 Hello friends and how are you all today? Let us integrate the following function. The function given to us is cos 2x divided by cos x plus sin x whole ways to the part 2 with respect to tx. Now cos 2x can be written as cos square x minus sin square x. So we have cos square x minus sin square x divided by cos x plus sin x whole ways to the part 2 into tx. Now we can write e square minus b square as a plus b where a is cos x and b is sin x into a minus b b. We will divide it by cos x plus sin x whole ways to the part 2 into tx. Now simplifying it we have integral of cos x minus sin x divided by cos x plus sin x into tx. Now let t be equal to sin x plus cos x which is the denominator. So dt will be equal to cos x minus sin x obviously into tx. So we have integral of dt which is the numerator divided by t. That is log mod of t plus c and on substituting the value of t as sin x plus cos x. So we have the answer as log mod of sin x plus cos x plus c and this is a required answer to the session. It ends and have a nice day.