CALCULUS HHHHEEEEEELLLPPPP!!!!!

Given the curves r=2sin(x) and r=2sin(2x), 0=x=pi/2, find the area of the region outside the first curve and inside the second curve.

r=2sin x
r=2sin2x

they cut when 2sinx=2sin2x
sinx=sin2x
sinx= 2sinx cos x
cosx=1/2 ie. they cut at x= pi/3 (60°)
So the area is between 0
In polar coodinates
dA= rdrdx
From 0 A= INT INT rdrdx
0 A=INT r^2/2 dx
A=(1/2) INT 4sin^2 x dx

A= 2INT sin^2x dx
A= 2 INT (1-cos2x) /2 dx
A= ( x- (1/2) sin2x )
A1=(pi/3 -(1/2) sin 2pi/3)

A2= INTr^2/2 dx

0 pi/3
A2= 2INT sin^2 2x dx
A2= INT (1-cos4x) dx
A2= x- (1/4) sin4x
A2=(pi/2-0)- (pi/3-(1/4)sin(4pi/3)
A2= pi/6+ (1/4) sin(4pi/3)

A= A1+A2 = (pi/3 -(1/2) sin 2pi/3) +pi/6+ (1/4)sin(4pi/3)
A= pi/2-(1/2) (sin 2pi/3) + (1/4)sin(4pi/3)

sin(2pi/3) =sin (pi/2+pi/6) =cos (pi/6) = (1/2)sqrt3
sin (4pi/3)= sin (pi+pi/3) = -sin pi/3 = -(1/2)sqrt3

A= pi/2- (1/2)(1/2)sqrt3 +(1/4) (-1/2)sqrt3
A=pi/2 -3/8 sqrt3