Short analysis of AC circuit

You won't (easily) get a closed form for \$\int e^{p(t)}dt\$.

The solution to the differential equation that you wrote is:

$$i(t)=\left(\frac{V}{L}\right)e^{\frac{-R_{mod}\sin(\omega t)}{L\omega}-\frac{R_{offset}t}{L}}\left(\int_0^t e^{\frac{R_{mod}\sin(\omega \tau)}{L\omega}+\frac{R_{offset}\tau}{L}} d\tau\right)+ Ce^{\frac{-R_{mod}\sin(\omega t)}{L\omega}-\frac{R_{offset}t}{L}}$$ where \$C\$ is an unknown determined by the initial condition. Obviously the term involving integration is the troublesome bit so lets focus on that piece.

The power series for \$e^{x}\$ converges uniformly on bounded sets so we can exchange the power series with the integral giving $$\int_0^t e^{\frac{R_{mod}\sin(\omega \tau)}{L\omega}+\frac{R_{offset}\tau}{L}} d\tau = \sum_{n=0}^{\infty}\frac{1}{n!}\int_0^t\left(\frac{R_{mod}\sin(\omega \tau)}{L \omega}+\frac{R_{offset} \tau}{L} \right)^n d\tau.$$

Integrating the inner terms involves integrating terms of the form \$\tau^k\sin(\omega \tau)^{n-k}\$ which have closed form solutions.

The real upshot is that if we only want approximate solutions we can bound the tail of this series by bounding the integrands and using that $$\sum_{n=N+1}^{\infty}\frac{1}{n!}x^n = e^x-\frac{e^x \Gamma(N+1,x)}{N!}.$$

Then we will have a closed formed approximate solution \$\tilde{i}_N(t)\$ which approximates \$i(t)\$ and we will know a bound for the error \$|\tilde{i}_N(t) - i(t)|\$.

When analyzing a circuit, you can put the arrows in either direction according to whim, a flipped coin, or Tarot cards.

After applying Kirchoff's laws to compute all the voltages and currents, you'll find some variables have negative values. Those correspond to arrows you drew backwards. Fix those, and then you know the directions of currents in all branches of the circuit.

It is perfectly normal for an experienced engineer to get a few initially backwards, when multiple different voltage sources are pushing in opposite directions. You can only guess, and let algebra tell you the net result.