\frac {8a}{4ac-b^{2}}}\right)}$
${\displaystyle \int {\frac {dx}{R^{2n+1}}}={\frac {2}{(2n-1)(4ac-b^{2})}}\left({\frac {2ax+b}{R^{2n-1}}}+4a(n-1)\int {\frac {dx}{R^{2n-1}}}\right)}$
${\displaystyle \int {\frac {x}{R}}\;dx={\frac {R}{a}}-{\frac {b}{2a}}\int {\frac {dx}{R}}}$
${\displaystyle \int {\frac {x}{R^{3}}}\;dx=-{\frac {2bx+4c}{(4ac-b^{2})R}}}$
${\displaystyle \int {\frac {x}{R^{2n+1}}}\;dx=-{\frac {1}{(2n-1)aR^{2n-1}}}-{\frac {b}{2a}}\int {\frac {dx}{R^{2n+1}}}}$
${\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\ln \left({\frac {2{\sqrt {c}}R+bx+2c}{x}}\right)}$
${\displaystyle \int {\frac {dx}{xR}}=-{\frac {1}{\sqrt {c}}}\operatorname {arsinh} \left({\frac {bx+2c}{|x|{\sqrt {4ac-b^{2}}}}}\right)}$
${\displaystyle \int S{dx}={\frac {2S^{3}}{3a}}}$
${\displaystyle \int {\frac {dx}{S}}={\frac {2S}{a}}}$
${\displaystyle \int {\frac {dx}{xS}}={\begin{cases}-{\frac {2}{\sqrt {b}}}\mathrm {arcoth} \left({\frac {S}{\sqrt {b}}}\right)&{\mbox{(for }}b>0,\quad ax>0{\mbox{)}}\\-{\frac {2}{\sqrt {b}}}\mathrm {artanh} \left({\frac {S}{\sqrt {b}}}\right)&{\mbox{(for }}b>0,\quad ax<0{\mbox{)}}\\{\frac {2}{\sqrt {-b}}}\arctan \left({\frac {S}{\sqrt {-b}}}\right)&{\mbox{(for }}b<0{\mbox{)}}\\\end{cases}}}$
${\displaystyle \int {\frac {S}{x}}\,dx={\begin{cases}2\left(S-{\sqrt {b}}\,\mathrm {arcoth} \left({\frac {S}{\sqrt {b}}}\right)\right)&{\mbox{(for }}b>0,\quad ax>0{\mbox{)}}\\2\left(S-{\sqrt {b}}\,\mathrm {artanh} \left({\frac {S}{\sqrt {b}}}\right)\right)&{\mbox{(for }}b>0,\quad ax<0{\mbox{)}}\\2\left(S-{\sqrt {-b}}\arctan \left({\frac {S}{\sqrt {-b}}}\right)\right)&{\mbox{(for }}b<0{\mbox{)}}\\\end{cases}}}$
${\displaystyle \int {\frac {x^{n}}{S}}dx={\frac {2}{a(2n+1)}}\left(x^{n}S-bn\int {\frac {x^{n-1}}{S}}dx\right)}$
${\displaystyle \int x^{n}Sdx={\frac {2}{a(2n+3)}}\left(x^{n}S^{3}-nb\int x^{n-1}Sdx\right)}$
${\displaystyle \int {\frac {1}{x^{n}S}}dx=-{\frac {1}{b(n-1)}}\left({\frac {S}{x^{n-1}}}+\left(n-{\frac {3}{2}}\right)a\int {\frac {dx}{x^{n-1}S}}\right)}$

${\displaystyle \int \sin ax\;dx=-{\frac {1}{a}}\cos ax+C\,\!}$
${\displaystyle \int \sin ^{2}{ax}\;dx={\frac {x}{2}}-{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}-{\frac {1}{2a}}\sin ax\cos ax+C\!}$
${\displaystyle \int \sin ^{3}{ax}\;dx={\frac {\cos 3ax}{12a}}-{\frac {3\cos ax}{4a}}+C\!}$
${\displaystyle \int x\sin ^{2}{ax}\;dx={\frac {x^{2}}{4}}-{\frac {x}{4a}}\sin 2ax-{\frac {1}{8a^{2}}}\cos 2ax+C\!}$
${\displaystyle \int x^{2}\sin ^{2}{ax}\;dx={\frac {x^{3}}{6}}-\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax-{\frac {x}{4a^{2}}}\cos 2ax+C\!}$
${\displaystyle \int \sin b_{1}x\sin b_{2}x\;dx={\frac {\sin((b_{1}-b_{2})x)}{2(b_{1}-b_{2})}}-{\frac {\sin((b_{1}+b_{2})x)}{2(b_{1}+b_{2})}}+C\qquad {\mbox{(for }}|b_{1}|\neq |b_{2}|{\mbox{)}}\,\!}$
${\displaystyle \int \sin ^{n}{ax}\;dx=-{\frac {\sin ^{n-1}ax\cos ax}{na}}+{\frac {n-1}{n}}\int \sin ^{n-2}ax\;dx\qquad {\mbox{(for }}n>2{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {dx}{\sin ax}}={\frac {1}{a}}\ln \left|\tan {\frac {ax}{2}}\right|+C}$
${\displaystyle \int {\frac {dx}{\sin ^{n}ax}}={\frac {\cos ax}{a(1-n)\sin ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\sin ^{n-2}ax}}\qquad {\mbox{(for }}n>1{\mbox{)}}\,\!}$
${\displaystyle \int x\sin ax\;dx={\frac {\sin ax}{a^{2}}}-{\frac {x\cos ax}{a}}+C\,\!}$
${\displaystyle \int x^{n}\sin ax\;dx=-{\frac {x^{n}}{a}}\cos ax+{\frac {n}{a}}\int x^{n-1}\cos ax\;dx=\sum _{k=0}^{2k\leq n}(-1)^{k+1}{\frac {x^{n-2k}}{a^{1+2k}}}{\frac {n!}{(n-2k)!}}\cos ax+\sum _{k=0}^{2k+1\leq n}(-1)^{k}{\frac {x^{n-1-2k}}{a^{2+2k}}}{\frac {n!}{(n-2k-1)!}}\sin ax\qquad {\mbox{(for }}n>0{\mbox{)}}\,\!}$
${\displaystyle \int _{\frac {-a}{2}}^{\frac {a}{2}}x^{2}\sin ^{2}{\frac {n\pi x}{a}}\;dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad {\mbox{(for }}n=2,4,6...{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {\sin ax}{x}}dx=\sum _{n=0}^{\infty }(-1)^{n}{\frac {(ax)^{2n+1}}{(2n+1)\cdot (2n+1)!}}+C\,\!}$
${\displaystyle \int {\frac {\sin ax}{x^{n}}}dx=-{\frac {\sin ax}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}\int {\frac {\cos ax}{x^{n-1}}}dx\,\!}$
${\displaystyle \int {\frac {dx}{1\pm \sin ax}}={\frac {1}{a}}\tan \left({\frac {ax}{2}}\mp {\frac {\pi }{4}}\right)+C}$
${\displaystyle \int {\frac {x\;dx}{1+\sin ax}}={\frac {x}{a}}\tan \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)+{\frac {2}{a^{2}}}\ln \left|\cos \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)\right|+C}$
${\displaystyle \int {\frac {x\;dx}{1-\sin ax}}={\frac {x}{a}}\cot \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)+{\frac {2}{a^{2}}}\ln \left|\sin \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)\right|+C}$
${\displaystyle \int {\frac {\sin ax\;dx}{1\pm \sin ax}}=\pm x+{\frac {1}{a}}\tan \left({\frac {\pi }{4}}\mp {\frac {ax}{2}}\right)+C}$
${\displaystyle \int \cos ax\;dx={\frac {1}{a}}\sin ax+C\,\!}$
${\displaystyle \int \cos ^{2}{ax}\;dx={\frac {x}{2}}+{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}+{\frac {1}{2a}}\sin ax\cos ax+C\!}$
${\displaystyle \int \cos ^{n}ax\;dx={\frac {\cos ^{n-1}ax\sin ax}{na}}+{\frac {n-1}{n}}\int \cos ^{n-2}ax\;dx\qquad {\mbox{(for }}n>0{\mbox{)}}\,\!}$
${\displaystyle \int x\cos ax\;dx={\frac {\cos ax}{a^{2}}}+{\frac {x\sin ax}{a}}+C\,\!}$
${\displaystyle \int x^{2}\cos ^{2}{ax}\;dx={\frac {x^{3}}{6}}+\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax+{\frac {x}{4a^{2}}}\cos 2ax+C\!}$
${\displaystyle \int x^{n}\cos ax\;dx={\frac {x^{n}\sin ax}{a}}-{\frac {n}{a}}\int x^{n-1}\sin ax\;dx\,=\sum _{k=0}^{2k+1\leq n}(-1)^{k}{\frac {x^{n-2k-1}}{a^{2+2k}}}{\frac {n!}{(n-2k-1)!}}\cos ax+\sum _{k=0}^{2k\leq n}(-1)^{k}{\frac {x^{n-2k}}{a^{1+2k}}}{\frac {n!}{(n-2k)!}}\sin ax\!}$
${\displaystyle \int {\frac {\cos ax}{x}}dx=\ln |ax|+\sum _{k=1}^{\infty }(-1)^{k}{\frac {(ax)^{2k}}{2k\cdot (2k)!}}+C\,\!}$
${\displaystyle \int {\frac {\cos ax}{x^{n}}}dx=-{\frac {\cos ax}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\int {\frac {\sin ax}{x^{n-1}}}dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {dx}{\cos ax}}={\frac {1}{a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C}$
${\displaystyle \int {\frac {dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n>1{\mbox{)}}\,\!}$
${\displaystyle \int {\frac {dx}{1+\cos ax}}={\frac {1}{a}}\tan {\frac {ax}{2}}+C\,\!}$
${\displaystyle \int {\frac {dx}{1-\cos ax}}=-{\frac {1}{a}}\cot {\frac {ax}{2}}+C\,\!}$
${\displaystyle \int {\frac {x\;dx}{1+\cos ax}}={\frac {x}{a}}\tan {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left|\cos {\frac {ax}{2}}\right|+C}$
${\displaystyle \int {\frac {x\;dx}{1-\cos ax}}=-{\frac {x}{a}}\cot {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left|\sin {\frac {ax}{2}}\right|+C}$
${\displaystyle \int {\frac {\cos ax\;dx}{1+\cos ax}}=x-{\frac {1}{a}}\tan {\frac {ax}{2}}+C\,\!}$
${\displaystyle \int {\frac {\cos ax\;dx}{1-\cos ax}}=-x-{\frac {1}{a}}\cot {\frac {ax}{2}}+C\,\!}$
${\displaystyle \int \cos a_{1}x\cos a_{2}x\;dx={\frac {\sin(a_{1}-a_{2})x}{2(a_{1}-a_{2})}}+{\frac {\sin(a_{1}+a_{2})x}{2(a_{1}+a_{2})}}+C\qquad {\mbox{(for }}|a_{1}|\neq |a_{2}|{\mbox{)}}\,\!}$

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

${\displaystyle \int \sec {ax}\,dx={\frac {1}{a}}\ln {\left|\sec {ax}+\tan {ax}\right|}+C}$
${\displaystyle \int \sec ^{2}{x}\,dx=\tan {x}+C}$
${\displaystyle \int \sec ^{n}{ax}\,dx={\frac {\sec ^{n-2}{ax}\tan {ax}}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{ax}\,dx\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,\!}$
${\displaystyle \int \sec ^{n}{x}\,dx={\frac {\sec ^{n-2}{x}\tan {x}}{n-1}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{x}\,dx}$
${\displaystyle \int {\frac {dx}{\sec {x}+1}}=x-\tan {\frac {x}{2}}+C}$
${\displaystyle \int {\frac {dx}{\sec {x}-1}}=-x-\cot {\frac {x}{2}}+C}$