Hello, Nath!
Your answrer is negative . . . otherwise, excellent work!
How would i find the area between the curves $\displaystyle y = \cos x$, the $\displaystyle y$axis
and the lines $\displaystyle y = \frac{1}{2}$ and $\displaystyle y = \frac{\sqrt{3}}{2}$
i keep getting: $\displaystyle \frac{\pi(2  \sqrt{3})  6(1 + \sqrt{3})}{12}$ . . . but i doubt this is right.
Code:

* *
_  *
√3/2 +    * 
::::::::::*
1/2 +      *  

+*
 π/2
I did it "sideways" . . . with respect to $\displaystyle y$.
We have: .$\displaystyle y = \cos x\quad\Rightarrow\quad x = \arccos y$
The area is: .$\displaystyle A \;= \;\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}} \arccos y\ dy$
This integral can be done byparts . . . or from a standard formula.
We get: .$\displaystyle A \;= \;y\cdot\arccos y  \sqrt{1  y^2}\:\bigg^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}} $
We have: .$\displaystyle \left[\frac{\sqrt{3}}{2}\cdot\arccos\frac{\sqrt{3}}{2}  \sqrt{1  \left(\frac{\sqrt{3}}{2}\right)^2} \right]$$\displaystyle  $$\displaystyle \left[\frac{1}{2}\cdot\arccos\frac{1}{2}  \sqrt{1  \left(\frac{1}{2}\right)^2}\right] $
. . . $\displaystyle = \;\left[\frac{\sqrt{3}}{2}\cdot\frac{\pi}{6}  \frac{1}{2}\right]  \left[\frac{1}{2}\cdot\frac{\pi}{3}  \frac{\sqrt{3}}{2}\right] \;=$ .$\displaystyle \frac{\pi\sqrt{3}}{12}  \frac{1}{2}  \frac{\pi}{6} + \frac{\sqrt{3}}{2}$
. . . $\displaystyle = \;\frac{\pi\sqrt{3}}{12}  \frac{6}{12}  \frac{2\pi}{12} + \frac{6\sqrt{3}}{12} \;=$ .$\displaystyle \frac{\pi\sqrt{3}  2\pi  6 + 6\sqrt{3}}{12}$
Therefore: .$\displaystyle A\;=\;\frac{\pi(\sqrt{3}  2) + 6(\sqrt{3}  1)}{12}$ . . . taDAA!