Integration Grandmaster - P12

$\begin{aligned} A & = \int_0^1 \frac {\sin 4x}{\sin^8 x + \cos^8 x} dx \\ & = \int_0^1 \frac {2\sin 2x \cos 2x}{(\sin^4 x + \cos^4 x)^2 - 2\sin^4 x \cos^4 x} dx \\ & = \int_0^1 \frac {2\sin 2x \cos 2x}{\left((\sin^2 x + \cos^2 x)^2-2\sin^2 x \cos^2 x\right)^2 - \frac 18 \sin^4 2 x} dx \\ & = \int_0^1 \frac {2\sin 2x \cos 2x}{\left(1-\frac 12 \sin^2 2 x \right)^2 - \frac 18 \sin^4 2 x} dx \\ & = \int_0^1 \frac {2\sin 2x \cos 2x}{1 - \sin^2 2x + \frac 18 \sin^4 2 x} dx & \small \blue{\text{Let }u = \sin 2x \implies du = 2 \cos 2x\ dx} \\ & = \int_0^{\sin 2} \frac {8u}{u^4 - 8u^2 + 8} du = \int_0^{\sin^2 2} \frac 4{t^2 - 8t + 8} dt & \small \blue{\text{Let }t = u^2 \implies dt = 2u \ du} \\ & = \int_0^{\sin^2 2} \frac 4{(t-4-2\sqrt 2)(t-4+2\sqrt 2)} dt \\ & = \frac 1{\sqrt 2} \int_0^{\sin^2 2} \left(\frac 1{t-4-2\sqrt 2} - \frac 1{t-4+2\sqrt 2} \right) dt \\ & = \frac 1{\sqrt 2} \ln \left(\frac {|t-4-2\sqrt 2|}{|t-4+2\sqrt 2|} \right) \bigg|_0^{\sin^2 2} \approx 0.773725069 \end{aligned}$

Therefore $\lfloor 10000 A\rfloor = \boxed{7737}$ .

Note that $\begin{aligned} \frac{\sin 4x}{\sin^8x + \cos^8x} & \equiv \; \frac{16\sin^4x}{(\cos2x-1)^4 + (\cos2x+1)^4} \; \equiv \; \frac{8\sin4x}{\cos^42x + 6\cos^22x + 1} \\ & \equiv \; \frac{32\sin4x}{(\cos4x+1)^2 + 12(\cos4x+1)+4} \; \equiv \; \frac{32\sin4x}{\cos^24x + 14\cos4x + 17} \; = \; 2\sqrt{2}\left(\frac{1}{\cos4x + 7 - 4\sqrt{2}} - \frac{1}{\cos4x + 7 + 4\sqrt{2}}\right)\sin4x \end{aligned}$ so that $\begin{aligned} A & = \; \frac{1}{\sqrt{2}}\left[ \ln\left(\frac{\cos4x + 7 + 4\sqrt{2}}{\cos4x + 7 - 4\sqrt{2}}\right)\right]_0^1 \; = \; \frac{1}{\sqrt{2}}\ln\left(\frac{(\cos4 + 7 + 4\sqrt{2})(\sqrt{2}-1)}{(\cos4 + 7 - 4\sqrt{2})(\sqrt{2}+1)}\right) \; = \; \frac{1}{\sqrt{2}}\ln\left(\frac{3\sqrt{2}+1 + (\sqrt{2}-1)\cos4}{3\sqrt{2}-1 + (\sqrt{2}+1)\cos4}\right) \end{aligned}$ which makes $\lfloor 10000A\rfloor = \boxed{7737}$ .