Question

integral 1 divided by cos(x-a)cos(x-b) ?

Answer 1

integration 1/cos(x -a).cos(x-b)

step1 :- sin{(x -b) -(x-a)} /sin(a-b) =1
put this in place of 1

e.g sin{(x-b) - (x -a)}/sin(a-b).cos(x-a).cos(x-b)

step2 :-
use sin(A-B) =sinA.cosB-cosA.sinB

e.g

{sin(x-b).cos(x-a) -sin(x-a).cos(x-b)}/sin(a-b).cos(x-a).cos(x-b)

1/sin(a-b) { tan(x-b) -tan(x-a) }

step3 :-
we know ,
according to basic integration ,
tan(x-a)dx =ln |sec(x-a) | +C
tab(x-b)dx =ln | sec(x-b)| + W
where C and W are constant.

put this
e.g.
1/sin(a -b) {ln |sec(x-b)| -ln |sec(x-a)| } +Q

=> 1/sin(a-b){ln(sec(x-b)/sec(x-a)}

we know ,
secx =1/cosx

so,
1/sin(a-b){ln.cos(x-a)/cos(x-b)}
( answer)

Answer 2

f(x) dx = dx / [cos (x-a) cos (x-b) ] 

Let (a+b)/2 = c,   (a-b)/2 = d

2 cos(x-a) cos(x-b) = cos[2x-(a+b)] + Cos(a-b) = Cos(2x-2c) + cos2d

Let    Tan(x-c) = z    => dx = dz/(1+z²),     cos2(x-c) = (1-z²)/(1+z²)

$f(x)dx=\frac{2dz}{(1+z^2)[ \frac{1-z^2}{1+z^2}+cos2d ]}=\frac{2dz}{(1+cos2d)-z^2(1-cos2d)}\\\\=\frac{dz}{cos^2d-z^2sin^2d}=\frac{cosec^2d\ dz}{cot^2d-z^2}\\\\\int f(x)dx=\frac{cosec^2d}{2cot d} Ln | \frac{cot d+z}{cot d-z} | +K\\\\=\frac{1}{sin2d}Ln| \frac{cosd\ cos(x-c)+sind\ sin(x-c)}{cosd\ cos(x-c)-sind\ sin(x-c)} | + K\\\\=cosec(a-b) Ln | \frac{cos(x-c-d)}{cos(x-c+d} |+K\\\\=cosec(a-b) Ln | \frac{cos(x-a)}{cos(x-b} |+K$

I = Cosec(a-b)  Ln | Cos(x-a) / Cos(x-b) | + K