Let $(W_t)$ be a Brownian motion with respect to a filtration $(\mathcal{F}_t)$. For all $t \geq 0 $ set $$X_t = \int_0^t W_s^2 \mathrm{d} W_s,\qquad Y_t = W_t^7.$$ Find the covariance process $\langle X,Y \rangle_t$ for all $t \geq 0$.
My idea is (since $X_t $ and $Y_t$ are cont. local martingales) to use that
$$\langle X,Y \rangle_t = \langle (W_s^2 \cdot W_s ), (1 \cdot W_t^7) \rangle_t = \int_0^t W_s^2 \mathrm{d} \langle W,W^7 \rangle_s $$
But I don´t know how to calculate the integral above. Should I find $\langle W,W^7 \rangle_t,$ where I use that $\langle W,W^7 \rangle_t,$ makes $W_tW_t^7 - \langle W,W^7 \rangle_t, = W_t^8 - \langle W,W^7 \rangle_t$ a continuous local martingale?
Why not going for the simplest? You are given that $\mathrm dX_t=\color{red}{W^2_t}\cdot\mathrm dW_t$ and you should be able to compute $$\mathrm dY_t=\color{red}{7W^6_t}\cdot\mathrm dW_t+21W^5_t\cdot\mathrm dt,$$ hence $$\mathrm d\langle X,Y\rangle_t=\color{red}{W^2_t}\cdot\color{red}{7W^6_t}\cdot\mathrm d\langle W,W\rangle_t=7W^8_t\cdot\mathrm dt,$$ or, equivalently, $$\langle X,Y\rangle_t=7\int_0^tW^8_s\mathrm ds.$$