Question

1/{cos (x - a)cos(x - b)}.dx"
Integrate the function

Answer 1

HELLO DEAR,

GIVEN FUNCTION IS:- 1/{cos (x - a)cos(x - b)}.dx

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

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

$\sf{\Rightarrow \frac{1}{sin(a - b)}\int{\frac{sin(x - b)cos(x - a) - cos(x - b)sin(x - a)}{cos(x - a)cos(x - b)}}\,dx}$

$\sf{\Rightarrow \frac{1}{sin(a - b)}[\int{\tan(x - b) - tan(x - a)}\,dx]}$

$\sf{\Rightarrow \frac{1}{sin(a - b)}[-log|cos(x - b)| + log|cos(x - a)|]}$

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

I HOPE ITS HELP YOU DEAR,
THANKS