Your are right...but I have been working on it.
I can get the proof for n = 2.
I thought I had a proof for n = 1, but I'm still working on it.
I haven't gotten any further, but I can tell you a little more.
Graph a couple of the integrands and you will immediately see that they fall into 2 groups:
the odd values of n which are symmetric with respect to x = pi/4 and and are shaped a little like y = x3 or y = tan x
and the even values which look a little like sinusoids.
Unless there is some trick which I have not yet discovered, each of these groups will have to be proved by induction, but I can't yet see how to go from one value to the value of n + 2...even if could prove for n = 1.
I will continue to work on it and let you know if I get any further. However, if you get the answer in class, I would be grateful to know how to do this. Thank you.
10/30/2018
OK, I solved the integral for n =1. The trick in this one is to multiply numerator and denominator by
(cos x - sin x), which splits the integral into 2 parts which can be integrated after a little trig manipulation.
For n = 2 the denominator is one which leaves only sin2 x to integrate which can be done by going to cos 2x.
I still do not know how to get no higher values of n for induction. I will keep working on it.
