# Choose the correct answer in $\Large \int \normalsize\frac{\large \sin^2x-\cos^2x}{\large \sin^2x\cos^2x}dx$ is equal to

$(a)\;tan x+cot x +C\qquad(b)\;tan x+cosec x+C\qquad(c)\;-tan x +cot x+C\qquad(d)\;tanx+sec x+C$

Toolbox:
• $(i)\;\int \sec^2xdx=\tan x+c.$
• $(ii)\;\int cosec^2xdx=-\cot x+c.$
Given $I=\int\frac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}dx.$

On separating the terms we get,

$I=\int\frac{\sin^2x}{\sin^2x\cos^2x}dx-\int\frac{\cos^2x}{\sin^2x\cos^2x}dx.$

Cancelling the common terms we get,

$\;\;\;=\int\frac{1}{\cos^2x}dx-\int\frac{1}{\sin^2x}dx.$