sorry if you cant see but its between pie/2 and 0, thanks.
Hello, Misa-Campo!
$\displaystyle \int^{\frac{\pi}{2}}_0 \sin^2(2x)\,dx$
Use the identity: .$\displaystyle \sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}$
And we have: .$\displaystyle \int^{\frac{\pi}{2}}_0 \sin^2(2x)\,dx \;=\;\frac{1}{2}\int^{\frac{\pi}{2}}_0(1 - \cos 4x)\,dx $
. . Got it?