Question

The value of the integral \(\displaystyle\int\dfrac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}\,dx\) is

• A. \(-\tan x+\cot x+C\)
• B. \(\tan x+\sec x+C\)
• C. \(\tan x-\cot x+C\)
• D. \(\tan x+\cot x+C\)

Hint:

Split into sec²x − cosec²x, then integrate each term.

Answer 1

The Correct Option is D

Step 1: Differentiating each option instead:
Check option D by differentiating: \(\dfrac{d}{dx}(\tan x+\cot x)=\sec^2x-\text{cosec}^2x\).

Step 2: Rewriting in sin, cos:
\(\sec^2x-\text{cosec}^2x=\dfrac{1}{\cos^2x}-\dfrac{1}{\sin^2x}=\dfrac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}\), which is exactly the given integrand.

Step 3: Confirming:
Since differentiating option D reproduces the integrand exactly, it is the correct antiderivative.

Final Answer:
\[ \boxed{\tan x+\cot x+C} \]