Work done by force field in moving particle along given path Path: $y=x^2$, from ($-\pi$, $\pi^2$) to ($\pi$,$\pi^2$) Field: $F(x,y) = e^y\sin(x)$ **i** $-(e^y\cos(x)-\sqrt {1+y})$ **j** I began this problem by parametrizing the path by $x=t$, $y=t^2$. Next, setting $r(t)=t$ **i** $+t^2$ **j** , found $r'(t)$. I then used the formula for work, $W=\int_a^bF\bullet r'(t)dt$, which forces me to take the integral of $e^{t^2}\sin(t)$. This is impossible as far as I know. I tried a different parametrization, $x=\sqrt{\ln(t)}, y=\ln(t)$ but also ran into an undeterminable integral. Where am I going wrong?

I got $$\int_{-\pi}^\pi(e^{t^2}\sin t-2te^{t^2}\cos t+2t\sqrt{1+t^2})\,dt.$$ I then thought about the derivative of $e^{t^2}\cos t$.