 Hello and how are you all today? My name is Priyanka and I shall be helping you with the question. It says, integrate the following. Here the function which is given to us is cos 2x-cos 2alpha divided by cos x-cos alpha. Now proceeding on with the solution, here we need to integrate this function with respect to dx. Now we know that we can write cos 2x as 1-2 cos square x-1, right? So we have minus, now here cos 2alpha can also be written as cos 2 cos square alpha-1 divided by cos x-cos alpha into dx. Now on opening brackets we have 2 cos square x-1-2 cos square alpha-1 the whole divided by cos x-cos alpha into dx. Now, unsimplifying we have 2 cos square x-cos square alpha divided by cos x-cos alpha into dx. Now a square minus b square can be written as a minus b where a is cos x and b is cos alpha into a plus b the whole divided by cos x-cos alpha into dx. Further simplifying we have further 2 into integral of cos x plus cos alpha into dx. So we have 2 integral of cos x into dx plus 2 integral of cos alpha into dx. Here cos alpha will be a constant so we have 2 into integral of cos x dx is sin x plus 2 cos alpha into integral of 1 dx will be into x. So we have the answer as taking out 2 common sin x plus x cos alpha plus c this is the required answer to the session. So hope you understood it well have a very nice day ahead.