Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/406

394

QUESTIONS

 Les corrections sont expliquées en page de discussion

$\operatorname {f} (a)=\int {\frac {\operatorname {d} a}{1+a^{2}}}\,;$

changeant donc, tour à tour, $a$ en $a\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right)$ et$a\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right),$ on aura, au moyen de notre théorème général, pour la somme de la suite proposée,

${\frac {1}{2}}\left\{\int {\frac {a\operatorname {d} .\left(\operatorname {Cos} .x+{\sqrt {-1}}\operatorname {Sin} .x\right)}{1+a^{2}\left(\operatorname {Cos} .2x+{\sqrt {-1}}\operatorname {Sin} .2x\right)}}+\int {\frac {a\operatorname {d} .\left(\operatorname {Cos} .x-{\sqrt {-1}}\operatorname {Sin} .x\right)}{1+a^{2}\left(\operatorname {Cos} .2x-{\sqrt {-1}}\operatorname {Sin} .2x\right)}}\right\},$

ou, en exécutant les différentiations,

${\frac {a}{2}}\int \operatorname {d} x\left\{{\frac {-\operatorname {Sin} .x+{\sqrt {-1}}\operatorname {Cos} .x}{\left(1+a^{2}\operatorname {Cos} .2x\right)+{\sqrt {-1}}a^{2}\operatorname {Sin} .2x}}-{\frac {\operatorname {Sin} .x+{\sqrt {-1}}\operatorname {Cos} .x}{\left(1+a^{2}\operatorname {Cos} .2x\right)-{\sqrt {-1}}a^{2}\operatorname {Sin} .2x}}\right\},$

ou bien

${\frac {1}{2}}\int -{\frac {2\left(1-a^{2}\right)a\operatorname {d} x\operatorname {Sin} .x}{\left(1-a^{2}\right)+4a^{2}\operatorname {Cos} .^{2}x}}={\frac {1}{2}}\int -{\frac {\frac {2a\operatorname {d} x\operatorname {Sin} .x}{1-a^{2}}}{1+\left({\frac {2a\operatorname {Cos} .x}{1-a^{2}}}\right)^{2}}}$

$={\frac {1}{2}}\operatorname {Arc} .\left(\operatorname {Tang} .={\frac {2a\operatorname {Cos} .x}{1-a^{2}}}\right),$

comme nous l’avions déjà trouvé.

2. Pour sommer la suite

${\frac {\operatorname {Cos} .x}{1}}+{\frac {1}{2}}.{\frac {\operatorname {Cos} .3x}{3}}+{\frac {1.3}{2.4}}.{\frac {\operatorname {Cos} .5x}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {\operatorname {Cos} .7x}{7}}+\ldots ,$

nous considérerons que

$\int {\frac {\operatorname {d} a}{\sqrt {1-a^{2}}}}={\frac {a}{1}}+{\frac {1}{2}}.{\frac {a^{3}}{3}}+{\frac {1.3}{2.4}}.{\frac {a^{5}}{5}}+{\frac {1.3.5}{2.4.6}}.{\frac {a^{7}}{7}}+\ldots$

d’où il suit que