Using the substitution u=cosx + 1, show that the integral of sinx e^cosx+1 is equal to e(e-1), for the values of x between x=π/2 and x=0

First we differentiate the substitution giving, du/dx=-sinx, which is rearanged to dx=du/-sinx. we can then substitute this into the integral to get sinx e^cosx+1 du/-sinx which can be simplified to -e^cosx+1 du. with this we can then use the substition to obtain -e^u du. Putting in the values of x in the substitution we get that the limits will be 1 and 2. Now when we integrate we get -(e^1 - e^2), which can be written as e^2 - e^1 or e(e-1).