Two challenging maths questions!?

Question

Hi, I have two problems to pose. The first one was problem 27
on the 2007 FAMAT( see http://mathgod.com/mathcomps/interschools/fall2007/
and the second one is related to a question recently posed by
(Ω)kaksi _guy.
Anyway, here they are:

1. If n = 4 I get 4 2 3 2 4 3 1 1
If n = 5 I get 3 5 2 3 2 4 5 1 1 4
I don't get anything for n = 6 or 7
What should I get for n = 8?

2. Compute the definite integral
∫(-π/2..π/2) cos 2x dx/(1 + e^(sin(x)).
The limits are -π/2 to π/2.
I ran this through an online program to
evaluate definite integrals and it seems the value of
this integral is 0. Prove it!

Answer 1

1. Still thinking
I'm begining to suspect that this has something to do with computing since I asked almost everyone who I know and they don't know but ... ...
n=4:
4232
4311
n=5:
35232
45114

It is a Skolem sequence.
n=8:


2. It is basically the same question but the numerator changed!!
It was a nice question and I liked it though.
Wish I could drain Dr d's brain power into my brain.

Let f(x) = cos 2x / (1 + e^sin(x))
and g(x) = cos 2x / (1 + e^{-sin(x)})

f(-x) = g(x)
This means that
I = ∫f(x) dx = ∫g(x) dx
where both integrals go from -π/2 to π/2

Now 2I = ∫[f(x) + g(x)] dx
= ∫ cos 2x * [1/(1+e^sinx) + 1/(1+e^(-sinx))]dx
=∫cos 2x*[2+e^sinx+e^-sinx] /[2+e^sinx+e^-sinx]dx
=∫cos 2x dx = 0

I =0 (Proven) Yay!

Answer 2

I worked out the second one in the comments of the original question.
http://answers.yahoo.com/question/index;_ylt=Ane5eES07D7WFfxck6yJbN_sy6IX;_ylv=3?qid=20071122035305AAqFPXr

Kaksi_guy also put there a link to another similar question which was worked out in a similar fashion. The idea is that you could find the definite integral without finding the indefinite integral, as long as the integral goes from -a to +a.

The difference with this question is that you end up having to integrate cos(2x) from -π/2 to +π/2, as pole did.