Let $\Gamma$ be the triangular path connecting the points (0,0), (2,2) and (0,2) in the counter-clockwise direction in $\mathbb R^2$. Then solve $$I=\oint_\Gamma\sin x^3\,dx+6xy\,dy$$
I tried this problem by taking different paths from (0,0) to (2,2) but I am not getting how to solve $$\int_0^2\sin x^3\,dx$$ if taking the path as $y=x$ from (0,0) to (2,2).
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Let $P=\sin x^3$ and $Q=6xy$. Green's theorem says that:$$\int_CLdx+Mdy=\iint_D\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}dxdy$$where $D$ is the area surrounded by closed path $C$. Also$$\dfrac{\partial M}{\partial x}=6y\\\dfrac{\partial L}{\partial y}=0$$therefore we can simplify the integral as following:$$I=\int_{0}^{2}\int_{0}^{y}6ydxdy=\int_{0}^{2}6y^2dy=16$$
You don't need to compute $$\int_0^2\sin x^3\,dx.$$ What you need to compute is $$\int_0^2\sin x^3\,dx+\int_2^0\sin x^3\,dx+\int_0^0\sin x^3\,dx$$ which is a much easier task...