이 기사는 검증 가능 한 참고 문헌이나 출전 이 전혀 나타나지 않은지, 불충분합니다. 출전을 추가 해 기사의 신뢰성 향상에 협력해 주십시오.(2016년 1월 )
본항은 역삼각함수 를 포함한 식의 원시 함수 의 일람이다.한층 더 완전한 원시 함수의 일람은, 원시 함수의 일람 을 참조.
이하의 모든 기술에 대하고, a 는 0이 아닌 실수로 한다.또, C 는 적분 정수로 한다.
목차
역사인 관수의 적분 ∫ arcsin x d x = x arcsin x + 1 − x 2 + C {\displaystyle \int \arcsin x\,dx=x\arcsin x+{\sqrt {1-x^{2}}}+C} ∫ arcsin a x d x = x arcsin a x + 1 − a 2 x 2 a + C {\displaystyle \int \arcsin ax\,dx=x\arcsin ax+{\frac {\sqrt {1-a^{2}x^{2}}}{a}}+C} ∫ x arcsin a x d x = x 2 arcsin a x 2 − arcsin a x 4 a 2 + x 1 − a 2 x 2 4 a + C {\displaystyle \int x\arcsin ax\,dx={\frac {x^{2}\arcsin ax}{2}}-{\frac {\arcsin ax}{4a^{2}}}+{\frac {x{\sqrt {1-a^{2}x^{2}}}}{4\,a}}+C} ∫ x 2 arcsin a x d x = x 3 arcsin a x 3 + ( a 2 x 2 + 2 ) 1 − a 2 x 2 9 a 3 + C {\displaystyle \int x^{2}\arcsin ax\,dx={\frac {x^{3}\arcsin ax}{3}}+{\frac {\left(a^{2}x^{2}+2\right){\sqrt {1-a^{2}x^{2}}}}{9\,a^{3}}}+C} ∫ x m arcsin a x d x = x m + 1 arcsin a x m + 1 − a m + 1 ∫ x m + 1 1 − a 2 x 2 d x ( m ≠ − 1 ) {\displaystyle \int x^{m}\arcsin ax\,dx={\frac {x^{m+1}\arcsin ax}{m+1}}\,-\,{\frac {a}{m+1}}\int {\frac {x^{m+1}}{\sqrt {1-a^{2}x^{2}}}}\,dx\quad (m\neq -1)} ∫ ( arcsin a x ) 2 d x = − 2 x + x ( arcsin a x ) 2 + 2 1 − a 2 x 2 arcsin a x a + C {\displaystyle \int (\arcsin ax)^{2}\,dx=-2\,x+x\,(\arcsin ax)^{2}+{\frac {2{\sqrt {1-a^{2}x^{2}}}\arcsin ax}{a}}+C} ∫ ( arcsin a x ) n d x = x ( arcsin a x ) n + n 1 − a 2 x 2 ( arcsin a x ) n − 1 a − n ( n − 1 ) ∫ ( arcsin a x ) n − 2 d x {\displaystyle \int (\arcsin ax)^{n}\,dx=x\,(\arcsin ax)^{n}\,+\,{\frac {n{\sqrt {1-a^{2}x^{2}}}\,(\arcsin ax)^{n-1}}{a}}\,-\,n\,(n-1)\int (\arcsin ax)^{n-2}\,dx} ∫ ( arcsin a x ) n d x = x ( arcsin a x ) n + 2 ( n + 1 ) ( n + 2 ) + 1 − a 2 x 2 ( arcsin a x ) n + 1 a ( n + 1 ) − 1 ( n + 1 ) ( n + 2 ) ∫ ( arcsin a x ) n + 2 d x ( n ≠ − 1 , − 2 ) {\displaystyle \int (\arcsin ax)^{n}\,dx={\frac {x\,(\arcsin ax)^{n+2}}{(n+1)\,(n+2)}}\,+\,{\frac {{\sqrt {1-a^{2}x^{2}}}\,(\arcsin ax)^{n+1}}{a\,(n+1)}}\,-\,{\frac {1}{(n+1)\,(n+2)}}\int (\arcsin ax)^{n+2}\,dx\quad (n\neq -1,-2)}
역여현 함수의 적분 ∫ arccos x d x = x arccos x − 1 − x 2 + C {\displaystyle \int \arccos x\,dx=x\arccos x-{\sqrt {1-x^{2}}}+C}
∫ arccos a x d x = x arccos a x − 1 − a 2 x 2 a + C {\displaystyle \int \arccos ax\,dx=x\arccos ax-{\frac {\sqrt {1-a^{2}x^{2}}}{a}}+C} ∫ x arccos a x d x = x 2 arccos a x 2 − arccos a x 4 a 2 − x 1 − a 2 x 2 4 a + C {\displaystyle \int x\arccos ax\,dx={\frac {x^{2}\arccos ax}{2}}-{\frac {\arccos ax}{4\,a^{2}}}-{\frac {x{\sqrt {1-a^{2}x^{2}}}}{4\,a}}+C} ∫ x 2 arccos a x d x = x 3 arccos a x 3 − ( a 2 x 2 + 2 ) 1 − a 2 x 2 9 a 3 + C {\displaystyle \int x^{2}\arccos ax\,dx={\frac {x^{3}\arccos ax}{3}}-{\frac {\left(a^{2}x^{2}+2\right){\sqrt {1-a^{2}x^{2}}}}{9\,a^{3}}}+C} ∫ x m arccos a x d x = x m + 1 arccos a x m + 1 + a m + 1 ∫ x m + 1 1 − a 2 x 2 d x ( m ≠ − 1 ) {\displaystyle \int x^{m}\arccos ax\,dx={\frac {x^{m+1}\arccos ax}{m+1}}\,+\,{\frac {a}{m+1}}\int {\frac {x^{m+1}}{\sqrt {1-a^{2}x^{2}}}}\,dx\quad (m\neq -1)} ∫ ( arccos a x ) 2 d x = − 2 x + x ( arccos a x ) 2 − 2 1 − a 2 x 2 arccos a x a + C {\displaystyle \int (\arccos ax)^{2}\,dx=-2\,x+x\,(\arccos ax)^{2}-{\frac {2{\sqrt {1-a^{2}x^{2}}}\arccos ax}{a}}+C} ∫ ( arccos a x ) n d x = x ( arccos a x ) n − n 1 − a 2 x 2 ( arccos a x ) n − 1 a − n ( n − 1 ) ∫ ( arccos a x ) n − 2 d x {\displaystyle \int (\arccos ax)^{n}\,dx=x\,(\arccos ax)^{n}\,-\,{\frac {n{\sqrt {1-a^{2}x^{2}}}\,(\arccos ax)^{n-1}}{a}}\,-\,n\,(n-1)\int (\arccos ax)^{n-2}\,dx} ∫ ( arccos a x ) n d x = x ( arccos a x ) n + 2 ( n + 1 ) ( n + 2 ) − 1 − a 2 x 2 ( arccos a x ) n + 1 a ( n + 1 ) − 1 ( n + 1 ) ( n + 2 ) ∫ ( arccos a x ) n + 2 d x ( n ≠ − 1 , − 2 ) {\displaystyle \int (\arccos ax)^{n}\,dx={\frac {x\,(\arccos ax)^{n+2}}{(n+1)\,(n+2)}}\,-\,{\frac {{\sqrt {1-a^{2}x^{2}}}\,(\arccos ax)^{n+1}}{a\,(n+1)}}\,-\,{\frac {1}{(n+1)\,(n+2)}}\int (\arccos ax)^{n+2}\,dx\quad (n\neq -1,-2)}
역탄젠트 함수의 적분 ∫ arctan x d x = x arctan x − ln ( x 2 + 1 ) 2 + C {\displaystyle \int \arctan x\,dx=x\arctan x-{\frac {\ln(x^{2}+1)}{2}}+C}
∫ arctan a x d x = x arctan a x − ln ( a 2 x 2 + 1 ) 2 a + C {\displaystyle \int \arctan ax\,dx=x\arctan ax-{\frac {\ln(a^{2}x^{2}+1)}{2\,a}}+C} ∫ x arctan a x d x = x 2 arctan a x 2 + arctan a x 2 a 2 − x 2 a + C {\displaystyle \int x\arctan ax\,dx={\frac {x^{2}\arctan ax}{2}}+{\frac {\arctan ax}{2\,a^{2}}}-{\frac {x}{2\,a}}+C} ∫ x 2 arctan a x d x = x 3 arctan a x 3 + ln ( a 2 x 2 + 1 ) 6 a 3 − x 2 6 a + C {\displaystyle \int x^{2}\arctan ax\,dx={\frac {x^{3}\arctan ax}{3}}+{\frac {\ln(a^{2}x^{2}+1)}{6\,a^{3}}}-{\frac {x^{2}}{6\,a}}+C} ∫ x m arctan a x d x = x m + 1 arctan a x m + 1 − a m + 1 ∫ x m + 1 a 2 x 2 + 1 d x ( m ≠ − 1 ) {\displaystyle \int x^{m}\arctan ax\,dx={\frac {x^{m+1}\arctan ax}{m+1}}-{\frac {a}{m+1}}\int {\frac {x^{m+1}}{a^{2}x^{2}+1}}\,dx\quad (m\neq -1)}
역코탄젠트 함수의 적분 ∫ arccot x d x = x arccot x + ln ( x 2 + 1 ) 2 + C {\displaystyle \int \operatorname {arccot} x\,dx=x\operatorname {arccot} x+{\frac {\ln \left(x^{2}+1\right)}{2}}+C}
∫ arccot a x d x = x arccot a x + ln ( a 2 x 2 + 1 ) 2 a + C {\displaystyle \int \operatorname {arccot} ax\,dx=x\operatorname {arccot} ax+{\frac {\ln \left(a^{2}x^{2}+1\right)}{2\,a}}+C} ∫ x arccot a x d x = x 2 arccot a x 2 + arccot a x 2 a 2 + x 2 a + C {\displaystyle \int x\operatorname {arccot} ax\,dx={\frac {x^{2}\operatorname {arccot} ax}{2}}+{\frac {\operatorname {arccot} ax}{2\,a^{2}}}+{\frac {x}{2\,a}}+C} ∫ x 2 arccot a x d x = x 3 arccot a x 3 − ln ( a 2 x 2 + 1 ) 6 a 3 + x 2 6 a + C {\displaystyle \int x^{2}\operatorname {arccot} ax\,dx={\frac {x^{3}\operatorname {arccot} ax}{3}}-{\frac {\ln \left(a^{2}x^{2}+1\right)}{6\,a^{3}}}+{\frac {x^{2}}{6\,a}}+C} ∫ x m arccot a x d x = x m + 1 arccot a x m + 1 + a m + 1 ∫ x m + 1 a 2 x 2 + 1 d x ( m ≠ − 1 ) {\displaystyle \int x^{m}\operatorname {arccot} ax\,dx={\frac {x^{m+1}\operatorname {arccot} ax}{m+1}}+{\frac {a}{m+1}}\int {\frac {x^{m+1}}{a^{2}x^{2}+1}}\,dx\quad (m\neq -1)}
역정할 함수의 적분 ∫ arcsec x d x = x arcsec x − arctanh 1 − 1 x 2 + C {\displaystyle \int \operatorname {arcsec} x\,dx=x\operatorname {arcsec} x-\operatorname {arctanh} \,{\sqrt {1-{\frac {1}{x^{2}}}}}+C}
∫ arcsec a x d x = x arcsec a x − 1 a arctanh 1 − 1 a 2 x 2 + C {\displaystyle \int \operatorname {arcsec} ax\,dx=x\operatorname {arcsec} ax-{\frac {1}{a}}\,\operatorname {arctanh} \,{\sqrt {1-{\frac {1}{a^{2}x^{2}}}}}+C} ∫ x arcsec a x d x = x 2 arcsec a x 2 − x 2 a 1 − 1 a 2 x 2 + C {\displaystyle \int x\operatorname {arcsec} ax\,dx={\frac {x^{2}\operatorname {arcsec} ax}{2}}-{\frac {x}{2\,a}}{\sqrt {1-{\frac {1}{a^{2}x^{2}}}}}+C} ∫ x 2 arcsec a x d x = x 3 arcsec a x 3 − 1 6 a 3 arctanh 1 − 1 a 2 x 2 − x 2 6 a 1 − 1 a 2 x 2 + C {\displaystyle \int x^{2}\operatorname {arcsec} ax\,dx={\frac {x^{3}\operatorname {arcsec} ax}{3}}\,-\,{\frac {1}{6\,a^{3}}}\,\operatorname {arctanh} \,{\sqrt {1-{\frac {1}{a^{2}x^{2}}}}}\,-\,{\frac {x^{2}}{6\,a}}{\sqrt {1-{\frac {1}{a^{2}x^{2}}}}}\,+\,C} ∫ x m arcsec a x d x = x m + 1 arcsec a x m + 1 − 1 a ( m + 1 ) ∫ x m − 1 1 − 1 a 2 x 2 d x ( m ≠ − 1 ) {\displaystyle \int x^{m}\operatorname {arcsec} ax\,dx={\frac {x^{m+1}\operatorname {arcsec} ax}{m+1}}\,-\,{\frac {1}{a\,(m+1)}}\int {\frac {x^{m-1}}{\sqrt {1-{\frac {1}{a^{2}x^{2}}}}}}\,dx\quad (m\neq -1)}
역코시컨트 함수의 적분 ∫ arccsc x d x = x arccsc x + ln | x + x 2 − 1 | + C = x arccsc x + arccosh ( x ) + C {\displaystyle \int \operatorname {arccsc} x\,dx=x\operatorname {arccsc} x\,+\,\ln \left|x+{\sqrt {x^{2}-1}}\right|\,+\,C=x\operatorname {arccsc} x\,+\,\operatorname {arccosh} (x)\,+\,C}
∫ arccsc a x d x = x arccsc a x + 1 a arctanh 1 − 1 a 2 x 2 + C {\displaystyle \int \operatorname {arccsc} ax\,dx=x\operatorname {arccsc} ax+{\frac {1}{a}}\,\operatorname {arctanh} \,{\sqrt {1-{\frac {1}{a^{2}x^{2}}}}}+C} ∫ x arccsc a x d x = x 2 arccsc a x 2 + x 2 a 1 − 1 a 2 x 2 + C {\displaystyle \int x\operatorname {arccsc} ax\,dx={\frac {x^{2}\operatorname {arccsc} ax}{2}}+{\frac {x}{2\,a}}{\sqrt {1-{\frac {1}{a^{2}x^{2}}}}}+C} ∫ x 2 arccsc a x d x = x 3 arccsc a x 3 + 1 6 a 3 arctanh 1 − 1 a 2 x 2 + x 2 6 a 1 − 1 a 2 x 2 + C {\displaystyle \int x^{2}\operatorname {arccsc} ax\,dx={\frac {x^{3}\operatorname {arccsc} ax}{3}}\,+\,{\frac {1}{6\,a^{3}}}\,\operatorname {arctanh} \,{\sqrt {1-{\frac {1}{a^{2}x^{2}}}}}\,+\,{\frac {x^{2}}{6\,a}}{\sqrt {1-{\frac {1}{a^{2}x^{2}}}}}\,+\,C} ∫ x m arccsc a x d x = x m + 1 arccsc a x m + 1 + 1 a ( m + 1 ) ∫ x m − 1 1 − 1 a 2 x 2 d x ( m ≠ − 1 ) {\displaystyle \int x^{m}\operatorname {arccsc} ax\,dx={\frac {x^{m+1}\operatorname {arccsc} ax}{m+1}}\,+\,{\frac {1}{a\,(m+1)}}\int {\frac {x^{m-1}}{\sqrt {1-{\frac {1}{a^{2}x^{2}}}}}}\,dx\quad (m\neq -1)} 
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