Ito Isometry and quadratic variation

Confusion regarding stochastic integral.Let

$Y_t=\int_0^tW_sds$ where $W_t$ is a Brownian Motion.Now $dY_t=W_tdt$.So from this expression one can conclude that $dY_t \cdot dY_t=d[Y_t,Y_t]$ where $[\ ]$ is the qudratic variation prcess and it must be zero as $dt \cdot dt =0$.But people apply Fubini-Stochastic integral exchange trick and see $Y_t=\int_0^t(t-u)dW_u$.So by applying Ito isometry one can see $E(Y_t^2)=\int_0^t(t-u)^2du \neq 0 .$Can someone tell me where I'm going wrong ?Why this discrepancy ..??

via Recent Questions - Mathematics - Stack Exchange http://math.stackexchange.com/questions/291290/ito-isometry-and-quadratic-variation