with respect to the first equation i'm thinking the following:

focusing on X int Sin (kX) .dk

as the integral is in respect to k, simple integration gives

X *[1/X* -cos(kX)] = -cos(kX) as over 0_1

-cos(1X) + cos(0X) = 1 - cos(X)

Plugging in to formula

Y(t) = [1 - 1 - cos(X) ] * e^(t/2)

gives

Y(t) = e^(t/2)cos(X)

as there is no drift term (.... .dt) this must mean its a martingale.

but i don't think i can perform simple integration on a function containing Brownian motion.