Line integral - Parametric representation the question is: C is the following area:
$$C = \left\lbrace (x,y) \; \middle| \;
         x^2 + y^2=\frac{\pi^2}{4}, \;  x \geq 0  \right\rbrace$$
Begins in: $\displaystyle \left( 0,\frac{\pi}{2} \right)$ end in $ \displaystyle \left( 0, -\frac{\pi}{2} \right)$
we want to calculate the integral : $$I = \int_{C} \left( e^x\sin{y} \; + 24 y \right) \, dx + e^x \cos{y} \,dy$$
First I'm required to use green's theorem, - I completed C to a complete circle and calculated :
$$ \displaystyle
  J   =  \int_{C\cup C_1 }(e^x\sin{y} \; + 24 y) \,dx + e^x\cos{y} \, dy
      =  \iint 24 dxdy = {3\pi ^3} $$
Now I'm told to subtract the following line integral with $y(t) = t$:
$$
\int_{C_1} ( e^x \sin{y} \; + 24 y ) \, dx + e^x \cos{y} \, dy = {} \rule{60pt}{0pt}
$$
Where $C_1$ is the "rest of the circle"
So its:
$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} *something* dt$
I Can't figure out what that something should be.. help would be greatly appreciated.
 A: 
First I'm required to use green's theorem, on a complete circle

Here is an alt approach, consider the half disc on right side and do greens on that,
$$ \int_{\partial C} \vec{F} \cdot ds = \int_{C} \nabla \times \vec{F} dA$$
Or,
$$ \int_C (e^x \sin y + 24 y) dx + (e^x \cos(y) ) dy = \int \nabla \times < e^x \sin y + 24 y, e^x cos y> dA = \int (e^x \cos(y) -e^x \cos(y) +24) dx dy = 24 \frac{\pi r^2}{2}$$
Now that you have this substract the vertical line integral from which goes from $(0 , \frac{\pi}{2} ) \to ( 0 , - \frac{\pi}{2})$ i.e: over the line $ x=0$ from y=$  \frac{\pi}{2} \to \frac{-\pi}{2}$. It's parameteric equaton is:$$ \vec{r(t)} =  \pi t \vec{j}$$
From $ t= \frac{1}{2}$ $\to$$ t= - \frac{1}{2}$
A: $\displaystyle
  J   =  \oint_{C\cup C_1 }(e^x\sin{y} \; + 24 y) \,dx + e^x\cos{y} \, dy
      =  \iint 24 \, dA = {6\pi ^3}$ (Area of the given circle is $\frac{\pi^3}{4}$).
(given it is clockwise orientation).
The part that you need to notice is that we can split the vector field into two parts to make it easier to work with -
$\vec{F} = \vec{F_1} + \vec {F_2} \, $ where
$\vec{F_1} = (e^x\sin{y}, e^x\cos{y}), \vec {F_2} = (24y, 0)$
Now the first vector field is a conservative vector field and is gradient of scalar function $e^x \, \sin y$. So the line integral of this vector field will be path independent and will only depend on start and end points. This part of the vector field would not contribute anything to the line integral over the full circle which we calculated earlier.
$\vec{F_1} = \nabla f$, where $f = e^x \sin y$.
Line integral of $\vec{F_1} \,$ over semicircular path $C_1$,
$J_1 = \displaystyle \int_{C_1} \vec{F_1} \cdot d\vec{r} = \int_{(0, \pi/2)}^{(0,3\pi/2)} (\nabla f) \cdot d\vec{r} = f(0,\frac{3\pi}{2}) - f(0,\frac{\pi}{2})$
Now let's calculate the line integral of the vector field $\vec{F_2}$ over $C_1$ oriented counterclockwise.
$\vec{r}(t) = (\frac{\pi}{2} \cos t, \frac{\pi}{2} \sin t) \,$ (parametric equation of the circle with radius $\frac{\pi}{2}$ and centered at the origin).
$\vec{r'}(t) = (-\frac{\pi}{2} \sin t, \frac{\pi}{2} \cos t)$
$J_2 = \displaystyle \int_{C_1} \vec{F_2} \cdot d\vec{r} = \int_{\pi/2}^{3\pi/2} \vec{F_2} \cdot \vec{r'}(t) \, dt$
The line integral you are looking for over the semicircle from point $(0, \frac{\pi}{2})$ to $(0,-\frac{\pi}{2})$ oriented clockwise should be $ = J + J_1 + J_2 \,$ as $J$ is line integral over the full circle oriented clockwise, $J_1$ is the line integral of a conservative vector field between given points and $J_2$ is the line integral over semicircle oriented counterclockwise.
