$$\int \frac{dx}{\sin^{2}x \cdot \cos^{2}x}$$ equals:

Correct option is C

Given:
​$$I = \int \frac{dx}{\sin^{2}x \cos^{2}x}$$​​

Concept used:
​$$\frac{1}{\sin^{2}x \cos^{2}x} = \sec^{2}x + \csc^{2}x$$​​

(Proof $$: \tan x + \cot x = \frac{\sin^{2}x + \cos^{2}x}{\sin x \cos x} = \frac{1}{\sin x \cos x};$$ differentiate both sides to confirm relation.)
Solution:
Let $$t = \tan x - \cot x$$​​
Then, $$\frac{dt}{dx} = \sec^{2}x + \csc^{2}x = \frac{1}{\sin^{2}x \cos^{2}x}$$​​
Thus,
​$$I = \int \frac{dx}{\sin^{2}x \cos^{2}x} = \int dt = t + C \\= \tan x - \cot x + C$$​​
Correct answer is (c) $$\tan x - \cot x + C$$