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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the rational functions\[\frac{\cos x}{(1-\sin x)(2-\sin x)}\qquad[Hint:Put\;\sin x=t]\]

$\begin{array}{1 1} \log\large \frac{|1-sin x|}{|2-sin x|}+c \\\log\large \frac{|2-sin x|}{|1-sin x|}+c \\\log\large \frac{|1+sin x|}{|2-sin x|}+c \\ \log\large \frac{|1-sin x|}{|2+sin x|}+c \end{array} $

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1 Answer

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Toolbox:
  • $(i)\;$Form of the ration function\[\frac{px+q}{(x-a)(x-b)}\]
  • $\;$Form of the partial function\[\frac{A}{(x-a)}+\frac{B}{(x-b)}\]
  • $(ii)\;\int\frac{dx}{(x-a)}=log|x-a|+c.$
Given:$I=\int\frac{\cos x}{(1-\sin x)(2-\sin x)}dx.$
 
Put sinx=t.
 
Therefore $\cos xdx=dt.$
 
$\int\frac{\cos x}{(1-\sin x)(2-\sin x)}=\int\frac{dt}{(1-t)(2-t)}$
 
Resolving into partial fraction
 
$\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}$
 
$\Rightarrow 1=A(2-t)+B(1-t).$
 
Equating the coefficient of t we get,
 
0=-A-B-----(1)
 
1=2A+B-----(2)
 
Equating the constant term,
 
1=-A-B------(3)
 
Add equ(3) with equ(2)
2A+B=0
-A-B=1
_____________
A=1
 
Therefore 2A+B=0$\Rightarrow B=-2.$
 
$I=\int\frac{-2}{(1-t)}+\int\frac{dt}{2-t}.$
 
$\;\;\;=-2\int\frac{dt}{(1-t)}+\int\frac{dt}{(2-t)}.$
 
On integrating we get,
 
$\;\;\;=-2log|1-t|+log|2-t|.$

substituting for t we get,

$\;\;\;=-2log|1-\sin x|+log|2-\sin x|.$

answered Feb 7, 2013 by sreemathi.v
edited Feb 7, 2013 by sreemathi.v
 
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