We get and this can be solved by standard methods
using the fact that
then (I used the negative to switch the limits of integration)
or, upon expanding,
If you look at your original integral and the second one of the right these are the same except for the variable . As this is only a dummy variable (it is used and never appears in the answer) then we can replace with giving
as others have said. Others can correct me if I'm wrong but I believe this question appears in Stewart's Calculus book.