Page:Annales de mathématiques pures et appliquées, 1817-1818, Tome 8.djvu/87

79

DES QUADRATURES.

Je suppose ensuite que $s,$ fonction de $\theta ,$ est un arc de courbe plane, terminé par les coordonnées rectangulaires $x,y,$ et faisant, à son extrémité, avec l’axe des $x,$ l’angle $\theta$ ; hypothèse exprimée par les relations

$\operatorname {d} x=\operatorname {d} s\operatorname {Cos} .\theta \,;\qquad \operatorname {d} y=\operatorname {d} s\operatorname {Sin} .\theta .\qquad$(16)

Mais on a, comme l’on sait, d’abord

$s.\operatorname {Sin} .\theta +{\frac {1}{2}}\Delta (s.\operatorname {Sin} .\theta )=s.\operatorname {Sin} .\theta +\Delta (s.\operatorname {Sin} .\theta )-{\frac {1}{2}}(s.\operatorname {Sin} .\theta )$

$=(s+\Delta s)\operatorname {Sin} .(\theta +\omega )-{\frac {1}{2}}\Delta (s.\operatorname {Sin} .\theta )$

$=(s+\Delta s)\operatorname {Cos} .{\frac {1}{2}}\omega \operatorname {Sin} .\left(\theta +{\frac {1}{2}}\omega \right)$

$+(s+\Delta s)\operatorname {Sin} .{\frac {1}{2}}\omega \operatorname {Cos} .\left(\theta +{\frac {1}{2}}\omega \right)-{\frac {1}{2}}\Delta (s.\operatorname {Sin} .\theta )\,;$

ensuite, parce que

$\Delta (s.\operatorname {Sin} .\theta )=\Delta s.\operatorname {Sin} .\theta +2(s+\Delta s)\operatorname {Sin} .{\frac {1}{2}}\omega \operatorname {Cos} .\left(\theta +{\frac {1}{2}}\omega \right),$

$\Delta (s.\operatorname {Cos} .\theta )=\Delta s.\operatorname {Cos} .\theta -2(s+\Delta s)\operatorname {Sin} .{\frac {1}{2}}\omega \operatorname {Sin} .\left(\theta +{\frac {1}{2}}\omega \right),$

on obtient

$(s+\Delta s)\operatorname {Cos} .{\frac {1}{2}}\omega \operatorname {Sin} .\left(\theta +{\frac {1}{2}}\omega \right)=-{\frac {1}{2}}\operatorname {Cot} .{\frac {1}{2}}\omega \Delta (s.\operatorname {Cos} .\theta )$

$+{\frac {1}{2}}\operatorname {Cot} .{\frac {1}{2}}\omega \Delta s.\operatorname {Cos} .\theta ,$

$(s+\Delta s)\operatorname {Sin} .{\frac {1}{2}}\omega \operatorname {Cos} .\left(\theta +{\frac {1}{2}}\omega \right)={\frac {1}{2}}\Delta (s.\operatorname {Sin} .\theta )-{\frac {1}{2}}\Delta s.\operatorname {Sin} .\theta \,;$

donc enfin on aura

$s.\operatorname {Sin} .\theta +{\frac {1}{2}}\Delta (s.\operatorname {Sin} .\theta )=-{\frac {1}{2}}\operatorname {Cot} .{\frac {1}{2}}\omega \Delta (s.\operatorname {Cos} .\theta )$

$+{\frac {1}{2}}\operatorname {Cot} .{\frac {3}{2}}\omega \Delta s.\operatorname {Cos} .\theta -{\frac {1}{2}}\Delta s.\operatorname {Sin} .\theta$

$=-{\frac {1}{2}}\operatorname {Cot} .{\frac {1}{2}}\omega \Delta (s.\operatorname {Cos} .\theta )+{\frac {1}{2}}\Delta s.{\frac {\operatorname {Cos} .\left(\theta +{\frac {1}{2}}\omega \right)}{\operatorname {Sin} .{\frac {1}{2}}\omega }}.\quad$(17)

D’ailleurs, à cause de

$\operatorname {d} (s.\operatorname {Cos} .\theta )=\operatorname {d} s.\operatorname {Cos} .\theta +s\operatorname {d} \operatorname {Cos} .\theta =\operatorname {d} s\operatorname {Cos} .\theta -s.\operatorname {Sin} .\theta \operatorname {d} \theta ,$

on a (16)