Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,arcsin)

0.1Bearbeiten

{displaystyle int _{0}^{1}{frac {arcsin x}{x}}\,dx={frac {pi }{2}}\,log 2}

Beweis

{displaystyle int _{0}^{1}{frac {arcsin x}{x}}\,dx} ist nach der Substitution {displaystyle xmapsto sin x} gleich {displaystyle int _{0}^{frac {pi }{2}}{frac {x}{sin x}}\,cos x\,dx=int _{0}^{frac {pi }{2}}x\,cot x\,dx}.

Und das ist nach partieller Integration {displaystyle underbrace {{Big [}xlog(sin x){Big ]}_{0}^{frac {pi }{2}}} _{=0}-int _{0}^{frac {pi }{2}}log(sin x)\,dx={frac {pi }{2}}\,log 2}.

 

0.2Bearbeiten

{displaystyle int _{0}^{1}left({frac {arcsin x}{x}} ight)^{2}dx=4\,G-{frac {pi ^{2}}{4}}}

Beweis

{displaystyle int _{0}^{1}left({frac {arcsin x}{x}} ight)^{2}\,dx} ist nach der Substitution {displaystyle xmapsto sin x} gleich {displaystyle int _{0}^{frac {pi }{2}}{frac {x^{2}}{sin ^{2}x}}\,cos x\,dx}.

Und das ist nach partieller Integration {displaystyle underbrace {left[x^{2}\,{frac {-1}{sin x}} ight]_{0}^{frac {pi }{2}}} _{-{frac {pi ^{2}}{4}}}+2underbrace {int _{0}^{frac {pi }{2}}{frac {x}{sin x}}\,dx} _{2G}=4G-{frac {pi ^{2}}{4}}}.

 

0.3Bearbeiten

{displaystyle int _{0}^{1}left({frac {arcsin x}{x}} ight)^{3}dx={frac {3pi }{2}}log 2-{frac {pi ^{3}}{16}}}

Beweis

{displaystyle I:=int _{0}^{1}left({frac {arcsin x}{x}} ight)^{3}dx} ist nach Substitution {displaystyle xmapsto sin x} gleich {displaystyle int _{0}^{frac {pi }{2}}x^{3}\,{frac {cos x}{sin ^{3}x}}\,dx}.

Das ist nach partieller Integration {displaystyle left[x^{3}\,{frac {-1}{2sin ^{2}x}} ight]_{0}^{frac {pi }{2}}+int _{0}^{frac {pi }{2}}3x^{2}\,{frac {1}{2\,sin ^{2}x}}\,dx=-{frac {pi ^{3}}{16}}+{frac {3}{2}}int _{0}^{frac {pi }{2}}x^{2}\,{frac {1}{sin ^{2}x}}\,dx}.

Nach wiederholter partieller Integration ist dabei {displaystyle int _{0}^{frac {pi }{2}}x^{2}\,{frac {1}{sin ^{2}x}}\,dx=underbrace {left[-x^{2}\,cot x ight]_{0}^{frac {pi }{2}}} _{=0}+int _{0}^{frac {pi }{2}}2xcot x\,dx}

{displaystyle =underbrace {{Big [}2x\,log(sin x){Big ]}_{0}^{frac {pi }{2}}} _{=0}-2int _{0}^{frac {pi }{2}}log(sin x)\,dx=pi log 2}. Also ist {displaystyle I=-{frac {pi ^{3}}{16}}+{frac {3}{2}}\,pi \,log 2}.

 

2.1Bearbeiten

{displaystyle int _{0}^{1}{frac {arcsin {sqrt {x}}}{1-left(2sin {frac {alpha }{2}} ight)^{2}\,x\,(1-x)}}\,dx={frac {pi }{4}}\,{frac {alpha }{sin alpha }}qquad -pi <alpha <pi }

Beweis

Nach Substitution {displaystyle xmapsto 1-x} lässt sich das Integral auch schreiben als {displaystyle int _{0}^{1}{frac {arcsin {sqrt {1-x}}}{1-left(2sin {frac {alpha }{2}} ight)^{2}\,x\,(1-x)}}\,dx}.

Addiert man beide Darstellungen, so ist {displaystyle 2I=int _{0}^{1}{frac {arcsin {sqrt {x}}+arcsin {sqrt {1-x}}}{1-left(2sin {frac {alpha }{2}} ight)^{2}\,x\,(1-x)}}\,dx}. Der Zähler ist konstant {displaystyle {frac {pi }{2}}}.

Somit ist {displaystyle I={frac {pi }{4}}int _{0}^{1}{frac {1}{1-left(2sin {frac {alpha }{2}} ight)^{2}\,x\,(1-x)}}\,dx={frac {pi }{4}}left[{frac {1}{sin alpha }}arctan left((2x-1)\, an {frac {alpha }{2}} ight) ight]_{0}^{1}={frac {pi }{4}}\,{frac {alpha }{sin alpha }}}.