Area between two polar curves $r = 2 \sin\theta$ and $r =2\cos\theta$ I am trying to find the area between the polar curves $r = 2 \sin θ$ and $r = 2 \cos θ$.

I set up the area equation as follows:

$$\frac12\int_0^{\pi/4}((2\sinθ)^2-(2\cosθ)^2)\,d\theta$$

I could not get the correct answer with this, which is $\frac\pi2-1$.

Any help with this problem would be appreciated :D
A picture might help: $r=2\sin t$ is in blue and $r=2\cos t$ is in red for $0\le t\le 2\pi$.



The intersection point is found by setting $2\sin t=2\cos t$ and solving; it occurs at $t=\pi/4$, so you want
$$
\underbrace{{1\over 2}\int_0^{\pi/4} (2\sin t)^2\,dt}_\text{area of blue region}
+
\underbrace{{1\over 2}\int_{\pi/4}^{\pi/2} (2\cos t)^2\,dt}_\text{area of red region}={\pi\over 2}-1.
$$
this is just a hint and you have to do it by yourself so see below.....
by using 12th method
                so 
we know by polar co-ordinates 
   $x= r \sin(x)$  and $y=r \cos(x)$
    here   in question 

$r=2\sin(x)$

multiply both side $r$

$r^2=2r\sin(x)$

$r^2 = 2x$   since $x=r \sin(x)$

since a point circle equation is $x^2+y^2=r^2$

therefore
$x^2+y^2=2x $ implies that   $y=\sqrt (2x-x^2)$
 similarly 
for $r=2\cos(x)$,
 $x^2+y^2=2y$  implies that   $y= \sqrt (2y-x^2)$
and integrate it by drawing the circle 

get the answer.
best of luck