We know we can use parametrization $t$ to solve the line integral (standard way for solving the line integral.)
But, We also know that, when we have the line integral that has path independent, we can considering the Exact ODE as an integrand and solve or find the function $f(x,y)$ that satisfies the Exact DE, then just applying substitution for the upper and lower bound of the integral.
For example:
$$\int_{(5,1)}^{(2,3)} (1+y^{-1})\,\mathbb dx -xy^{-2}\,\mathbb dy=\left[x+xy^{-1}+C\right]_{(5,1)}^{(2,3)}$$
Which is $f(x,y)=x+xy^{-1}+C$ is a function that satisfies the Exact DE (The Integrand).
My question is:
What if i have the integrand, which is Non-Exact DE? If i found the integrating factor and makes it Exact, could it be work for solving the Regular Line Integral using that Path Independent Integral?
For example, i have:
$$\int_{(5,1)}^{(2,3)} (y^2+y)\,\mathbb dx -x\,\mathbb dy$$
(We know that integrand is Non-Exact DE)
In other words, is it possible solving line integral with path independece of integral method if the integrand is Non-Exact DE?
Sorry if my question unclear, and please to tell me if you have a question for this.
No, this won't work. If the differential is not exact, the line integral will depend on the path. You will need a parameterization of the path to perform the integral.
Note also that even if you have an exact differential, the integral can still depend on the path. For example,
$$ \int_{(5,0)}^{(3,4)}\frac{-y\,dx + x\,dy}{x^2+y^2}. $$
Although the integrand is an exact differential of $\tan^{-1}(y/x)$, the integral still depends on the path taken between the points. To see this for yourself, the circle of radius $5$ passes through both points. Try integrating one direction on it, then the other.