quantum teleportation with shared state and teleported state in bloch representation

Suppose that two parties (Alice and Bob) share an entangled state

$$\rho_F = F \lvert \phi^+ \rangle \langle \phi^+ \rvert + \frac{1-F}{3} \left( I \otimes I - \lvert \phi^+ \rangle \langle \phi^+ \rvert \right).$$

Alice wants to teleport a state $$\lvert \psi(\theta, \phi)\rangle = \mathrm{cos}\,\theta \, \lvert 0 \rangle\ + \mathrm{e}^{\mathrm{i}\phi} \cdot \mathrm{sin}\,\theta \, \lvert1\rangle$$ by using the shared state and let $$\rho(\theta, \phi)$$ denote the resulting state. Compute the following

$$\int \mathrm{sin} \, \theta \, \mathrm{d}\theta \, \mathrm{d} \phi \, \langle\psi(\theta, \phi) \rvert\rho(\theta, \phi)\lvert\psi(\theta, \phi)\rangle$$

My approach is

• Calculate first $$\lvert \psi \rangle \langle \psi \rvert$$ to later compute the partial trace over the shared state to obtain the resulting state $$\rho$$, i.e. $$\rho = \mathrm{Tr}_{\psi_f} \left[ \left( I \otimes \lvert \psi \rangle \langle \psi \rvert\right) \cdot \rho_F \cdot \left( I \otimes \lvert \psi \rangle \langle \psi \rvert\right)^\dagger\right]$$
• put the obtained $$\rho$$ in the integral and compute it

However even while calculating the density matrix I get nasty terms with trigonomic mixtures in bloch sphere representation even before starting matrixmultiplying this with $$\rho_F$$, similar to here. So this will go even nastier when $$\rho_F$$ comes in.

Am I missing some obvious simplification here? Or is my approach wrong? Should I maybe not calculate in Bloch representation (I'd need to go back for it for the integral anyways later)

The problem comes from an online Coursera lecture in a course I do for work. I know the result of the integral should be $$F$$ but I really do not see the point where everything would cancel out to only leave this.