Evaluate: `int1/(cos(x-a)cos(x-b))\ dx`

Explanation

`int (1)/(cos(x-a).cos(x-b))dx`
`=(1)/(sin(a-b))int (sin (a-b))/(cos(x-a).cos(x-b))dx`
`=(1)/(sin(a-b)) int [(sin(x-b)-(x-a))/(cos(x-a).cos(x-b))]dx`
`=(1)/(sin(a-b)).[inttan(x-b)dx-inttan(x-a)dx]`
`=(1)/(sin(a-b)).[log|sec(x-b)|-log|sec(x-a)|]+c`
`=(1)/(sin(a-b)).log|(sec(x-b))/(sec(x-a))|+c`
`=(1)/(sin(a-b)).log|(cos(x-a))/(cos(x-b))|+c`
Remark: If the denominator contains one sine and one cosine, we adjust cos (a-b).