Let $\omega=(x+\cos{y})dx + (y\ln{(1+y^2)}-x\sin{y})dy$. Curve $C$ is given by the parametric equation: $\varphi(t)=(\sin{(\ln{(1+t^2)})}, e^{t^3})$, where $t \in [0, \sqrt{e-1}].$ Calculate $\int_C \omega$.
Since the curve isn't closed ($\varphi(0) \neq \varphi(\sqrt{e-1})$) and isn't a boundary of any area, I suppose I shouldn't try to use the Stokes' theorem here. However, by definition of the line integral and defining $F := (F_1, F_2) = (x+\cos{y}, y\ln{(1+y^2)}-x\sin{y})$, I get:
$\int \limits_C \omega = \int \limits_{0}^{\sqrt{e-1}} (F_1(\varphi(t))\varphi_1'(t) + F_2(\varphi(t))\varphi_1'(t)) dt=$
$=\int \limits_{0}^{\sqrt{e-1}} \left ( \sin{(\ln{(1+y^2)})} + \cos{(e^{t^3})} \right ) \left(\frac{2t}{1+t^2} \cdot \cos({\ln{(1+t^2)}}) \right) + $
$ + \left ( e^{t^3}\ln{(1+e^{2t^3})} - \sin{(e^{t^3})}\right)3t^3e^{t^3} dt$
which seems quite painful to calculate. I was therefore wondering if there was any easier way to approach this problem, given that I cannot use the Stokes' theorem.
Hint:
Observe the vector field
$$F(x,y):=\left(\,x+\cos y\,,\,\,y\log(1+y^2)-x\sin y\,\right)$$
is conservative, so the line integral only depends on its extreme points. Try to calculate its potential, say $\;\Phi\;$ :
$$\Phi_x'=x+\cos y\implies\Phi=\frac12x^2+x\cos y+C(y)\implies$$
$$\Phi_y'=-x\sin y+C'(y)\stackrel?=y\log(1+y^2)-x\sin y\implies C'(y)=y\log(1+y^2)\implies$$
$$C(y)=\int y\log(1+y^2)dy=\frac12\left[(1+y^2)\log(1+y^2)-(1+y^2)\right]+K\ldots$$
Note: The final result still seems to be monstruous...but still easier this way, I think, than the other one...