Zum Inhalt springen

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

Zurück zu Bestimmte Integrale

0.1
${\displaystyle \int _{0}^{1}{\frac {\arcsin x}{x}}\,dx={\frac {\pi }{2}}\,\log 2}$
1. 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 ${\displaystyle \underbrace {{\Big [}x\log(\sin x){\Big ]}_{0}^{\frac {\pi }{2}}} _{=0}-\int _{0}^{\frac {\pi }{2}}\log(\sin x)\,dx={\frac {\pi }{2}}\,\log 2}$.

2. Beweis

Für den nun folgenden Beweis wird der Satz von Fubini verwendet:

${\displaystyle \int _{0}^{1}{\frac {1}{x}}\arcsin(x)\,\mathrm {d} x=\int _{0}^{1}\int _{0}^{1}{\frac {{\sqrt {1-x^{2}}}\,y}{(1-x^{2}y^{2})\,{\sqrt {1-y^{2}}}}}\,\mathrm {d} y\,\mathrm {d} x=}$

${\displaystyle =\int _{0}^{1}\int _{0}^{1}{\frac {{\sqrt {1-x^{2}}}\,y}{(1-x^{2}y^{2})\,{\sqrt {1-y^{2}}}}}\,\mathrm {d} x\,\mathrm {d} y=\int _{0}^{1}{\frac {\pi \,y}{2{\sqrt {1-y^{2}}}\,(1+{\sqrt {1-y^{2}}})}}\mathrm {d} y={\frac {\pi }{2}}\ln(2)}$

0.2
${\displaystyle \int _{0}^{1}\left({\frac {\arcsin x}{x}}\right)^{2}dx=4\,G-{\frac {\pi ^{2}}{4}}}$
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 ${\displaystyle \underbrace {\left[x^{2}\,{\frac {-1}{\sin x}}\right]_{0}^{\frac {\pi }{2}}} _{-{\frac {\pi ^{2}}{4}}}+2\underbrace {\int _{0}^{\frac {\pi }{2}}{\frac {x}{\sin x}}\,dx} _{2G}=4G-{\frac {\pi ^{2}}{4}}}$.

0.3
${\displaystyle \int _{0}^{1}\left({\frac {\arcsin x}{x}}\right)^{3}dx={\frac {3\pi }{2}}\log 2-{\frac {\pi ^{3}}{16}}}$
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 ${\displaystyle I=-{\frac {\pi ^{3}}{16}}+{\frac {3}{2}}\,\pi \,\log 2}$.

2.1
${\displaystyle \int _{0}^{1}{\frac {\arcsin {\sqrt {x}}}{1-\left(2\sin {\frac {\alpha }{2}}\right)^{2}\,x\,(1-x)}}\,dx={\frac {\pi }{4}}\,{\frac {\alpha }{\sin \alpha }}\qquad -\pi <\alpha <\pi }$
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 ${\displaystyle I={\frac {\pi }{4}}\int _{0}^{1}{\frac {1}{1-\left(2\sin {\frac {\alpha }{2}}\right)^{2}\,x\,(1-x)}}\,dx={\frac {\pi }{4}}\left[{\frac {1}{\sin \alpha }}\arctan \left((2x-1)\,\tan {\frac {\alpha }{2}}\right)\right]_{0}^{1}={\frac {\pi }{4}}\,{\frac {\alpha }{\sin \alpha }}}$.