Zurück zu Bestimmte Integrale
Für 0 ≤ x < π {\displaystyle 0\leq x<\pi } ist tan ( 1 2 x ) = 1 − cos x 1 + cos x {\displaystyle \tan \left({\frac {1}{2}}\,x\right)={\sqrt {\frac {1-\cos x}{1+\cos x}}}} . ⇒ tan ( 1 2 ⋅ arccos ( 1 1 + 2 cos x ) ) = 1 − 1 1 + 2 cos x 1 + 1 1 + 2 cos x = 2 cos x 2 + 2 cos x = cos x 1 + cos x {\displaystyle \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}}}} ⇒ arccos ( 1 1 + 2 cos x ) = 2 arctan cos x 1 + cos x {\displaystyle \Rightarrow \,{\text{arccos}}\left({\frac {1}{1+2\cos x}}\right)=2\arctan {\sqrt {\frac {\cos x}{1+\cos x}}}} . Also ist I := ∫ 0 π 3 arccos ( 1 1 + 2 cos x ) d x = 2 ∫ 0 π 3 arctan cos x 1 + cos x d x {\displaystyle 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 arcsin u 2 2 ⋅ 1 1 + u 2 ( u = 2 sin x 2 cos x ) {\displaystyle 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 ∫ 0 1 arctan 1 2 + u 2 2 ( 1 + u 2 ) 2 + u 2 d u {\displaystyle 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 ∫ 0 1 2 a 2 a 2 + x 2 arctan 1 2 a 2 + x 2 2 a 2 + x 2 d x = ( arctan 1 a ) 2 {\displaystyle \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 {\displaystyle a=1\,} den Wert ( π 4 ) 2 {\displaystyle \left({\frac {\pi }{4}}\right)^{2}} . Also ist I = 2 ⋅ ( π 4 ) 2 = π 2 8 {\displaystyle I=2\cdot \left({\frac {\pi }{4}}\right)^{2}={\frac {\pi ^{2}}{8}}} .
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