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Integrate the function\[\int\frac{1}{\cos(x+a)\cos(x+b)}\]

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  • (i)$\sin (A+B)=\sin A\cos B+\cos A\sin B$
  • (ii)$\cos (A+B)=\cos A\cos B-\sin A\sin B$
  • (iii)$\tan x dx=-log |\cos x|+c$
Given $\int\frac{dx}{\cos(x+a)\cos(x+b)}$
Multiply and divide by, $\sin (a-b)$
$I=\int\frac{\sin (a-b)\;dx}{\cos(x+a)\cos(x+b)(\sin (a-b))}$
Let $a-b=(x+a)-(x+b)$
$I=\large\frac{1}{\sin (a-b)}\int\frac{\sin[(x+a)-(x+b)]}{\cos(x+a)\cos(x+b)}dx$
But we know $\sin (A-B)=\sin A\cos B-\cos A\sin B$.
$\sin[(x+a)-(x+b)]$ is of the form $\sin(A-B),$ hence
$I=\large\frac{1}{\sin (a-b)}\int\frac{\sin(x+a)-\cos (x+b)-\cos (x+a)\sin (x+b)}{\cos(x+a)\cos(x+b)}dx$
Seperating the terms:
$=\large\frac{1}{\sin (a-b)}\int\frac{\sin(x+a)-\cos(x+b)}{\cos(x+a)\cos(x+b)}-\frac{\cos (x+a)\sin (x+b)}{\cos(x+a)\cos(x+b)}dx$
On simplifying we get,
$\large\frac{1}{\sin (a-b)}\bigg[\int\frac{\sin(x+a)}{\cos(x+a)}-{\sin (x+b)}{\cos(x+b)}dx\bigg]$
But $\large\frac{\sin(x+a)}{\cos(x+a)}\;and \;\frac{\sin (x+b)}{\cos(x+b)}=\tan{x+a}$
and $\tan (x+b)$ respectively
But $\frac{1}{\sin(a-b)}[-log|\cos(x+a)|+log|\cos(x+b)|+c]$
But we know $log a-log b=log \frac{a}{b}$
Therefore$ \int \frac{1}{\cos(x+a)\cos(x+b)}=\frac{1}{\sin(a-b)} log\bigg|\frac{\cos (x+b)}{\cos (x+a)}\bigg|+c$



answered Feb 17, 2013 by meena.p
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