Question:
The value of the integral \(\displaystyle\int\dfrac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}\,dx\) is
The value of the integral \(\displaystyle\int\dfrac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}\,dx\) is
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Split into sec²x − cosec²x, then integrate each term.
Updated On: Sep 23, 2026
- \(-\tan x+\cot x+C\)
- \(\tan x+\sec x+C\)
- \(\tan x-\cot x+C\)
- \(\tan x+\cot x+C\)
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The Correct Option is D
Solution and Explanation
Step 1: Splitting the fraction:
\(\dfrac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}=\dfrac{\sin^2x}{\sin^2x\cos^2x}-\dfrac{\cos^2x}{\sin^2x\cos^2x}=\dfrac{1}{\cos^2x}-\dfrac{1}{\sin^2x}=\sec^2x-\text{cosec}^2x\).
Step 2: Integrating term by term:
\(\displaystyle\int\sec^2x\,dx=\tan x\) and \(\displaystyle\int\text{cosec}^2x\,dx=-\cot x\), so \(\displaystyle\int(\sec^2x-\text{cosec}^2x)\,dx=\tan x-(-\cot x)=\tan x+\cot x\).
Final Answer:
\[ \boxed{\tan x+\cot x+C} \]
\(\dfrac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}=\dfrac{\sin^2x}{\sin^2x\cos^2x}-\dfrac{\cos^2x}{\sin^2x\cos^2x}=\dfrac{1}{\cos^2x}-\dfrac{1}{\sin^2x}=\sec^2x-\text{cosec}^2x\).
Step 2: Integrating term by term:
\(\displaystyle\int\sec^2x\,dx=\tan x\) and \(\displaystyle\int\text{cosec}^2x\,dx=-\cot x\), so \(\displaystyle\int(\sec^2x-\text{cosec}^2x)\,dx=\tan x-(-\cot x)=\tan x+\cot x\).
Final Answer:
\[ \boxed{\tan x+\cot x+C} \]
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