Evaluate $\int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} \, dx$

$\tan x - \cot x - 3x + C$

The correct answer is Option (1) → $\tan x - \cot x - 3x + C$

Let $I = \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} \, dx = \int \frac{(\sin^2 x)^3 + (\cos^2 x)^3}{\sin^2 x \cos^2 x} \, dx$

$= \int \frac{1 - 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} \, dx \quad \left[ ∵\sin^2 x + \cos^2 x = 1 \right]$

$= \int \left[ \frac{(\sin^2 x + \cos^2 x) - 3 \sin^2 x \cos^2 x}{\sin^2 x \cdot \cos^2 x} \right] dx$

$= \int \frac{\sin^2 x}{\sin^2 x \cdot \cos^2 x} dx + \int \frac{\cos^2 x}{\sin^2 x \cdot \cos^2 x} dx - \int \frac{3 \sin^2 x \cos^2 x}{\sin^2 x \cdot \cos^2 x} dx$

$= \int \sec^2 x \, dx + \int \text{cosec}^2 x \, dx - \int 3 \, dx$

$= \int \sec^2 x \, dx + \int \text{cosec}^2 x \, dx - 3 \int dx$

$= \tan x - \cot x - 3x + C$