Donc on aura
![{\displaystyle {\begin{aligned}d\mathrm {A} =&{\frac {dn+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}}}},\\d\mathrm {B} =&{\frac {\cos(n+\varphi )(dn+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}}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ac5cebce2d5dda4187952265400d60326c6e444)
Supposons, pour plus de simplicité,
![{\displaystyle \mathrm {F} =\int {\frac {d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}},\quad \mathrm {G} =\int {\frac {\cos(n+\varphi )d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ae31425040c7785a8c4aa1f2c687c087d591755)
et comme on ne peut pas trouver les valeurs de 
et 
par l’intégration, il faut tâcher de les déterminer par le moyen des quantités 
Pour cela, je remarque que l’on a
1o ![{\displaystyle {\frac {d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}}={\frac {\cos nd\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}-{\frac {\cos(n+\varphi )d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/509da4dec9897fd10cf7c47b4dfda53ef2d83b50)
d’où, en intégrant, on aura
2o ![{\displaystyle {\frac {\cos(n+\varphi )d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}}={\frac {\cos n\cos(n+\varphi )d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}-{\frac {\cos ^{2}(n+\varphi )d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc6b123c837e8e407f08b81dadce13df5e884440)
et
![{\displaystyle d{\frac {\sin(n+\varphi )}{\sqrt {\cos n-\cos(n+\varphi )}}}={\frac {\cos(n+\varphi )d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}}-{\frac {\sin ^{2}(n+\varphi )d\varphi }{2\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94197481c872800d98a0fd88d0273cbb82a82c24)
par conséquent
![{\displaystyle {\begin{aligned}2d{\frac {\sin(n+\varphi )}{\sqrt {\cos n-\cos(n+\varphi )}}}&={\frac {\cos n\cos(n+\varphi )d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}\\+&{\frac {\cos(n+\varphi )d\varphi }{\sqrt {\cos n-\cos(n+\varphi )}}}-{\frac {d\varphi }{\left[\cos n-\cos(n+\varphi )\right]^{\frac {3}{2}}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f2eee0d90e3d011d92771e6638c76492075fc1a)
Donc, en intégrant, on aura
