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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x}\]

$\begin{array}{1 1}sec x-cosec x+c \\ sec x+cosec x+c \\2sec x+c \\-2cosec x+c \end{array}$

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1 Answer

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Toolbox:
  • $(i)\;\int\sec x\tan xdx=\sec x+c.$
  • $(ii)\;\int \cot xcosec xdx=-cosec x+c$
Given $I=\int\frac{\sin^3x+\cos ^3x}{\sin^2x\cos^2x}dx.$
 
Separating the terms we get,
 
$I=\int\frac{\sin^3x}{\sin^2x\cos^2x}dx+\int\frac{\cos^3x}{\sin^2x\cos^2x}dx.$
 
$\;\;\;=\int\frac{\sin x}{\cos x}\frac{1}{\cos x}dx+\frac{\cot x}{\sin x}\frac{1}{\sin x}dx.$
 
$\;\;\;=\int\tan x\sec xdx+\int\cot xcosec xdx.$
 
On integrating we get,
 
$\;\;\;=\sec x-cosec x+c.$
 
$\int\frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x}dx=\sec x-cosec x+c.$

 

answered Feb 3, 2013 by sreemathi.v
 
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