I have a line integral to evaluate:
$$ F(x,y,z)=(9x^8\ln(5y^2+3)+5z^3) i + (\frac{10x^9y}{5y^2+3}+4z) j+(15xz^2+4y-6\pi\sin(\pi z)) k \\ r(t) =(t^3+1)i + (t^2+2)j + t^3k \\ \text{Evaluate} \int_C F\cdot dr \text{, from 0}\leq t \leq 1$$
My progress towards a solution:
I found dr $=3t^2i+2tj+3t^2dt$ and F by substituting $x=(t^3+1)$, $y=(t^2+2)$, and $z=(t^3)$. I applied the dot product, but I have no idea how to evaluate the integral. The integral is too difficult (for me) to evaluate.. How should I go about trying to simplify the problem?
$\nabla\times \mathbf F=0$ and $\mathbf F$ is defined on all of $\mathbb R^3$, so $\mathbf F$ is conservative. The value of the line integral is therefore $\phi(\mathbf r(1))-\phi(\mathbf r(0))$, where $\phi$ is any antiderivative of $\mathbf F$—a scalar function such that $\nabla\phi=\mathbf F$.
There are a couple of standard ways to find such a function. The one commonly taught is a process of alternating integration and differentiation with respect to each variable in succession, or you can compute $\phi$ in one fell swoop: $\phi(x,y,z)=\int_0^1 \mathbf F(tx,ty,tz)\cdot(x,y,z)\,dt$, i.e., the line integral of $\mathbf F$ along the line segment from the origin to $(x,y,z)$ with the obvious parameterization. Depending on $\mathbf F$, it might be easier to avoid computing $\phi$ explicitly and instead evaluate $$\phi(\mathbf r(1))-\phi(\mathbf r(0)) = \int_0^1 \mathbf F(t\,\mathbf r(1))\cdot\mathbf r(1)-\mathbf F(t\,\mathbf r(0))\cdot\mathbf r(0)\,dt.$$ Again, depending on $\mathbf F$, there might be some other, more convenient path from the origin to $(x,y,z)$ that makes for a simpler integral.