Given the vector field $$\vec{F}(x, y)=\left(-x y^{2} \sin \left(x^{2}\right), y \cos \left(x^{2}\right)\right)$$ $$\gamma:[0,1] \mapsto \mathbb{R}^{2}$$
I am asking to calculate $\int_{\gamma} \vec{F} \cdot \overrightarrow{d r}$ while $$\vec{\gamma}(t)=( t, tan (\pi t))$$
So I notice that this field is conservative and I think I should use that, I just not sure how should I close the path by making the calculation easier. any one has sugestion how should I close it?
Hint: Note that $\int_{\gamma}\vec{F}\cdot d\vec{r}=\int_{0}^{1}\langle F(\gamma(t)), \gamma'(t)\rangle dt$ can you continue from here?
Also if $F$ is field conservative, so there exists a function $f$, such that $F=\nabla f$. Then $$\int_{\gamma}\vec{F}\cdot d\vec{r}=\int_{\gamma}\nabla F\cdot d \vec{r}=f(\gamma(1))-f(\gamma(0))$$can you continue from here?
Partial Solution: Let $P(x,y):=-xy^{2}\sin(x^{2})$ and $Q(x,y):=y\cos(x^{2})$, so you can see that $$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$so $F$ is a field conservative. Since $F$ is a field conservative, there exists a function $f$, such that $F=\nabla f$. Now, we can find $f$ using "partial integration". Indeed since $F=\nabla f$, so $$\frac{\partial f}{\partial x}=P(x,y) \quad {and} \quad \frac{\partial f}{\partial y}=Q(x,y)$$
Now if $$ \frac{\partial f}{\partial x}=P(x,y) \implies f(x,y)=\int P(x,y)dx+g(y)$$
Now, $$\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left( \int P(x,y)dx \right)+g'(y)$$
Since $\frac{\partial f}{\partial y}=Q(x,y)$, so $$Q(x,y)=\frac{\partial }{\partial y}\left( \int P(x,y)dx \right)+g'(y)$$
Finally, you can find $g$ for integration add any constant $K$.
When you find function $f$, you can choose for example $K=0$, and then you can use $$\int_{\gamma}\vec{F}\cdot d\vec{r}=\int_{\gamma}\nabla F\cdot d \vec{r}=f(\gamma(1))-f(\gamma(0))$$