∫(sin^6x+cos^6x)/(sin^2xcos^2x) dx is equal to

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

Let \(s = \sin^2x\) and \(c = \cos^2x\), so \(s+c = 1\). By the sum of cubes \(s^3+c^3 = (s+c)^3-3sc(s+c)\):

    \(\sin^6x+\cos^6x = 1-3\sin^2x\cos^2x\)

Dividing by \(\sin^2x\cos^2x\):

    \( \displaystyle \frac{\sin^6x+\cos^6x}{\sin^2x\cos^2x} = \frac{1}{\sin^2x\cos^2x}-3 \)

Writing \(1 = \sin^2x+\cos^2x\) in the first fraction:

    \( \displaystyle \frac{\sin^2x+\cos^2x}{\sin^2x\cos^2x} = \frac{1}{\cos^2x}+\frac{1}{\sin^2x} \)

    \( = \sec^2x+\mathrm{cosec}^2x \)

So the integrand is \(\sec^2x+\mathrm{cosec}^2x-3\). Integrating each term:

    \( \displaystyle \int\left(\sec^2x+\mathrm{cosec}^2x-3\right)dx = \tan x-\cot x-3x+C \)

Why the other options are incorrect

• Option (1) \(\tan x+\cot x-3x+C\) — Incorrect. \(\int\mathrm{cosec}^2x\,dx = -\cot x\), so the cot term is negative.

• Option (2) \(\tan x-\cot x+3x+C\) — Incorrect. The constant part of the integrand is \(-3\), so the x term is \(-3x\).

• Option (4) \(\sec x-\mathrm{cosec}\,x-3x+C\) — Incorrect. The integral of \(\sec^2x\) is \(\tan x\), not \(\sec x\); the derivative of \(\sec x\) is \(\sec x\tan x\).