Lassen
ICH= ∫Sünde3( θ / 2 )cos( θ / 2 )cos3θ +cos2θ + cosθ−−−−−−−−−−−−−−−−−√Dθ =12∫2Sünde2θ2⋅ 2 Sündeθ2⋅ cosθ22cos2θ2cos3θ +cos2θ + cosθ−−−−−−−−−−−−−−−−−√Dθ
Also bekommen wir
ICH=12∫( 1 − cosθ ) ⋅ Sündeθ( 1 + cosθ )cos3θ +cos2θ + cosθ−−−−−−−−−−−−−−−−−√Dθ
Jetzt setzencosθ = t,
DannSündeθ dθ = − dT
Also integral
ICH= −12∫( 1 - t )( 1 + t )T3+T2+ T−−−−−−−−√Dt = −12∫( 1 -T2)( 1 + T)2T3+T2+ T−−−−−−−−√DT
Also bekommen wir
ICH=12∫( 1 -1T2)( t +1T+ 2 )t +1T+ 1−−−−−−−−√DT
Nun lass( t +1T+ 1 ) =u2,
Dann( 1 -1T2) dt = 2 u du
Also integral
ICH=12∫2 uu2+ 1⋅1uDu =bräunen− 1( u ) + C
Also bekommen wir
ICH=bräunen− 1( t +1T+ 1 )−−−−−−−−−−−√+ C
Also bekommen wir
ICH=bräunen− 1( weilθ + sekθ + 1 )−−−−−−−−−−−−−−√+ C
MathematikPhysiker
Claude Leibovici
Abhishek Bakshi
Benutzer64494