integration; big heads!!! plzzzzz heeeeelp, wailing and pulling my hair out?

Question

LOL; As you can see our ungrateful kids have crάpped the whole math category to frustrating boredom! As a change I propose a bit more interesting stuff.
1) this problem was posted 2 weeks ago
http://answers.yahoo.com/question/index;_ylt=AnlJEhvRg1XOB4qLEB5R76jsy6IX;_ylv=3?qid=20071105142649AAumw5p
I wonder who can do it besides Scythian;
evaluate ∫dx sin(x*2007^207) /sin(x) {x=0 until pi};
2) here I shall not mention the source as it contains solution;
evaluate ∫dx (cos x)^2 /(1+ exp(sinx)) {for x from –pi/2 to +pi/2};

-- wow Steiner! This formula
cos x + cos 2x + ... + cos nx =
is also familiar to me! but it did not come to my mind to be applied here! I used another way! Thank you for the useful link! I’ve been integrating in Excel;
--Liverpool, strive in formal way please!

Answer 1

Well, Scythian gave the answer to the first integral as π.
He was right, but never showed how he got the answer.
We will show, more generally, that
∫(0..π) sin(2n+1)x dx/ sin x = π for n any positive integer.
I browsed through my trig book(Trigonometry,
Functions and Applications, by Paul Foerster, p.146)
and found an outline of a proof of the following identity:
1+ cos x + cos 2x + ... + cos nx = ½(1+sin(2n+1)x /2 / sin(x/2) )
Replacing x by 2x in this identity and doubling both sides,
we get
1 + 2(cos 2x + cos 4x + ... + cos 2nx )= sin(2n+1)x/ sin(x)
So,
∫(0..π) sin(2n+1)x/ sin x dx=
(x +2( sin 2x/2 + sin 4x/4 + ... + sin 2n x/2n))(0..π)
which just equals x(0..π) = π.
I haven't made any headway on the second integral yet.
I tried using substitution u = sin x
and parts, letting u = cos x and v = cos x/(1+ e^sin x),
but these just lead to dead ends.
I did run it through the link below and
it gave the answer as π/4, but I have no idea
how to prove it right now. Will keep trying!
Yes, let's keep posting these challenging
problems. I enjoy them!

Answer 2

what use a calculator