∫ cos2x/ cos^2x.sin^2x dx =

Question

\(\int \frac{\cos 2 x}{\cos ^{2} x . \sin ^{2} x} d x=\)

(A) \(\cot x-\tan x+c\)

(B) \(\tan x-\cot x+c\)

(C) \(-\cot x-\tan x+c\)

(D) \(-\tan x+c\)

Answer 1

Correct option is (C) \(-\cot x-\tan x+c\)

\(\int \frac{\cos 2 x}{\cos ^{2} x . \sin ^{2} x} d x\)

\(= \int \frac{\cos^2x - \sin^2 x}{\cos^2 x.\sin ^2 x}dx\)

\(= \int \left(\frac 1{\sin^2 x} - \frac 1{\cos^2 x}\right)dx\)

\(= \int \text{cosec}^2x dx- \int \sec^2 x dx\)

\(= - \cot x - \tan x + c\)