Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 9.djvu/91

en differentiant présentement l’équation

$\sin(\varpi '-\varpi )={\frac {2\sin p\cos p\sin q}{\sin \theta '}}$

par rapport à $\varpi '$ et $p,$ on aura

$\cos(\varpi '-\varpi )d\varpi '$

$={\frac {2\sin q\left(\cos ^{2}p-\sin ^{2}p\right)dp}{\sin \theta '}}-{\frac {2\sin p\cos p\sin q{\frac {\partial \theta '}{\partial p}}\cos \theta 'dp}{\sin ^{2}\theta '}}\,;$

partant,

${\frac {\partial \varpi '}{\partial p}}={\frac {2\left(\cos ^{2}p-\sin ^{2}p\right)\sin q}{\sin \theta '\cos(\varpi '-\varpi )}}+{\frac {8\sin ^{2}p\cos ^{2}p\sin ^{2}q\sin(\theta -q)\cos \theta '}{\sin ^{3}\theta '\cos(\varpi '-\varpi )}}\,;$

on aura donc

${\frac {\mathrm {B} }{2}}=\alpha \int _{0}^{\pi }\int _{0}^{\pi }$

$\left\{{\begin{aligned}{\frac {\partial \mu '}{\partial \theta '}}&{\frac {2\sin p\cos ^{2}p\cos q\sin(\theta -q)}{\sin \theta '}}\\-&{\frac {\partial \mu '}{\partial \varpi '}}\left[{\frac {\cos p\cos q\left(\cos ^{2}p-\sin ^{2}p\right)}{\sin \theta '\cos(\varpi '-\varpi )}}\right.\\&\left.\quad \qquad +{\frac {4\sin ^{2}p\cos ^{3}p\sin q\cos q\sin(\theta -q)\cos \theta '}{\sin ^{3}\theta '\cos(\varpi '-\varpi )}}\right]\end{aligned}}\right\}dpdq,$

d’où l’on tirera facilement

${\frac {\mathrm {B} }{2}}=\alpha \int _{0}^{\pi }\int _{0}^{\pi }$

$\left\{{\begin{aligned}{\frac {\partial \mu '}{\partial \theta '}}&{\frac {\sin p\cos ^{2}p\left[\sin \theta +\sin(\theta -q)\right]}{\sin \theta '}}\\-&{\frac {\partial \mu '}{\partial \varpi '}}\left[{\frac {\cos p\cos q\left(\cos ^{2}p-\sin ^{2}p\right)}{\sin \theta '\cos(\varpi '-\varpi )}}\right.\\&\left.\quad \qquad +{\frac {2\sin ^{2}p\cos ^{3}p\sin q\cos \theta '\left[\sin \theta +\sin(\theta -2q)\right]}{\sin ^{3}\theta '\cos(\varpi '-\varpi )}}\right]\end{aligned}}\right\}dpdq\,;$

si l’on différentie l’expression de $\mathrm {A} ,$ par rapport à $\theta ,$ on aura

${\frac {\partial \mathrm {A} }{\partial \theta }}=-{\frac {2}{3}}\alpha \pi {\frac {\partial \mu }{\partial \theta }}+\alpha \int _{0}^{\pi }\int _{0}^{\pi }\sin p\left({\frac {\partial \mu '}{\partial \theta '}}{\frac {\partial \theta '}{\partial \theta }}+{\frac {\partial \mu '}{\partial \varpi '}}{\frac {\partial \varpi '}{\partial \theta }}\right)dpdq.$

Or, en differentiant l’équation

$\cos \theta '=\cos \theta \cos ^{2}p+\sin ^{2}p\cos(\theta -2q)$

par rapport à $\theta '$ et $\theta ,$ on a

$\sin \theta 'd\theta '=\sin \theta \cos ^{2}pd\theta +\sin ^{2}p\sin(\theta -2q)d\theta ,$