Integration Grandmaster - P9

$A = \int_{0}^{1} \frac{\sin{x} \cos{x}}{\sin^4{x} + \cos^4{x}} \ dx$ $A= \int_{0}^{1} \frac{\sin{x} \cos{x}}{(\sin^2{x} + \cos^2{x})^2-2\sin^2{x}\cos^2{x}} \ dx$ $A= \int_{0}^{1} \frac{2\sin{x} \cos{x}}{2(\sin^2{x} + \cos^2{x})^2-4\sin^2{x}\cos^2{x}} \ dx$ $A= \int_{0}^{1} \frac{\sin(2x)}{2-\sin^2(2x)} \ dx$ $A= \int_{0}^{1} \frac{\sin(2x)}{1+\cos^2(2x)} \ dx$

$\cos(2x) = z$ $\implies -2\sin(2x) \ dx = dz$ $A= \frac{1}{2} \int_{\cos{2}}^{1}\frac{dz}{1 + z^2}$ $A= \frac{1}{2} \left(\frac{\pi}{4} - \arctan(\cos{2})\right)$ $\boxed{\lfloor 10000A \rfloor = 5898}$

Let's evaluate this integral by the following method: First take

$u = \sin(x), du = \cos(x) dx, u \in [0, \pi/2] \Rightarrow \int_{0}^{\pi/2} \frac{u}{u^4 + (1-u^2)^2} du =\int_{0}^{\pi/2} \frac{u}{2u^4 -2 u^2+1} du = \frac{1}{2} \cdot \int_{0}^{\pi/2} \frac{u}{(u^2 - 1/2)^2 + (1/2)^2} du$ (i)

Now take:

$v = u^2 - 1/2, dv = 2u du, v \in [-1/2, (\pi^2-2)/4] \Rightarrow \frac{1}{4} \cdot \int_{-1/2}^{(\pi^2-2)/4} \frac{1}{v^2 + (1/2)^2} dv = \frac{1}{2}\arctan(2v)|_{-1/2}^{(\pi^2-2)/4} \approx 0.58983$ (ii)

which gives us $\lfloor 10000 \cdot 0.58983 \rfloor = \boxed{5898}.$