Work done by $\vec F=\sin(x^2)\hat x+(3x-y)\hat y$ needs problematic integral? We attempt to find the work done by the force
$$\vec F=\sin(x^2)\hat x+(3x-y)\hat y$$
in moving a particle from $O\to A=(3,0)\to B=(0,4)$ in straight lines. This is rather problematic as far as I can tell. How can I evaluate the integral of $\sin(x^2)$? I assume I am not meant to obtain a number as this integral is a special function. For example I obtain the following for $O\to A$
$$\int_0^1\sin(3t^2)dt=\left[\sqrt{\frac{\pi}{6}}\operatorname{S}\left(\sqrt\frac{6}{\pi}x\right)\right]_0^1$$
Is this a legitimate answer to give? This is an exam style question so I am not sure.

The force can be decomposed as two forces:
$$
F_1=(\sin(x^2),-y)\,\quad F_2=(0,3x)
$$
The work, thus, can be computed as two parts. The work done by $F_2$ is very easy.
To handle $F_1$, note that
$$
\textrm{curl } F_1=0\;
$$
This condition should remind you something about the Green's theorem. It follows that you can calculate the work done by $F_1$ using the path of the line segment from $O$ to $B$, where you don't need to worry about the $\sin(x^2)$ term anymore.
To summarize,
$$
\int_{O\to A\to B} F\cdot dr = \int_{O\to B}F_1\cdot dr+\int_{O\to A\to B} F_2\cdot dr.
$$
