Evaluate the Integral?

Question

Of cosx from (-pi/27) to (pi/27)... I need the answer in radical form, can't figure it out. thanks :)

Answer 1

2 sin pi/27 is certainly correct, but the asker wanted
the answer in radical form.
Don't we have to get the solution to x^27-1 =0
in radical form?
Here's a partial answer:
x^27-1 = 0
gives
(x^9 -1)(x^18 + x^9 + 1) =0.
The first factor gives the 9th roots of unity.
By the quadratic formula, the second factor
gives
x^9 = (1 +- i*sqrt 3)/2.
By Galois theory, that's as far as we can go.
x = ((1 +- i*sqrt 3)/2)^1/9

Answer 2

The answer 2sin(pi/27) is correct. You won't be able to express this as a real radical. Only values of sine and cosine corresponding to constructible polygons can be solved in terms of real radicals. If the limits were -pi/17 and pi/17, it would be possible, but very messy.

Answer 3

Integral cos x dx = sin x[-pi/27, pi/27 = 2sin pi/27

Answer 4

sin(pi/27)-sin(-pi/27)
sin(pi/27)+sin(pi/27)
=2sin(pi/27)
=2sin(180/27)=2sin6*40'
2(0.1161)
=0.2322