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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Find the integrals of the functions\[\frac{1}{\cos(x-a)\cos(x-b)}\]

$\begin{array}{1 1} \frac{log\mid \frac{\cos(x-a)}{\cos(x-b)}\mid}{\sin(a-b)}+c \\ \frac{log\mid \frac{\cos(x-a)}{\sin (x-b)}\mid}{\sin(a-b)}+c \\ \frac{log\mid \frac{\cos(x-a)}{\cos(x-b)}\mid}{\cos (a-b)}+c \\ \frac{log\mid \frac{\sin (x-a)}{\cos(x-b)}\mid}{\cos (a-b)}+c\end{array}$

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1 Answer

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Toolbox:
  • $(i)\;\sin(A+B)=\sin A\cos B+\cos A\sin B.$
  • $(ii)\;\int\tan xdx=-log|\cos x|+c.$
Given:$I=\int\frac{1}{\cos (x-a)\cos(x-b)}dx.$
 
Multiply and divide by $\sin(a-b)$
 
$I=\frac{1}{\sin(a-b)}\int\frac{\sin(a-b)}{\cos (x-a)\cos(x-b)}dx.$
 
Add and subtract x in $\sin(a-b)$
 
$I=\frac{1}{\sin(a-b)}\int\frac{\sin[(x-b)-(x-a)]}{\cos(x-a)\cos(x-b)}dx.$
 
On expanding we get,
 
$I=\frac{1}{\sin(a-b)}\int\frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)}dx.$
 
Now separating the terms we get,
 
$I=\frac{1}{\sin(a-b)}\int[\tan(x-b)-\tan(x-a)]dx.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{\sin(a-b)}[log|\cos(x-b)|+log|\cos(x-a)|+c.$
 
$log a-log b=log\mid\frac{a}{b}\mid.$
 
In the same way we get,
 
$\;\;\;=\frac{1}{\sin(a-b)}log\mid\frac{\cos(x-a)}{\cos(x-b)}\mid+c.$

 

answered Feb 3, 2013 by sreemathi.v
 
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