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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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  • $(i)\;\sin(A+B)=\sin A\cos B+\cos A\sin B.$
  • $(ii)\;\int\tan xdx=-log|\cos x|+c.$
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 $\sin(a-b)$
On expanding we get,
Now separating the terms we get,
On integrating we get,
$log a-log b=log\mid\frac{a}{b}\mid.$
In the same way we get,


answered Feb 3, 2013 by sreemathi.v
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