Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/97

ou bien, en substituant pour $a$ et $b$ les valeurs du § IV,

${\begin{aligned}&{\frac {l{\sqrt {p}}}{\mathrm {K} }}=\mathrm {A} ,\\\\&{\frac {r{\sqrt {p}}}{\mathrm {K} }}\cos q=\mathrm {B} ,\\\\&{\frac {r{\sqrt {p}}}{2\mathrm {K} }}\sin q={\sqrt {\cos(q+\alpha -m)-\cos(q+\alpha )}}.\end{aligned}}$

Maintenant, puisque l’on a

${\begin{aligned}\mathrm {A} =&\int {\frac {d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}},\\\mathrm {B} =&\int {\frac {\cos(n+\varphi )d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}},\end{aligned}}$

si l’on fait varier dans ces expressions tant $\varphi$ que $n,$ on aura

${\begin{aligned}d\mathrm {A} =&{\frac {d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}}+{\frac {\sin ndn}{2}}\int {\frac {d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}\\&-{\frac {dn}{2}}\int {\frac {\sin(n+\varphi )d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}},\\d\mathrm {B} =&{\frac {\cos(n+\varphi )d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}}+{\frac {\sin ndn}{2}}\int {\frac {\cos(n+\varphi )d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}\\&-dn\int {\frac {\sin(n+\varphi )d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}}-{\frac {dn}{2}}\int {\frac {\sin(n+\varphi )\cos(n+\varphi )d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}.\end{aligned}}$

Or

${\begin{aligned}{\frac {1}{2}}&\int {\frac {\sin(n+\varphi )d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}={\frac {-1}{\sqrt {\cos n-\cos(n+\varphi )}}},\\\\&\int {\frac {\sin(n+\varphi )d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}}-{\frac {1}{2}}\int {\frac {\sin(n+\varphi )\cos(n+\varphi )d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}\\\\&=-{\frac {\cos(n+\varphi )}{\sqrt {\cos n-\cos(n+\varphi )}}}.\end{aligned}}$