Integrate : $\int \large\frac{\cos 2 x}{\sin ^2 x \cos ^2 x }$$dx$

Question

$(a)\;-\cot x -\tan x +c \\(b)\;\cos x +\tan x +c \\(c)\;\cos x - \tan x +c \\ (d)\;\sin x - cosec x +c $

Answer 1

$\int \large\frac{\cos ^2 x -\sin ^2 x }{\sin ^2 x \cos ^2 x }$$dx$

$=\int cosec ^2 x dx-\int \sec^2 x dx $

$=-\cot x -\tan x +c$

Hence a is the correct answer.