Calculate the integral independent of the path:
$\displaystyle \int_C tg(y) dx + ysec^2(y) dy$, C -> $\displaystyle (1,0)$ to $\displaystyle (2, \frac{\pi}{4})$
Answer:
2
Parametrize the curve.
$\displaystyle r(t) = (1-t)(1,0) + t(2, \frac {\pi}{4})= (1+t, \frac {\pi t}{4}), \ 0 \le t \le 1$
$\displaystyle \int_{C} tg(y) \, \mathrm{d}x + ysec^{2}(y) \ \mathrm{d}y = \int_{C} tg(y) \ \mathrm{d}x + \int_{C} sec^{2}(y) \ \mathrm{d}y$
$\displaystyle \int^{1}_{0} tg( \frac {\pi t}{4})\ \mathrm{d}t + \frac {\pi}{4}\int^{1}_{0} sec^{2}(\frac { \pi t}{4}) \ \mathrm{d}t$
But what's the function g?
Oh.
Anyways, since the line integral is independent of the path taken, you're integrating along a curve in a gradient vector field.
And I'm going to go out on a limb and say you meant $\displaystyle xsec^{2}y \ \mathrm{d}y $. This would mean that the vector field is the gradient of the function $\displaystyle f(x,y) = x \ tan(y) $.
Therefore the line integral is equal to $\displaystyle f(2, \frac {\pi}{4}) -f(1,0) = 2*1 - 0*0 = 2$