= ∏_{j=1}^∞ [1/2 * cos(x/(2j-1))] * ∏_{j=1}^∞ [1/2 * cos(x/(2j+1))]

= (1/2)^∞ * cos(x/1) * cos(x/3) * cos(x/5) *

∫_{0}^{π/2} x * ∏_{j=1}^∞ cos(x/2j) dx

= ∫_{0}^{π/2} x * (1/2)^∞ * cos(x/1) * cos(x/3) * cos(x/5) * ... dx

= (1/2)^∞ * ∫_{0}^{π/2} x * cos(x/1) * cos(x/3) * cos(x/5) * ... dx

cos(x/1) * cos(x/3) * cos(x/5) * ... = (1/2) * (sin(x) + sin(3x)/3 + sin(5x)/5 + ...)

∫_{0}^{π/2} x * cos(x/1) * cos(x/3) * cos(x/5) * ... dx

= (1/2)^∞ * ∫_{0}^{π/2} x * (sin(x) + sin(3x)/3 + sin(5x)/5 + ...) dx

= (1/2)^∞ * [(-x*cos(x))/1 + (1/3)*x*cos(3x) - (1/5)*x*cos(5x) + ...]_{0}^{π/2}

= (1/2)^∞ * [(-π/2)*cos(π/2)/1 + (1/3)*(π/2)*cos(3*(π/2)) - (1/5)*(π/2)*cos(5*(π/2)) + ...]

= (1/2)^∞ * [(π/2)/3 - (π/2)/5 + (π/2)/7 - ...]

= (1/2)^∞ * π/4 * [1/1 - 1/3 + 1/5 - ...]

1 - 1/3 + 1/5 - 1/7 + ... = π/4

Therefore,

∫_{0}^{π/2} x * ∏_{j=1}^∞ cos(x/2j) dx

= (1/2)^∞ * π/4

= 0

oh that’s not—