The quickest method would probably be to just expand the integrand:
[tex](\cos4x+\sin4x)^2=\cos^24x+2\cos4x\sin4x+\sin^24x=1+\sin8x[/tex]
Then
[tex]\displaystyle\int_0^{\pi/3}(1+\sin8x)\,\mathrm dx=\left(x-\dfrac18\cos8x\right)\bigg|_0^{\pi/3}[/tex]
[tex]=\left(\dfrac\pi3-\dfrac18\cos\dfrac{8\pi}3\right)-\left(0-\dfrac18\right)=\boxed{\dfrac3{16}+\dfrac\pi3}[/tex]
