allery$\int \dfrac{dx}{\sin^2 x\cos^2 x}$ equals
A$\tan x+\cot x+C$
B$\tan x\cot x+C$
C$\tan x-\cot x+C$
D$\tan x-\cot 2x+C$
Answer & Solution
Correct answer: C. $\tan x-\cot x+C$
1. Write $\dfrac{1}{\sin^2 x\cos^2 x}=\dfrac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x}$.
2. Split into $\dfrac{1}{\cos^2 x}+\dfrac{1}{\sin^2 x}=\sec^2 x+\operatorname{cosec}^2 x$.
3. $\int \sec^2 x\,dx=\tan x$ and $\int \operatorname{cosec}^2 x\,dx=-\cot x$.
4. Sum: $\tan x-\cot x+C$. Option A wrongly adds $\cot x$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.304_
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