 Hello and how are you all today? Let us integrate the following function with respect to dx. We have 1 over cos x minus a cos x minus b into dx. Now multiplying the numerator and denominator by sin a minus b we have the integral sin a minus b divided by cos x minus a cos x minus b into dx. Now further we have 1 over sin a minus b integral o. Now here on adding subtracting x we have sin x minus b minus x minus a. We have added and subtracting x in this. We have the angle in numerator. So divided by x minus a cos x minus a cos x minus b into dx. Further 1 over sorry a minus b integral o. Now using the identity that is sin a minus b we have sin a cos minus sin b the whole divided by cos x minus a cos x minus b where a is x minus b and b is x minus a into dx. Now on simplifying we can write the function as integral. Now here we are left with sin x minus b divided by cos x minus b into dx minus integral of sin x minus a divided by cos x minus a dx. Now let t be equal to cos x minus b. So dt will be equal to minus sin x minus b whereas let z be cos x minus a. So dz will be equal to minus sin x minus a. Substituting we have 1 upon sin a plus b into integral of minus dt over t minus integral of minus dz over z that is further equal to 1 over sin a plus sorry sin a minus b into minus log of plus log of mod z plus which is 1 over sin a minus b into log of mod z divided by log of mod t plus c. Now on substituting the value of z and t we have the answer as 1 over sin a minus b into log mod z was taken as cos x minus a whereas t was taken as cos x minus b plus c. So this is the required answer to the session. Hope you understood it well. Have a very nice day ahead.