Man berechne:
∫02π/3sin2x+11+x2dx=∫02π/3sin2x+∫02π/311+x2dx{\displaystyle \int _{0}^{2\pi /3}\sin ^{2}x+{\frac {1}{\sqrt {1+x^{2}}}}\,dx=\int _{0}^{2\pi /3}\sin ^{2}x+\int _{0}^{2\pi /3}{\frac {1}{\sqrt {1+x^{2}}}}\,dx}
Mal das erste Integral mittels partieller Integration
∫sin2xdx=−cosxsinx+∫cos2xdx{\displaystyle \int \sin ^{2}x\,dx=-\cos x\sin x+\int \cos ^{2}x\,dx}
sin2x+cos2=1{\displaystyle \sin ^{2}x+\cos ^{2}=1}
∫sin2xdx=−cosxsinx+∫1−sin2xdx{\displaystyle \int \sin ^{2}x\,dx=-\cos x\sin x+\int 1-\sin ^{2}x\,dx}
∫sin2xdx=−cosxsinx+x−∫sin2xdx{\displaystyle \int \sin ^{2}x\,dx=-\cos x\sin x+x-\int \sin ^{2}x\,dx}
2∫sin2xdx=−cosxsinx+x{\displaystyle 2\int \sin ^{2}x\,dx=-\cos x\sin x+x}
∫sin2xdx=−cosxsinx+x2{\displaystyle \int \sin ^{2}x\,dx={\frac {-\cos x\sin x+x}{2}}}
Dann nehmen wir uns den zweiten Teil vor.
Dazu gibt es die Formel
∫dxa2+x2=arsinhxa=ln|x2+(x2+a2)|{\displaystyle \int {\frac {dx}{\sqrt {a^{2}+x^{2}}}}=arsinh{\frac {x}{a}}=\ln \left|x^{2}+{\sqrt {(}}x^{2}+a^{2})\right|}
Somit können wir schreiben:
∫dx1+x2=arsinh(x)=ln|x2+(x2+1)|{\displaystyle \int {\frac {dx}{\sqrt {1+x^{2}}}}=arsinh(x)=\ln \left|x^{2}+{\sqrt {(}}x^{2}+1)\right|}
Die Lösung lautet
∫02π/3sin2x+11+x2dx=−cos2π3sin2π3+2π32+ln|2π32+(2π32+1)|+cos0sin02−ln|1|{\displaystyle \int _{0}^{2\pi /3}\sin ^{2}x+{\frac {1}{\sqrt {1+x^{2}}}}\,dx={\frac {-\cos {\frac {2\pi }{3}}\sin {\frac {2\pi }{3}}+{\frac {2\pi }{3}}}{2}}+\ln \left|{\frac {2\pi }{3}}^{2}+{\sqrt {(}}{\frac {2\pi }{3}}^{2}+1)\right|+{\frac {\cos 0\sin 0}{2}}-ln\left|1\right|}
Wenn ich das am TR nun für die Grenzen einsetze komme ich auf 3,167.
