Question
The function $$f\left( \theta \right) = \frac{d}{{d\theta }}\int\limits_0^\theta {\frac{{dx}}{{1 - \cos \,\theta \,\cos \,x}}} $$ satisfies
the differential equation :
A.
$$\frac{{df}}{{d\theta }} + 2f\left( \theta \right)\cot \,\theta = 0$$
B.
$$\frac{{df}}{{d\theta }} - 2f\left( \theta \right)\cot \,\theta = 0$$
C.
$$\frac{{df}}{{d\theta }} + 2f\left( \theta \right) = 0$$
D.
$$\frac{{df}}{{d\theta }} - 2f\left( \theta \right) = 0$$
Answer :
$$\frac{{df}}{{d\theta }} + 2f\left( \theta \right)\cot \,\theta = 0$$
Solution :
$$\eqalign{
& {\text{We have,}}\,\,f\left( \theta \right) = \frac{d}{{d\theta }}\int\limits_0^\theta {\frac{{dx}}{{1 - \cos \,\theta \,\cos \,x}}} \cr
& \Rightarrow f\left( \theta \right) = \frac{1}{{1 - {{\cos }^2}\theta }} \cr
& \Rightarrow f\left( \theta \right) = {\text{cose}}{{\text{c}}^2}\theta \,\,\,\,\,\,\left( {{\text{using Leibnitz's Rule}}} \right) \cr
& \Rightarrow \frac{{df\left( \theta \right)}}{{d\theta }} = - 2\,{\text{cose}}{{\text{c}}^2}\theta \,\cot \,\theta \cr
& \Rightarrow \frac{{df\left( \theta \right)}}{{d\theta }} + 2\,f\left( \theta \right)\cot \,\theta = 0 \cr} $$