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Let $X_t=\int_0^tsW_s^2dW_s$. How to set $<\int_0^tW_scos(s)dX_s>$? is it: $<\int_0^tW_scos(s)sW_s^2dW_s>=\int_0^ts^2W_s^6cos^2(s)ds$

wiwnes691
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1 Answers1

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$$dX_t=t\,W_t^2\,dW_t$$ we have $$Y_t=\int_{0}^{t}\cos(s)\,W_s\,dX_T=\int_{0}^{t}s\cos(s)W_s^3\,dW_s$$ The quadratic variation of Ito integral $Y_t$ is given by $$\langle Y_t\rangle=\left[Y,Y\right](t)=\left[\int_{0}^{t}s\cos(s)W_s^3\,dW_s,\int_{0}^{t}s\cos(s)W_s^3\,dW_s\right](t)=\int_{0}^{t}s^2\cos^2(s)W_s^6ds$$