Area in Polar Coordinates 2

Ok so the equation is

r= 6 +4 sin theta

Its is a limacon so the area will be symmetrical from 0 to pi/2

Plugging this into the area formula I get

Integral (6+4 sin theta)^2

Integral (36 + 48 sin theta + 16 sin^2 theta)

Integral (36 + 48 sin theta + 16 [(1-cos 2 theta)/2]

Integral (36 + 48 sin theta + 8 (1 - cos 2 theta)

Integral (36 + 48 sin theta + 8 - 8 cos 2 theta)

Integral (44 + 48 sin theta - 8 cos 2 theta)

which gives me

44 theta - 48 cos theta - 8 sin 2 theta + C

I worked it out and I am still unable to arrive at the answer. Can anyone tell me what I'm doing wrong?