# Line integral $\int_{\gamma}e^xcos(y)dx+e^xsin(y)dy$

First of all, I'm having trouble evaluating $$\int_{\gamma}e^xcos(y)dx+e^xsin(y)dy$$ where $\gamma$ is the triangle with vertices $(0,0), (1,0), (1,\frac\pi2)$

This is what I've done so far:

1) With Green's theorem:

$$\int_{\gamma}e^xcos(y)dx+e^xsin(y)dy=\iint_D2e^xsin(y)dydx=\int_0^1\int_0^{\frac \pi 2 x}2e^xsin(y)dydx$$

2) Using regular line integration: as $F=(e^xcos(y), e^xsin(y))$

$$\int_{\gamma}e^xcos(y)dx+e^xsin(y)dy=$$ $$\int_0^1F(t,0)(1,0)dt+\int_0^{\frac \pi 2} F(1,t)(0,1)+\int_1^0F(t,\frac\pi2t)(1,\frac\pi2)$$ $$\int_0^1e^t+\int_0^{\frac \pi 2}esin(t)+\int_1^0e^tcos(\frac\pi2t)+e^tsin(\frac\pi2t)\frac\pi2$$

I've computed both in wolfram, and they are different. They shouldn't! I don't get why. If someone could please explain me where I went wrong, I'd really appreciate it.

Secondly, are those the only two methods we can use in order to evaluate a line integral? I mean, if I have a line integral, then it's either Green or regular integration right?

Finally, when we use the formula $\int_{\gamma}F=\int F(\gamma(t))\gamma'(t)dt$, are we assuming that the curve is positively oriented?

• I believe your limits of integration are incorrect, for the inner integral, it should be $0$ to $\pi x/2$, shouldn't it? – SuperAbound Aug 4 '14 at 13:12