I've been stuck on a problem for a few hours, and I think it's no longer productive to keep trying the same methods. I need to find $\int_\gamma(F+G)dl$, where $\gamma$ is a positively-oriented circle of radius $R$ centered in $(0,0)$. I've already calculated $\int_\gamma F dl$, which equals zero, but I'm stuck trying to find out $\int_\gamma G dl$, where $G:\mathbb{R}^2 \rightarrow \mathbb{R}^2$, $G(x,y)=(2x\cos(y),-x^2\sin(y)+ye^y)$.
My approach has been the following: the positively-oriented circle with radius R can be parametrized as $(R\cos(\theta),R\sin(\theta))$, so what I'm looking for is: $$\int_{\theta=0}^{\theta=2\pi}(2R\cos\theta\cos(R \sin\theta),-R^2 \cos^2(\theta) \sin(R\sin\theta)+R\sin\theta e^{R\sin\theta})\cdot(R\cos\theta,R\sin\theta)d\theta=$$ $$\int_{\theta=0}^{\theta=2\pi}(2R^2\cos^2(\theta)\cos(R\sin\theta)-R^3\cos^2(\theta)\sin(\theta)\sin(R\sin\theta)+R^2\sin^2(\theta)e^{R\sin\theta})d\theta$$
And that's where I get lost. Can anyone help me out, or give me a hint? Is there a way to simplify this? Thanks in advance :)
It’s a good idea when presented with a complicated-looking integrand to see if it’s conservative. The partial derivatives are often very easy to compute and even if the integrand isn’t conservative, when the path is a simple closed loop you’ve now got the pieces that you need to use Green’s theorem. In this case we have $-2x\sin y = -2x\sin y$ and a simple loop, so you’re done.