Q.$\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$ is equal to (A) $\tan x + \cot x + C$ (B) $\tan x + \operatorname{cosec} x + C$ (C) $-\tan x + \cot x + C$ (D) $\tan x + \sec x + C$

Concept: Trigonometric Simplification — rewrite the numerator using $\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x$, but splitting term-by-term is faster.

Step 1: Separate the fraction:

Step 2: Integrate term by term: …

The problem looks like a trigonometric integral, but the real trick is noticing that the denominator is a product of squares. Many students try to use double-angle identities immediately, but the cleanest path is to split the fraction first.

When you have a sum (or difference) in the numerator and a product in the denominator, always check if you can break it into separate terms. Here:

Each fraction simplifies beautifully:

So the integrand becomes $\sec^2 x - \csc^2 x$.

Now integrate term by term:

1. $\int \sec^2 x \, dx = \tan x + C_1$ — this is a standard result, since the derivative of $\tan x$ is $\sec^2 x$.
2. $\int \csc^2 x \, dx = -\cot x + C_2$ — because the derivative of $\cot x$ is $-\csc^2 x$. …

Method: Split $\dfrac{\sin^2 x-\cos^2 x}{\sin^2 x\cos^2 x}$ into $\sec^2$ and $\csc^2$

When a difference of squared trig terms sits over their product, divide each numerator term by the whole denominator — each piece becomes a standard $\sec^2$ or $\csc^2$.

Steps

Step 1: Split the fraction.

Step 2: Cancel to standard forms.

Step 3: Integrate. …

Common Mistakes

Mistake 1: Sign confusion giving $-\tan x+\cot x$ (option C) instead of $\tan x+\cot x$.

Why it's wrong: the fraction splits as $\sec^2 x-\csc^2 x$; integrating gives $\tan x-(-\cot x)=\tan x+\cot x$, because $\int\csc^2 x=-\cot x$ and there is a leading minus. Correct approach: track both signs carefully to land on option (A).

Mistake 2: Not splitting the numerator and trying a direct substitution. …