Question

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

Answer 1

The correct answer is: $=\frac{1}{sin(a-b)}[log|\frac{cos(x-a)}{cos(x-b)}|]+C$
$\frac{1}{cos(x-a)cos(x-b)} = \frac{1}{sin(a-b)}[\frac{sin(a-b)}{cos(x-a)cos(x-b)}]$
$=\frac{1}{sin(a-b)}[\frac{sin[(x-b)-(x-a)]}{cos(x-a)cos(x-b)}]$
$=\frac{1}{sin(a-b)}\frac{[sin(x-b)cos(x-a)-cos(x-b)sin(x-a)]}{cos(x-a)cos(x-b)}$
$=\frac{1}{sin(a-b)}[tan(x-b)-tan(x-a)]$
$⇒ ∫\frac{1}{cos(x-a)cos(x-b)}dx = \frac{1}{sin(a-b)} ∫[tan(x-b)-tan(x-a)]dx$
$=\frac{1}{sin(a-b)}[-log|cos(x-b)|+log|cos(x-a)|]$
$=\frac{1}{sin(a-b)}[log|\frac{cos(x-a)}{cos(x-b)}|]+C$