 Hello and welcome to the session. Let's solve the following problem on integration. It says integrate the following function The given function is sin x upon 1 plus cos x. Let us now proceed on with the solution and let I be the integral sin x upon 1 plus cos x dx Now we see that the derivative of 1 plus cos x is minus sin x So put equal to 1 plus cos x so dt by dx is equal to minus sin x and this implies dt is equal to minus sin x dx and this implies sin x dx is equal to minus dt. So sin x dx is equal to minus dt and 1 plus cos x is t. So substituting all these values in this integral the integral becomes minus dt Which is further equal to minus integral 1 upon t dt now the integral of 1 upon dt is log mod t So this becomes minus log mod t plus c where c is the constant of the integration Now t is 1 plus cos x so put t is equal to 1 plus cos x and this can be further written as log mod 1 plus cos x to the power minus 1 plus c and this is further written as log 1 upon mod 1 plus cos x plus c Hence the integral of the given function is log 1 upon mod 1 plus cos x plus c So this completes the question and the session. Hope you'll be able to solve more of such problems Keep practicing such questions. Use the substitution method. Goodbye and take care