Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,cos,arccos)

Coxeter Integrale

0.1

\int _{0}^{\frac {\pi }{2}}\arccos \left({\frac {\cos x}{1+2\cos x}}\right)dx={\frac {5\pi ^{2}}{24}}

ohne Beweis

0.2

\int _{0}^{\frac {\pi }{3}}\arccos \left({\frac {\cos x}{1+2\cos x}}\right)dx={\frac {2\pi ^{2}}{15}}

ohne Beweis

0.3

\int _{0}^{\frac {\pi }{2}}\arccos \left({\frac {1}{1+2\cos x}}\right)dx={\frac {\pi ^{2}}{6}}

ohne Beweis

0.4

\int _{0}^{\frac {\pi }{3}}\arccos \left({\frac {1}{1+2\cos x}}\right)dx={\frac {\pi ^{2}}{8}}

Beweis

Für $0\leq x<\pi$ ist $\tan \left({\frac {1}{2}}\,x\right)={\sqrt {\frac {1-\cos x}{1+\cos x}}}$ .

$\Rightarrow \,\tan \left({\frac {1}{2}}\cdot {\text{arccos}}\left({\frac {1}{1+2\cos x}}\right)\right)={\sqrt {\frac {1-{\frac {1}{1+2\cos x}}}{1+{\frac {1}{1+2\cos x}}}}}={\sqrt {\frac {2\cos x}{2+2\cos x}}}={\sqrt {\frac {\cos x}{1+\cos x}}}$

$\Rightarrow \,{\text{arccos}}\left({\frac {1}{1+2\cos x}}\right)=2\arctan {\sqrt {\frac {\cos x}{1+\cos x}}}$ .

Also ist $I:=\int _{0}^{\frac {\pi }{3}}\arccos \left({\frac {1}{1+2\cos x}}\right)dx=2\int _{0}^{\frac {\pi }{3}}\arctan {\sqrt {\frac {\cos x}{1+\cos x}}}\,dx$ .

Nach der Substitution $x=2\,{\text{arcsin}}{\sqrt {{\frac {u^{2}}{2}}\cdot {\frac {1}{1+u^{2}}}}}\quad \left(u={\frac {{\sqrt {2}}\,\sin {\frac {x}{2}}}{\sqrt {\cos x}}}\right)$

ist $I=2\int _{0}^{1}\arctan {\sqrt {\frac {1}{2+u^{2}}}}\,{\frac {2}{(1+u^{2})\,{\sqrt {2+u^{2}}}}}\,du$ .

Das Ahmed'sche Integral $\int _{0}^{1}{\frac {2a^{2}}{a^{2}+x^{2}}}\,{\frac {\arctan {\frac {1}{\sqrt {2a^{2}+x^{2}}}}}{\sqrt {2a^{2}+x^{2}}}}\,dx=\left(\arctan {\frac {1}{a}}\right)^{2}$ liefert im Fall $a=1\,$ den Wert $\left({\frac {\pi }{4}}\right)^{2}$ .

Also ist $I=2\cdot \left({\frac {\pi }{4}}\right)^{2}={\frac {\pi ^{2}}{8}}$ .