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robim kango Grade:
        

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


 


 


and while registering on askiitians.com i am writing correct emailid robin.kango@yahoo.com but why it is returning a message thet this id is not valid plzzzzzzzzzzz tell me on my id bcoz i want to register on askiitians.com


 

7 years ago

Answers : (2)

SAGAR SINGH - IIT DELHI
879 Points
										

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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7 years ago
vikas askiitian expert
510 Points
										

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

7 years ago
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