This problem is mostly about remembering the trigonometric identities, in particular, the double-angle formulae. Recall that $$ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta,\qquad \sin(2\theta) = 2\sin\theta \cos\theta; $$ with this, the radicand in the numerator becomes $$ \sqrt{1 - \cos(4x)} = \sqrt{1 - \cos^2 (2x) + \sin^2 (2x)} = \sqrt{2\sin^2(2x)} = 2\sqrt2 \sin x \cos x, $$ and therefore, also completing the square in the denominator $$ {\rm I} = \lint_0^{\pi/2} \frac{2\sqrt2 \sin x\cos x}{1 + (\sin^2 x - 1)^2} \diff x = \left\{\begin{aligned} t &= \sin^2 x - 1\\ \diff t &= 2\sin x \cos x\diff x \end{aligned}\right\} = \lint_{-1}^0 \frac{\sqrt2 \diff t}{1 + t^2} = \sqrt2 \arctan 1 = \color{green}\boxed{\frac{\sqrt2 \pi}{4}.} $$