The problem requires me to find the area of the region cut from the first quadrant by the curve $r=2(2-\sin2\theta)^{1/2}$
Since it is in the first quadrant, I set $0\le\theta\le\frac{\pi}{2}$.
Also I set $0\le r\le 2(2-\sin2\theta)^{1/2}$.
Now I have that
$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2(2-\sin2\theta)^{1/2}}rdrd\theta$$
It seems that I have to convert the double integral from polar to Cartesian, since I was in trouble with that power $\frac{1}{2}$.
$$ \begin{split} r &= 2(2-\sin2\theta)^{1/2} \\ \implies r &= 2(2-2\sin\theta\cos\theta)^{1/2} \\ \implies r^2 &= 8(1-\sin\theta\cos\theta) \\ \implies x^2+y^2 &= 8(1-\sin\theta\cos\theta) \end{split} $$ and then I stuck here.
Am I on the wrong track?
You can see the graph of your curve below:
As you wrote, this sets up the integral as $$ \begin{split} I &= \int_0^{\pi/2} \int_0^{2(2-\sin(2\theta))^{1/2}}rdrd\theta \\ &= \int_0^{\pi/2} \left[\frac{r^2}{2} \right]_0^{2(2-\sin(2\theta))^{1/2}} d\theta \\ &= \int_0^{\pi/2} \frac{2(2-\sin(2\theta)}{2} d\theta \\ &= \int_0^{\pi/2} (2-\sin(2\theta) d\theta \\ \end{split} $$
Can you now finish this?