\(\int\displaystyle\frac{dx}{\sin^2 x\cos^2 x} =\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\tan x - \cot x + C\)
Since \(\sin^2 x + \cos^2 x = 1\), the integrand splits into two parts:
\( \displaystyle \frac{1}{\sin^2 x\cos^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x\cos^2 x} \)
\( \displaystyle = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \)
\( = \sec^2 x + \mathrm{cosec}^2 x \)
Using \(\int\sec^2 x\,dx = \tan x\) and \(\int\mathrm{cosec}^2 x\,dx = -\cot x\):
\( \displaystyle \int\frac{dx}{\sin^2 x\cos^2 x} = \tan x - \cot x + C \)
NCERT (2026-27 edition):
Integration by Substitution
