Question

$\int \frac{dx}{\sin^{2}x \cos^{2}x} =$

• A. $\tan x + \cot x + c$
• B. $\tan x - \cot x + c$
• C. $\tan x \cot x + c$
• D. $\tan x + \sec x + c$

Hint:

When you see $\sin^2 x \cos^2 x$ in the denominator, adding "1" (as $\sin^2 + \cos^2$) in the numerator is the standard trick.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To integrate this trigonometric fraction, we can use the identity \( \sin^2 x + \cos^2 x = 1 \) to split the integral into two simpler parts.
Step 2: Key Formula or Approach:
1. Replace \( 1 \) in the numerator with \( \sin^2 x + \cos^2 x \).
2. Use standard integrals: \( \int \sec^2 x \, dx = \tan x \) and \( \int \csc^2 x \, dx = -\cot x \).
Step 3: Detailed Explanation:
\[ \int \frac{1}{\sin^2 x \cos^2 x} \, dx = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \, dx \] Split the fraction: \[ = \int \left( \frac{\sin^2 x}{\sin^2 x \cos^2 x} + \frac{\cos^2 x}{\sin^2 x \cos^2 x} \right) \, dx \] \[ = \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) \, dx \] \[ = \int (\sec^2 x + \csc^2 x) \, dx \] Integrating term by term: \[ = \tan x + (-\cot x) + c = \tan x - \cot x + c \]
Step 4: Final Answer:
The integral is \( \tan x - \cot x + c \).