Question Number 149625 by puissant last updated on 06/Aug/21

1)∫_0 ^(2π) (1/(a+sin(t)))dt , a>0 2)∫_(2π) ^(4π) (1/(2+sin(t)))dt..

Answered by ArielVyny last updated on 06/Aug/21

∫_0 ^(2π) (1/(a+sint))dt posons t=tan((x/2))→sint=((2t)/(1+t^2 )) 2dt=(1+t^2 )dx→dx=(2/(1+t^2 ))dt 2π=tan((x/2))→x=2arctg(2π) ∫_0 ^(2arctg(2π)) (1/(a+((2t)/(1+t^2 ))))×(2/(1+t^2 ))dt ∫_0 ^(2arctg(2π)) (2/(a(1+t^2 )+2t))dt=2∫_0 ^(2arctg(2π)) (1/(at^2 +2t+a)) 2∫_0 ^(2arctg(2π)) (1/(a[(t+(2/a))^2 −(4/a^2 )+(a^2 /a^2 )]))dt (2/a)∫_0 ^(2arctg(2π)) (1/([(t+(2/a))^2 +((a^2 −4)/a^2 )]))dt (2/a)∫_0 ^(2arctg(2π)) (1/((((a^2 −4)/a^2 ))[1+(a^2 /(a^2 −4))(t+(2/a))^2 ]))dt (2/a)×(a^2 /(a^2 −4))∫_0 ^(2arctg(2π)) (1/(1+(a^2 /(a^2 −4))(t+(2/a))^2 )) (2/a)∫_0 ^(2arctg(2π)) (a^2 /(a^2 −4))(1/(1+[(√((a^2 /(a^2 −4))(t+(2/a))))]^2 ))dt (2/a)arctg[((a^2 /(a^2 −4)))(2arctg(2π)+(2/a))] a>0

Commented bypuissant last updated on 06/Aug/21

Desolee mais moi je trouve ((2π)/( (√(a^2 −1)))) comment on a des resultats differents..?

Commented bypuissant last updated on 06/Aug/21

merci je pense que c′est mieux

Commented byArielVyny last updated on 06/Aug/21

peut-etre ma methode est un peu drastique je vais voir une autre approche

Commented byArielVyny last updated on 06/Aug/21

et je pense que a>1 pour que ta reponse soit valide

Commented bypuissant last updated on 06/Aug/21

oui