Can I compute $\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}\cos{x}~~dx}{\cos^4{x}+\sin^4{x}}}$ using only properties of definite integrals?

Question

I have already computed the integral by considering the indefinite integral, and then putting the limits. The answer was $\frac{\pi}{4}$ which is half of the upper limit, so I am thinking that it should easily come to the answer, without computing a hard indefinite integral and just using properties of definite integrals (like the integral should probably just come to $\frac{1}{2}\int_{0}^{\frac{\pi}{2}}dx$ or something of the same sort). But I'm still unable to find the method to do so.

Answer 1

Another way: $$I=\int_{0}^{\pi/2} \frac{\sin x \cos x dx}{\sin^4 x+\cos^4x}=\int_{0}^{\pi/2} \frac{2\sin x \cos x dx}{2[(\sin^2x+\cos^2 x)^2-2\sin^2 x \cos^2 x]}=\int_{0}^{\pi/2} \frac{\sin 2x dx}{1+\cos^2 2x}$$ Use $t=\cos 2x$ $$\implies I=-\frac{1}{2}\int_{1}^{-1}\frac{ dt}{1+t^2}=\frac{\pi}{4}. $$

Answer 2

Hint: Under $u=\tan x$, one has \begin{eqnarray} &&\int_{0}^{\frac{\pi}{2}}{\frac{\tan{x}}{1+\tan^4{x}}}\frac{dx}{\cos^2x}\\ &=&\int_{0}^{\infty}{\frac{udu}{1+u^4}}\\ &=&\frac12\arctan (u^2)\bigg|_0^\infty\\ &=&\frac{\pi}{4}. \end{eqnarray}

Answer 3

\begin{align} \int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}\cos{x}}{\cos^4{x}+\sin^4{x}}}dx = \int_{0}^{\frac{\pi}{2}}{\frac{\sin{2x}}{1+\cos^2{2x}}} \overset{\tan t=\cos 2x}{dx} =\frac12\int_{-\pi/4}^{\pi/4}dt=\frac\pi4 \end{align}