I am answering my own question since two members of the Physics StackExchange (see the comment section) kindly provided me with enough information to derive the equation. Because of the somewhat intricate nature of the derivation, it is easy to make errors. If any are present, they fortuitously cancelled each other out.
Merzbacher indicates that the desired relationship
$\sin\delta_l = -k\int_0^\infty j_l(kr')U(r')u_{l,k}(r')r'dr'\tag{11.83}$
can be obtained from the two equations below by utilizing a partial wave analysis. $ U=\frac {2\mu V}{\hbar^2}$, and $\delta_l$ is understood to be a function of k.
$$f_k( \mathbf {\hat r}) = -\frac{\sqrt{2\pi}\mu}{\hbar^2}\int exp(-ik\mathbf {\hat r}\cdot \mathbf r')V(r')\psi_k(\mathbf r’)d \mathbf{\tau}' \tag{11.35a}$$
$$f_k(\theta)=\frac{1}{k}\sum_{l=0}^\infty(2l+1)exp(i\delta_l)\sin\delta_lP_l(\cos\theta)\tag{11.59}$$
One starts by setting the two equations equal to one another but with only a partial wave taken from Equation 11.59. For consistency, this decision necessitates the use of only a partial wave from the wave equation on the RHS later in this derivation. Alternatively, the summation in 11.59 can be left in place (and in the wave equation) and removed at the end of the derivation, but omitting it now is less cumbersome.
$$\frac{1}{k}exp(i\delta_l)\sin\delta_l P_l(\cos\theta)= -\frac{\sqrt{2\pi}}{2} \int exp(-ik\mathbf {\hat r}\cdot \mathbf r’)U(r’)\psi_k(\mathbf r’) d \mathbf {\tau}' $$
Next, an important identity is substituted for $exp(-ik \mathbf {\hat r}\cdot \mathbf r') $ on the RHS.
$$ = -\frac{\sqrt{2\pi}}{2} \int \sum_{l'=0}^\infty(2l'+1)i^{3l'}j_{l'}(kr')P_{l'}(\mathbf {\hat r}\cdot \mathbf {\hat r’})U(r')\psi_k(\mathbf r’)d \mathbf{\tau}'$$
Another identity is substituted for $P_{l'}(\mathbf {\hat r}\cdot \mathbf {\hat r’})$ (shown in the square brackets below), and a partial wave from the wave equation $\psi(\mathbf r')=\sum _{l=0}^\infty(2l+1)\frac {i^lexp(i\delta_l)}{(2\pi)^{3/2}}P_l(\cos\theta')u_{l,k}(r')/r' $ is substituted for $\psi.$
$$ = -\frac{1}{4\pi} \int \sum_{l'=0}^\infty(2l'+1)i^{3l'}j_{l'}(kr')\mathbf[\sum_{m=-l'}^{l'}\frac{4\pi}{2l'+1}Y_{l'}^{m*}(\theta',\phi')Y_{l'}^m(\theta,\phi) \mathbf]U(r')i^l exp(i\delta_l)P_l(cos \theta')u_{l,k}(r')/r'd \mathbf{\tau}'$$
The integral is now expanded in spherical polar coordinates, $P_l$ and $Y_{l'}^m$ are respectively converted to $Y_{l}^0$ and $P_{l'}^m$, and some rearranging is done.
$$ = -\frac{1}{4\pi} \int_0^\infty \sum_{l'=0}^\infty(2l'+1)i^{3l'+l}j_{l'}(kr')\sum_{m=-l'}^{l'}U(r')exp(i\delta_l)\int_{\phi'=0}^{2\pi} \int_{\theta'=0}^{\pi} \frac{4\pi}{2l'+1}Y_{l'}^{m*}(\theta',\phi')Y_{l}^0(\theta',\phi')[(-1)^m exp(im\phi)\sqrt{\frac{(2l'+1)(l'-m)!}{(2l+1)(l'+m)!}}] P_{l'}^m (cos\theta)sin(\theta')u_{l,k}(r')r'd{\phi'}d{\theta'}dr'$$
After taking orthogonality of terms into account, one obtains
$$ = -\int_0^\infty \sum_{l'=0}^\infty \sqrt {\frac {2l'+1}{2l+1}}i^{3l'+l}\delta_{ll'}j_{l'}(kr')U(r')exp(i\delta_l)P_{l'}(\cos \theta) u_{l,k}(r')r'dr'$$
$$ = -\int_0^\infty j_{l}(kr')U(r')exp(i\delta_l)P_{l}(\cos\theta)u_{l,k}(r')r' dr'$$
The LHS of the equation at the beginning of the derivation and the RHS above are now compared to get $$\sin\delta_l = -k\int_0^\infty j_l(kr')U(r')u_{l,k}(r')r'dr'\tag{11.83}$$
