1. ## Integral

Calculate the integral independent of the path:
$\int_C tg(y) dx + ysec^2(y) dy$, C -> $(1,0)$ to $(2, \frac{\pi}{4})$

2

2. Parametrize the curve.

$r(t) = (1-t)(1,0) + t(2, \frac {\pi}{4})= (1+t, \frac {\pi t}{4}), \ 0 \le t \le 1$

$\int_{C} tg(y) \, \mathrm{d}x + ysec^{2}(y) \ \mathrm{d}y = \int_{C} tg(y) \ \mathrm{d}x + \int_{C} sec^{2}(y) \ \mathrm{d}y$

$\int^{1}_{0} tg( \frac {\pi t}{4})\ \mathrm{d}t + \frac {\pi}{4}\int^{1}_{0} sec^{2}(\frac { \pi t}{4}) \ \mathrm{d}t$

But what's the function g?

3. Originally Posted by Random Variable
Parametrize the curve.

$r(t) = (1-t)(1,0) + t(2, \frac {\pi}{4})= (1+t, \frac {\pi t}{4}), \ 0 \le t \le 1$

$\int_{C} tg(y) \, \mathrm{d}x + ysec^{2}(y) \ \mathrm{d}y = \int_{C} tg(y) \ \mathrm{d}x + \int_{C} sec^{2}(y) \ \mathrm{d}y$

$\int^{1}_{0} tg( \frac {\pi t}{4})\ \mathrm{d}t + \frac {\pi}{4}\int^{1}_{0} sec^{2}(\frac { \pi t}{4}) \ \mathrm{d}t$

But what's the function g?
It's tg, and it's the tangent function

4. 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 $xsec^{2}y \ \mathrm{d}y$. This would mean that the vector field is the gradient of the function $f(x,y) = x \ tan(y)$.

Therefore the line integral is equal to $f(2, \frac {\pi}{4}) -f(1,0) = 2*1 - 0*0 = 2$

5. Ok. Thank you