If you know what the curve looks like it becomes a bit easiar to evaluate the area. Look at cos^2 x - Wolfram|Alpha and let us know how you want to approach this now.
Hello, Riazy!
$\displaystyle \text{A polar curve is given: }\:r \,=\,\cos^2\!\theta,\;0 \le \theta \le 2\pi$
$\displaystyle \text{Calculate the area enclosed by the curve.}$
$\displaystyle \displaystyle \text{Formula: }\:A \;=\;\tfrac{1}{2}\!\!\int^{\beta}_{\alpha}\!\! r^2\,d\theta $
$\displaystyle \displaystyle \text{We have: }\;A \;=\;\tfrac{1}{2}\!\!\int^{2\pi}_0\!\! \cos^4\!\theta\,d\theta $
Why involve complex variables? . . . Why not standard Trig identities?
$\displaystyle \displaystyle \tfrac{1}{2}\cos^4\!\theta \;=\;\tfrac{1}{2}(\cos^2\!\theta)^2 \;=\;\tfrac{1}{2}\left(\frac{1+\cos2\theta}{2}\rig ht)^2 \;=\;\tfrac{1}{8}(1 + 2\cos2\theta + \cos^2\!2\theta)$
. . $\displaystyle =\;\frac{1}{8}\left(1 + 2\cos2\theta + \dfrac{1 + \cos4\theta}{2}\right) \;=\;\frac{1}{8}\left(\frac{3}{2} + 2\cos2\theta + \frac{1}{2}\cos4\theta\right) $
. . $\displaystyle =\;\frac{1}{16}(3 + 4\cos2\theta + \cos4\theta) $
$\displaystyle \displaystyle \text{We have: }\:A \;=\;\tfrac{1}{16}\!\int^{2\pi}_0 \!\!\left(3 + 4\cos2\theta + \cos4\theta \right)\,d\theta $
. . $\displaystyle =\;\frac{1}{16}\!\bigg[3\theta + 2\sin2\theta + \frac{1}{4}\sin4\theta\bigg]^{2\pi}_0$
. . $\displaystyle =\;\frac{1}{16}\bigg[\left(6\pi + 2\sin4\pi + \frac{1}{4}\sin8\pi\right) - \left(0 + 2\sin0 + \frac{1}{4}\sin0\right)\bigg] $
. . $\displaystyle =\;\;\dfrac{3\pi}{8}$
Okay there is one thing I didnt understand, sorry but i dont consider this to be a bump as I have got my answer, but I would like to ask
how you got 1/8 into 1/16 I can see that you multiplied everything within the brackets with 2. but what about 1/16?
Thanks for answers!