The function $\text{⨍ (θ)=}\frac{\mathrm{d}}{\mathrm{d\theta }}\underset{0}{\overset{\mathrm{\theta }}{\int }}\frac{\mathrm{dx}}{1-\mathrm{cos\theta } \mathrm{cosx}}$ satisfies the differential equation

$\begin{array}{l}f\left(\theta \right)=\frac{d}{d\theta }\underset{0}{\overset{\theta }{\int }}\frac{dx}{1-\cos \theta \cos x}\\ =\frac{1}{1-\cos \theta \cos \theta }\\ =\frac{1}{1-\cos \theta }\\ =\frac{1}{{\sin }^2\theta }\\ f\left(\theta \right)=\cos e{c}^2\theta \\ \Rightarrow \frac{d}{d\theta }\left(f\left(\theta \right)\right)=-2\cos e{c}^2\theta .\cot \theta \\ \Rightarrow \frac{d}{d\theta }\left(f\left(\theta \right)\right)+2f\left(\theta \right)\cot \theta =0\end{array}$