Compute the following integrals:
- $$\int_{\overline {AB}}^{ }\left(\cos\left(2y\right)dx-2x\sin\left(2y\right)dy\right)$$
- $$\int_{\overline {AB}}^{ }\left(\tan\left(y\right)dx+x\sec^{2}\left(y\right)dy\right)$$
Where $A=(1,\frac{\pi}{6}), B=(2,\frac{\pi}{4})$.
The line can be parameterized as $$C: r(t)=(t,\frac{\pi}{12}(t+1)), \;\;\; t \in [1,2]$$
So:$$\int_{\overline {AB}}^{ }\left(\cos\left(2y\right)dx-2x\sin\left(2y\right)dy\right)$$$$=\frac{6}{\pi}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\left(\cos\left(t\right)+\left(-t+\frac{\pi}{6}\right)\sin\left(t\right)\right)dt$$$$=\frac{6}{\pi}\left(1-\frac{\sqrt{3}}{2}-\frac{\pi}{6}-1+\frac{\sqrt{3}}{2}+\frac{\pi}{12}\right)$$$$=-1/2$$
$$\int_{\overline {AB}}^{ }\left(\tan\left(y\right)dx+x\sec^{2}\left(y\right)dy\right)$$$$=\int_{1}^{2}\left(\tan\left(\frac{\pi}{12}\left(t+1\right)\right)+t\sec^{2}\left(\frac{\pi}{12}\left(t+1\right)\right)\frac{\pi}{12}\right)dt$$$$=\frac{12}{\pi}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\left(\tan\left(t\right)+\left(t-\frac{\pi}{12}\right)\sec^{2}\left(t\right)\right)dt$$$$=\frac{\sqrt{12}-1}{\sqrt{3}}$$
I want to know how much of my work is correct.
Please note that both vector fields are conservative and are gradients of a scalar function.
$\vec F = \nabla f(x,y)$
So by Fundamental Theorem of Line Integral,
$\displaystyle \int_C (\nabla f) \cdot dr = f(b) - f(a), \ $ where $a$ and $b$ are start and end points respectively.
For the first one, $\vec F = \langle\cos(2y), -2x\sin(2y)\rangle = \nabla (x \cos (2y))$
Hence the line integral for the first one is $ \ 2 \cos(\frac{\pi}{2}) - \cos (\frac{\pi}{3}) = - \dfrac{1}{2}$
For the second one, note that $\vec F = \nabla (x \tan y)$ so you can again apply fundamental theorem of line integral.