Integral about lengh o f an arc I can't find a way to solve this: $$ \int_{\pi/2}^{\pi} \sqrt{8sen^2(t)cos^2(t)}dt $$ The integral is to calculate the length of an arc, by parametric equations. The answer is $\sqrt{2}$, but i'm finding $-\sqrt{2}$ instead. Can you solve this? Thanks!

On the second quadrant $\;\frac\pi2
$$\sin x>0\,,\,\,\,\cos x<0\implies \sqrt{8\sin^2x\cos^2x}=\sqrt8\sin x(-\cos x)=-2\sqrt2\sin x\cos x\implies$$

$$-2\sqrt2\int\limits_{\frac\pi2}^\pi\sin x\cos x\;dx=\left.-2\sqrt2\frac12\sin^2x\right|_{-\frac\pi2}^\pi=-\sqrt2\left(\sin\pi-\sin\frac\pi2\right)=\sqrt2$$