The instructions are to use the fundamental theory of line integrals to evaluate $$\int_C \cos x \sin y dx + \sin x \cos y dy$$
Where $C:$ Line segment from $(0, -\pi)$ to $(\dfrac{3\pi}{2}, \dfrac{\pi}{2})$
When I have an integral of the form $\int_C g(x, y)$i $ + h(x, y)$j I can find the potential function by integrating $g$ with respect to $x$ and integrating $h$ with respect to $y$, but I don't know how to find it when I have differentials in the integrand.
You want to calculate $\int \boldsymbol f(x,y)\cdot\mathrm d\boldsymbol l$, where $$ \boldsymbol f(x,y)=(\cos x\sin y,\sin x\cos y) $$ and $\mathrm d\boldsymbol l=(\mathrm dx,\mathrm dy)$. We would like to find some $\phi(x,y)$ such that $\boldsymbol f=\nabla\phi$, that is, such that $$ \frac{\partial \phi}{\partial x}=\cos x\sin y $$ and $$ \frac{\partial \phi}{\partial y}=\sin x\cos y $$
It is easy to see that $\phi=\sin x\sin y$ works just fine. You can find this $\phi$ by integrating, as you said. Integrate the first equation: $$ \phi(x,y)=\int\frac{\partial\phi}{\partial x}\mathrm dx=\sin x\sin y+c(y) $$ and plug into the second one: $$ \sin x\cos y=\frac{\partial \phi}{\partial y}=\sin x\cos y+c'(y) $$ so that $c(y)$ is any constant (irrelevant to your integral).
NOTE
How can we know that such a $\phi$ may or may not exist? Well, $\boldsymbol f$ is contiuous, and $\text{curl}\boldsymbol f=0$:
$$ \text{curl}\boldsymbol f=\frac{\partial f_x}{\partial y}-\frac{\partial f_y}{\partial x}=\cos x\cos y-\cos x\cos y=0 $$
So that $\phi$ is guaranteed to exist.
NOTE II
I believe you know how to use this $\phi$ to calculate your integral, but just in case, you need the formula $$ \int \nabla\phi\cdot\mathrm d\boldsymbol l=\phi(P_2)-\phi(P_1) $$ where $P_1$ and $P_2$ are the boundary points of your segment.