 Hello friends and how are you all today? The question says Integrate the following rational functions now the head the function which is given to us is Cos x upon 1 minus sin x into 2 minus sin x now when we are dealing with trigonometric functions It is always good to see if the numerator in the fraction can be obtained by differentiating a component in the denominator So try to observe that cos x in the numerator and sine x in the denominator So let us proceed on with our solution We have the trigonometric function as cos x upon 1 minus sin x Into 2 minus sin x now put sin x Is equal to t Therefore, we will have cos x dx is equal to dt right now. Let us integrate The function we have cos x dx upon 1 minus sin x Into 2 minus sin x now on substituting the values. We will have dt upon 1 minus t into 2 minus t now The above fraction can be solved by using partial fraction method now Let us solve it by partial fraction method separately. So this function can be written as h upon 1 minus t Plus b upon 2 minus t on taking the LCM Of the right-hand side Equating the numerator we have 1 is equal to a into 2 minus t plus b into 1 minus t we have 2 a plus b equal to 1 and Minus a minus b Equal to 0 so on solving these two equations. We have the value of a and b equal to And be the value of a is obtained to be 1 whereas b is minus 1 so we have now the function written as Integral 1 upon 1 minus t minus 1 upon 2 minus t dt now Taking integration sign separately with each term we have integral Dt upon 1 minus t minus Dt upon 2 minus t Which can be written as So we have the answer as minus log mod 1 minus t plus log 2 minus t plus c which can be written as log 2 minus t upon 1 minus t Plus c now on substituting the value of t as sin x we have log Mod 2 minus sin x upon 1 minus sin x plus This is the required answer to the session. Hope you understood the whole concept Well, and have a very nice day ahead