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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Choose the correct answer in $\Large \int \normalsize\frac{\large \sin^2x-\cos^2x}{\large \sin^2x\cos^2x}dx$ is equal to

$(a)\;tan x+cot x +C\qquad(b)\;tan x+cosec x+C\qquad(c)\;-tan x +cot x+C\qquad(d)\;tanx+sec x+C$

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • $(i)\;\int \sec^2xdx=\tan x+c.$
  • $(ii)\;\int cosec^2xdx=-\cot x+c.$
Given $I=\int\frac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}dx.$
 
On separating the terms we get,
 
$I=\int\frac{\sin^2x}{\sin^2x\cos^2x}dx-\int\frac{\cos^2x}{\sin^2x\cos^2x}dx.$
 
Cancelling the common terms we get,
 
$\;\;\;=\int\frac{1}{\cos^2x}dx-\int\frac{1}{\sin^2x}dx.$
 
But $\frac{1}{\cos^2x}=\sec^2x$ and $\frac{1}{\sin^2x}=cosec^2xdx.
 
$\qquad=\int\sec^2xdx-\int cosec ^2xdx.$
 
On integrating,
 
$\;\;\;=\tan x+\cot x+c.$
 
Hence A is the correct answer.

 

answered Feb 3, 2013 by sreemathi.v
 
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