`int(dx)/(sin^2xcos^2x)`equals(A) `tanx+cotx+C` (B) `tanx-cotx+C`(C) `tanxcotx+C` (D) `tanx-cot2x+C`

Explanation

To solve the integral

\[ \int \frac{dx}{\sin^2 x \cos^2 x} \]

we can start by rewriting the integrand. We know that

\[ \sin^2 x + \cos^2 x = 1. \]

We can multiply the integrand by \(1\) in a clever way:

\[ 1 = \frac{\sin^2 x + \cos^2 x}{\sin^2 x + \cos^2 x}. \]

This allows us to express the integrand as:

\[ \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x}. \]

Now, we can split the integral:

\[ \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) dx = \int \frac{1}{\cos^2 x} \, dx + \int \frac{1}{\sin^2 x} \, dx. \]

The first integral can be simplified as:

\[ \int \frac{1}{\cos^2 x} \, dx = \int \sec^2 x \, dx = \tan x + C_1, \]

and the second integral can be simplified as:

\[ \int \frac{1}{\sin^2 x} \, dx = \int \csc^2 x \, dx = -\cot x + C_2. \]

Combining these results, we have:

\[ \int \frac{dx}{\sin^2 x \cos^2 x} = \tan x - \cot x + C, \]

where \(C = C_1 + C_2\) is the constant of integration.

Now, we can compare our result with the given options:

- (A) \( \tan x + \cot x + C \)
- (B) \( \tan x - \cot x + C \)
- (C) \( \tan x \cot x + C \)
- (D) \( \tan x - \cot^2 x + C \)

The correct answer is:

\[ \text{(B) } \tan x - \cot x + C. \]