# Find the integrals of the functions$\frac{1}{\cos(x-a)\cos(x-b)}$

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

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