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.
 A: We consider the triangle $T$ formed by the points $(3,0)$ , $(0,4)$ , $(0,0)$.
Denote $L_{1}:(0,0)\to (3,0)\\ L_{2}:(3,0)\to (0,4)\\L_{3}:(4,0)\to(0,0)$.
$F= P(x,y)\hat{i}+Q(x,y)\hat{j}$
and
$L=L_{1}+L_{2}+L_{3}$
Then we have by Green's Theorem in a plane(Stokes theorem):-
$$\int_{L} F\cdot d\vec{r} = \iint_{T}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy $$.
But you have $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=3$.
And the double integral of a constant over a region is just the constant times the area of that region. So:-
$$\int_{L} F\cdot d\vec{r}=3\cdot\text{Area of triangle} = 3\cdot 6 = 18$$.
So you have :-
$$\int_{L_{1}} + \int_{L_{2}}+\int_{L_{3}}=18$$
$$\int_{L_{1}} + \int_{L_{2}} = 18-\int_{L_{3}}$$.
So you have $$\int_{L_{3}}F\cdot \vec{r} =\int_{4}^{0}(F\cdot\hat{j})dy= \int_{4}^{0} -y\,dy = 8$$ (As along $L_{3}$ we have $x=0$)
So our answer is $$18 - (8) = 10$$
A: 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,
$$
\int_{O\to A\to B} F\cdot dr = \int_{O\to B}F_1\cdot dr+\int_{O\to A\to B} F_2\cdot dr.
$$
