Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/407

395

RÉSOLUES.

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

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

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

ou encore

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

ou bien

$\int -{\frac {{\frac {1}{2}}\operatorname {d} x}{\sqrt {\operatorname {Sin} .x}}}.{\sqrt {1+\operatorname {Sin} .x}}.$

Or, soit ${\sqrt {\operatorname {Sin} .x}}=t$, on aura

${\frac {{\frac {1}{2}}\operatorname {d} x\operatorname {Cos} .x}{\sqrt {\operatorname {Sin} .x}}}=\operatorname {d} t,\qquad {\frac {{\frac {1}{2}}\operatorname {d} x}{\sqrt {\operatorname {Sin} .x}}}={\frac {\operatorname {d} t}{\sqrt {1-t^{4}}}}\,;$

donc

$\int -{\frac {{\frac {1}{2}}\operatorname {d} x}{\sqrt {\operatorname {Sin} .x}}}.{\sqrt {1+\operatorname {Sin} .x}}=\int -{\frac {\operatorname {d} t}{\sqrt {1-t^{4}}}}{\sqrt {1+t^{2}}}=\int -{\frac {\operatorname {d} t}{\sqrt {1-t^{2}}}}$

$=\operatorname {Arc} .(\operatorname {Cos} .=t)$

donc enfin la somme de la série proposée sera

$\operatorname {Arc} .\left(\operatorname {Cos} .={\sqrt {\operatorname {Sin} .x}}\right)={\frac {1}{2}}\operatorname {Arc} .(\operatorname {Sin} .=2\operatorname {Sin} .x\operatorname {Cos} .x-1)\,;$

comme nous l’avions également trouvé.