RE: Can you solve this integral from MIT's Integration Bee?

I see that cos(x/2^1) = cos(2*x/2^2) = 1 - 2sin^2(x/22).
When k = 1, the product is cos(x/2^1).
When k = 2, the product is (1 - 2sin^2(x/22))*cos(x/2^2).
When k = 3, the product is (1 - 4sin^2(x/23)cos^2(x/23))*(1 - sin^2(x/23)*cos(x/2^3).
Does this lead to finding any useful pattern?