Question

Find :

$\int \frac{1}{\cos (x-a)\cos (x-b)}dx$

Answer 1

$\int \frac{1}{\cos (x-a)\cos (x-b)}dx$

$=\frac{1}{sin(a-b)}\int \frac{sin(a-b)}{cos(x-a).cos(x-b)}$ [$\because$ multiply & divide by $sin(a-b)$]

$=\frac{1}{sin(a-b)}\int \frac{sin(a-b+x-x)}{cos(x-a).cos(x-b)}dx$

$=\frac{1}{sin(a-b)}\int \frac{sin(x-b)+(a-x)}{cos(x-a).cos(x-b)}dx$

$=\frac{1}{sin(a-b)}\int \frac{sin(x-b)-(x-a)}{cos(x-a).cos(x-b)}dx$

$=\frac{1}{sin(a-b)}\int \frac{sin(x-b).(x-a)-cos(x-b).sin(x-a)}{cos(x-a).cos(x-b)}dx$

$=\frac{1}{sin(a-b)}\int [\frac{sin(x-b)}{cos(x-b)}-\frac{sin(x-a)}{cos(x-a)}dx]$

$=\frac{1}{sin(a-b)}\int \left(tan\right(x-b)-tan(x-a\left)\right)dx$

$=\frac{1}{sin(a-b)}[-log|cos(x-b)-(-log|cos(x-a)|\left)\right]+c$

$=\frac{1}{sin(a-b)}[-log|cos(x-b)+log|cos(x-a\left)|\right]+c$

$=\frac{1}{sin(a-b)}\left[log|\frac{cos(x-a)}{cos(x-b)}|\right]+c$