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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\frac{\cos x-\sin x}{1+\sin2x}\]

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Toolbox:
  • $(i)\sin^2x+\cos^2x=1.$
  • $(ii)\sin2x=2\sin x\cos x.$
  • $(iii)method\;of\;substitution:$
  • If f(x)=t.Then f'(x)dx=dt.
  • Thus $I=\int f(x)dx=\int t.dt.$
Given $I=\int\frac{\cos x-\sin x}{1+\sin2x}dx.$
 
This can be written as,(since $\sin^2x+\cos^2x=1)$
 
$I=\int\frac{\cos x-\sin x}{\sin^2x+\cos^2x+2\sin x\cos x}dx.$
 
$\;\;\;=\int\frac{\cos x-\sin x}{(\sin x+\cos x)^2}dx.$
 
Put $\sin x+\cos x=t.$
 
On differentiating we get,
 
$(\cos x-\sin x)dx=dt.$
 
Now substituting t and dt we get,
 
$I=\int\frac{dt}{t^2}.$
 
On integrating we get,
 
$\;\;\;=\frac{t^{-1}}{-1}+c.$
 
$\;\;\;=\frac{-1}{t}+c.$
 
Substituting for t we get,
 
$\;\;\;=-\frac{1}{(\sin x+\cos x)}+c.$
 
$\;\;\;=\frac{-1}{(\sin x+\cos x)}+c.$
 
 
 
 
 
 

 

answered Jan 31, 2013 by sreemathi.v
 
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