Q.$\int \frac{dx}{\sin^2 x \cos^2 x}$ equals (A) $\tan x + \cot x + C$ (B) $\tan x - \cot x + C$ (C) $\tan x \cot x + C$ (D) $\tan x - \cot 2x + C$

Concept: Sine Double Angle Integration — rewrite the integrand using $\sin x \cos x = \frac{1}{2}\sin 2x$.

Step 1:

$\sin^2 x \cos^2 x = (\sin x \cos x)^2 = \left(\frac{1}{2}\sin 2x\right)^2 = \frac{1}{4}\sin^2 2x$.

Step 2:

So the integral becomes

Step 3:

Recall $\int \csc^2 u \, du = -\cot u + C$. Let $u = 2x$, $du = 2\,dx$:

Step 4:

Use $\cot 2x = \frac{\cot^2 x - 1}{2\cot x}$ or the identity $\cot 2x = \frac{1 - \tan^2 x}{2\tan x}$. A simpler approach: …

The key insight here is that the denominator $\sin^2 x \cos^2 x$ is a perfect square of a product — and that product is exactly $\frac{1}{2} \sin 2x$. This is the classic sine double-angle relationship: $\sin 2x = 2 \sin x \cos x$. Squaring it gives us a clean way to rewrite the integrand in terms of a single trigonometric function, making the integral straightforward.

Let’s walk through it.

1. Rewrite the denominator using the double-angle identity. We know $\sin 2x = 2 \sin x \cos x$, so squaring both sides:

Therefore,

This is the central simplification.

2. Substitute into the integral.

And $\frac{1}{\sin^2 2x} = \csc^2 2x$, so:

3. Integrate $\csc^2 2x$. Recall that $\int \csc^2 u \, du = -\cot u + C$. Here $u = 2x$, so $du = 2\,dx$, meaning $dx = \frac{du}{2}$.

4. Now check which option matches. The options are given in terms of $\tan x$ and $\cot x$, not $\cot 2x$. So we need to rewrite $-2 \cot 2x$ using the double-angle formula for cotangent:

But a more direct path: recall $\cot 2x = \frac{\cot^2 x - 1}{2 \cot x}$. However, the simplest is to use the identity:

$\cot 2x = \frac{\cos 2x}{\sin 2x} = \frac{\cos^2 x - \sin^2 x}{2 \sin x \cos x}$ …

Method: Use $\sin^2 x+\cos^2 x=1$ to split a single trig fraction into standard pieces

When the denominator is a product like $\sin^2 x\cos^2 x$ and the numerator is just $1$ (or a constant), replace that $1$ with $\sin^2 x+\cos^2 x$ so the fraction breaks into terms you can integrate directly.

Steps

Step 1: Rewrite the constant numerator using the Pythagorean identity.

so $\dfrac{1}{\sin^2 x\cos^2 x}=\dfrac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x}$.

Step 2: Split the fraction term by term.

$\frac{\sin^2 x}{\sin^2 x\cos^2 x}+\frac{\cos^2 x}{\sin^2 x\cos^2 x}=\sec^2 x+\csc^2 x$ …

Common Mistakes

Mistake 1: Getting the sign wrong on $\displaystyle\int\csc^2 x\,dx$.

Why it's wrong: $\dfrac{d}{dx}\cot x=-\csc^2 x$, so $\int\csc^2 x\,dx=-\cot x$; writing $+\cot x$ flips the sign and lands on the wrong option. Correct approach: $\tan x-\cot x+C$ (option B), since $\int\sec^2 x=\tan x$ and $\int\csc^2 x=-\cot x$.

Mistake 2: Not thinking to rewrite $1$ as $\sin^2 x+\cos^2 x$. …