Update cheatsheet.fr.md

12.temporary_ins/20.magnetostatics-vacuum/40.ampere-theorem-applications/30.cylindrical-current-distributions/10.rectilinear-current/10.ampere-integral/20.overview/cheatsheet.fr.md

@@ -198,8 +198,8 @@ Image à faire
 
 * Si **$`\mathbf{\overrightarrow{dl}=+\,\rho_M\,\overrightarrow{e_{\varphi}}}`$** :   
 <br>
-**$`\mathbf{\oint_{\Gamma_A}\overrightarrow{B}\cdot\overrightarrow{dl}}`$**
-$`\displaystyle\quad=\oint_{\varphi=0}^{\varphi=2\pi}\big(B_{\varphi}(\rho_M)\,\overrightarrow{e_{\varphi}}\big)\cdot \big(+\rho_M\,d\varphi\,\overrightarrow{e_{\varphi}}\big)   
+**$`\mathbf{\oint_{\Gamma_A}\overrightarrow{B}\cdot\overrightarrow{dl}}`$**   
+$`\quad=\oint_{\varphi=0}^{\varphi=2\pi}\big(B_{\varphi}(\rho_M)\,\overrightarrow{e_{\varphi}}\big)\cdot \big(+\rho_M\,d\varphi\,\overrightarrow{e_{\varphi}}\big)`$   
 $`\displaystyle\quad=\oint_{\varphi=0}^{\varphi=2\pi}\big(B_{\varphi}(\rho_M)\,\overrightarrow{e_{\varphi}}\big)\cdot \big(\rho_M\,d\varphi\,\overrightarrow{e_{\varphi}}\big)`$   
 $`\displaystyle\quad=\oint_{\varphi=0}^{\varphi=2\pi}B_{\varphi}(\rho_M)\,\rho_M\,\big(\overrightarrow{e_{\varphi}}\cdot \overrightarrow{e_{\varphi}}\big) d\varphi`$   
 $`\displaystyle\quad=\rho_M\,B_{\varphi}(\rho_M)\,\oint_{\varphi=0}^{\varphi=2\pi}d\varphi`$