Med samma substitution som i videon får vi\[\int\dfrac{dx}{2 - \cos x}=\Bigg[ \begin{array}{rl} x=&2\arctan t\\ t=&\tan \dfrac{x}{2}\\ dx=&\dfrac{2 dt}{1+t^2} \end{array} \Bigg] = \]\[ \int\dfrac{\dfrac{2 dt}{1+t^2}}{2 - \dfrac{1-t^2}{1+t^2}}= \int\dfrac{\dfrac{2 dt}{1+t^2}}{\dfrac{1+3t^2}{1+t^2}}= \int\dfrac{2 dt}{1+3t^2}= \]\[ \Bigg[ \begin{array}{rl} t=&\dfrac1{\sqrt3}s\\ s=&\sqrt3 t\\ dt=&\dfrac{1}{\sqrt3}ds \end{array} \Bigg] = \dfrac{2}{\sqrt3}\int\dfrac{ds}{1+s^2}= \]\[ \dfrac{2}{\sqrt3}\arctan s+C=\dfrac{2}{\sqrt3}\arctan \left(\sqrt3 \tan \dfrac{x}{2}\right)+C. \]