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 x\mapsto \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 [}x\log(\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}}\right)^{2}dx=4\,G-{\frac {\pi ^{2}}{4}}}

Beweis

{\displaystyle \int _{0}^{1}\left({\frac {\arcsin x}{x}}\right)^{2}\,dx} ist nach der Substitution {\displaystyle x\mapsto \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}}\right]_{0}^{\frac {\pi }{2}}} _{-{\frac {\pi ^{2}}{4}}}+2\underbrace {\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}}\right)^{3}dx={\frac {3\pi }{2}}\log 2-{\frac {\pi ^{3}}{16}}}

Beweis

{\displaystyle I:=\int _{0}^{1}\left({\frac {\arcsin x}{x}}\right)^{3}dx} ist nach Substitution {\displaystyle x\mapsto \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}{2\sin ^{2}x}}\right]_{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\right]_{0}^{\frac {\pi }{2}}} _{=0}+\int _{0}^{\frac {\pi }{2}}2x\cot x\,dx}

{\displaystyle =\underbrace {{\Big [}2x\,\log(\sin x){\Big ]}_{0}^{\frac {\pi }{2}}} _{=0}-2\int _{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(2\sin {\frac {\alpha }{2}}\right)^{2}\,x\,(1-x)}}\,dx={\frac {\pi }{4}}\,{\frac {\alpha }{\sin \alpha }}\qquad -\pi <\alpha <\pi }

Beweis

Nach Substitution {\displaystyle x\mapsto 1-x} lässt sich das Integral auch schreiben als {\displaystyle \int _{0}^{1}{\frac {\arcsin {\sqrt {1-x}}}{1-\left(2\sin {\frac {\alpha }{2}}\right)^{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(2\sin {\frac {\alpha }{2}}\right)^{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(2\sin {\frac {\alpha }{2}}\right)^{2}\,x\,(1-x)}}\,dx={\frac {\pi }{4}}\left[{\frac {1}{\sin \alpha }}\arctan \left((2x-1)\,\tan {\frac {\alpha }{2}}\right)\right]_{0}^{1}={\frac {\pi }{4}}\,{\frac {\alpha }{\sin \alpha }}}.