CAN U PLZ INTEGRATE 1/COS(x-a)cos(x-b)

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Dear robim,

I= ∫ { 1 / [ cos ( x - a ). cos ( x - b ) ] } dx

I = csc(a-b) • ∫ { sin(a-b) / [ cos(x-a). cos(x-b) ] } dx


= csc(a-b) • ∫ { sin [ ( x-b) - (x-a) ] / [ cos(x-a).cos(x-b) ] } dx


= csc(a-b) • ∫ { [ sin(x-b).cos(x-a) - cos(x-b).sin(x-a) ] / [ cos(x-a).cos(x-b) ] } dx

..= csc(a-b) • ∫ [ tan(x-b) - tan(x-a) ] dx

..= csc(a-b) • { ln | sec(x-b) | - ln | sec(x-a) | } + C

..= csc(a-b) • ln | sec(x-b) / sec(x-a) | + C

..= csc(a-b) • ln | cos(x-a) / cos(x-b) | + C

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multiply and divide expression by sin(b-a)

 integration of sin(b-a)/sin(b-a)cos(x-a).cos(x-b)

 sin(b-a)=sin{(x-a)-(x-b)}=sinx-a .cosx-b  - cosx-a .sinx-b

  now expression becomes {sinx-a .cosx-b - sinx-b.cosx-a}/sin(b-a).cosx-a .cosx-b

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

   =1/sin(b-a). {logsec(x-a)-logsec(x-b)} +c

  =1/sin(b-a). {logsec(x-a)/sec(x-b)} +c