Mais on a
soit donc 
et 
partant
et
![{\displaystyle {\begin{aligned}\Pi &\left[\cos \varphi +2\sin ^{2}p\sin q(\cos q\sin \varphi -\sin q\cos \varphi )\right]\\&=y+2\sin ^{2}p\sin q(\cos q\sin \varphi -\sin q\cos \varphi ){\frac {dy}{dx}}\\&\quad +4\sin ^{4}p\sin ^{2}q(\cos q\sin \varphi -\sin q\cos \varphi )^{2}{\frac {d^{2}y}{1.2.dx^{2}}}+\ldots \,;\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f546983bac872ac712b4a7bdcd6ec2ca2d2c33f)
de plus, on aura
donc
![{\displaystyle (\Lambda )\left\{{\begin{aligned}-{\frac {4}{3}}\pi {\frac {dy}{dx}}&+\int _{0}^{\pi }\int _{0}^{\pi }{\frac {1}{\sin \varphi }}\left[2\sin ^{3}p\cos q(\cos q\sin \varphi -\sin q\cos \varphi ){\frac {dy}{dx}}\right.\\&+4\sin ^{5}p\sin q\cos q(\cos q\sin \varphi -\sin q\cos \varphi )^{2}{\frac {d^{2}y}{1.2.dx^{2}}}\\&+8\sin ^{7}p\sin ^{2}q\cos q(\cos q\sin \varphi -\sin q\cos \varphi )^{3}{\frac {d^{3}y}{1.2.3.dx^{3}}}\\&+\left.\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots {\begin{aligned}\\\\\end{aligned}}\right]dpdq=hx.\end{aligned}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdf1c5a01b5a4c1d882177f00a3bec61b83117df)
Or il est aisé de voir, par ce qui précède, que l’on ne doit admettre dans le développement des puissances de 
que les termes dans lesquels 
se trouve élevé à une puissance impaire, d’où l’on tire
![{\displaystyle {\begin{aligned}hx=&-{\frac {4}{3}}\pi {\frac {dy}{dx}}\\&+\int _{0}^{\pi }\int _{0}^{\pi }\sin ^{3}p\cos q\\&\qquad \times \left\{2{\frac {dy}{dx}}\cos q+4{\frac {d^{2}y}{1.2.dx^{2}}}\sin ^{2}p\sin q(-2x\sin q\cos q)\right.\\&\qquad +8{\frac {d^{3}y}{1.2.3.dx^{3}}}\sin ^{4}p\sin ^{2}q\left[\left(1-x^{2}\right)\cos ^{3}q+3x^{2}\cos q\sin ^{2}q\right]\\&\qquad +\left.\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots {\begin{aligned}\\\\\end{aligned}}\right\}dpdq.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f23196b4337427a4f2e7719743e34ca3eb86197)
