According to http://integrals.wolfram.com, the answer is (1/2) (sin x cos x).
I believe these integrals are calculated by converting the sine or cosine square with some trigonometric identity first. See this reference: http://en.wikipedia.org/wiki/Trigonometric_identity#Power-reduction_formulas
In this case, the relevant formula is cos2x = (1 + cos 2x) / 2.
According to Mathcad the answer is (1/2)*(sin(x)*cos(x)+x). The proof by derivation confirms this solution.
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Cosine to the negative first power and cosine cancel each other out because cosine to the negative first power is one over cosine, and one over anything times anything is just one.
No
The cosine of 2pi is 1. In fact, for every integer N, the cosine of 2 N pi is 1.
If x has the power 2 then you want the integral of x2, I think. When you integrate this you get : x3/3 , plus a constant.
Cosine(30) = sqrt(3)/2