Putnam definite integral evaluation $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$

Evaluate $$\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$$ Source : Putnam

By the property $\displaystyle \int_0^af(x)\,dx=\int_0^af(a-x)\,dx$:

$$=\int_0^{\pi/2}\frac{(\pi/2-x)\sin x\cos x}{\sin^4 x+\cos^4 x}dx=\frac{\pi}{2}\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}dx-\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$$

$$\Longleftrightarrow\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx=\frac{\pi}{4}\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx$$

Now I'm stuck. WolframAlpha says the indefinite integral of $\dfrac{\sin x\cos x}{\sin^4 x+\cos^4x}$ evaluates nicely to $-\frac12\arctan(\cos(2x))$.

I already factored $\sin^4 x+\cos^4 x$ into $1-\left(\frac{\sin(2x)}{\sqrt{2}}\right)^2$, but I don't know how to continue.. I suggest a substitution $u=\frac{\sin(2x)}{\sqrt{2}}$?

Could someone provide me a hint, or maybe an easier method I can refer to in the future?

Solution 1:

Here is one line proof

$$\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx=\int_0^{\pi/2}\frac{\tan(x) (\tan(x))'}{\tan^4(x)+1}\ dx=\int_0^{\infty} \frac{x}{x^4+1}\ dx=\left[\frac{\arctan(x^2)}{2}\right]_0^{\infty}=\frac{\pi}{4}$$

Q.E.D.

Solution 2:

$1 - \dfrac{\sin^2(2x)}{2} = \dfrac{1+\cos^2(2x)}{2}$, and $\sin x\cos x = \dfrac{\sin (2x)}{2} \Rightarrow \displaystyle \int \dfrac{\sin x\cos x}{\cos^4x+\sin^4x}dx = \displaystyle \int -\dfrac{1}{2}\dfrac{d(\cos(2x))}{1+\cos^2(2x)}dx = -\dfrac{1}{2}\arctan(\cos (2x)) + C$

Solution 3:

\begin{align} \int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}\mathrm dx&=\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\left(1-\sin^2x\right)^2}\mathrm dx\\[7pt] &=\int_0^{\pi/2}\frac{\sin x\cos x}{2\sin^4x-2\sin^2x+1}\mathrm dx\\[7pt] &=\frac14\int_0^1\frac{\mathrm dt}{t^2-t+\frac12}\qquad\color{blue}{\implies}\qquad t=\sin^2x\\[7pt] &=\frac14\int_0^1\frac{\mathrm dt}{\left(t-\frac12\right)^2+\frac14}\qquad\color{blue}{\implies}\qquad \frac u2=t-\frac12\\[7pt] &=\frac12\int_{-1}^1\frac{\mathrm du}{u^2+1}\\[7pt] &=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large\frac \pi4}} \end{align}

Solution 4:

\begin{aligned} I&= \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x}{1-\frac{\sin ^{2} 2 x}{2}} d x\\&=\int_{0}^{\frac{\pi}{2}} \frac{x \sin 2 x}{1+\cos ^{2} 2 x} d x \\&= \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin 2 x}{1+\cos ^{2} 2 x} d x \quad \text{ (By }x \mapsto \frac{\pi}{2}-x.)\\ &=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2 x}{1+\cos ^{2} 2 x} d x \\ &=-\frac{\pi}{8} \cdot\left[\tan ^{-1}(\cos 2 x)\right]_{0}^{\frac{\pi}{2}} \\ &=-\frac{\pi}{8}\left(-\frac{\pi}{4}-\frac{\pi}{4}\right) \\ &=\frac{\pi^{2}}{16} . \end{aligned}