Evaluate $\int_0^{π/2} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx$

Question

Find the value of $$\int_0^{\pi/2} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx$$

I'm having a very tough time solving this question, can any body give me some hints?

Answer 1

Note that after lettimg $x=\pi/2-t$, we obtain $$I:=\int_{0}^{\frac{\pi}{2}} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx =\int_{0}^{\frac{\pi}{2}} \frac{(\frac{\pi}{2}-t)\sin t\cos t}{\sin^4 t+\cos^4 t}\,dt=\int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2}\sin t\cos t}{\sin^4 t+\cos^4 t}\,dt-I$$ which implies that \begin{align*} I&=\frac{\pi}{4}\int_{0}^{\frac{\pi}{2}} \frac{\sin t\cos t}{\sin^4 t+\cos^4 t}\,dt\quad (s:=\sin(t))\\ &=\frac{\pi}{4}\int_{0}^{1}\frac{s}{s^4+(1-s^2)^2}\,ds \\ &=\frac{\pi}{8}\int_{0}^{1}\frac{4s}{(2s^2-1)^2+1}\,ds \\ &=\frac{\pi}{8}\left[\arctan(2s^2-1)\right]_0^1=\frac{\pi^2}{16}. \end{align*}

Answer 2

Let $f(x) = \dfrac{\sin x\cos x}{\sin^4 x + \cos^4 x}$ and let (capital) $F$ by some antiderivative of (lower-case) $f.$ Then $$ \int_0^{\pi/2} x f(x)\,dx = \overbrace{\int x\,dv = xv-\int v\,dx}^{\Large\text{integration by parts}} = \left. xF(x)\vphantom{\frac 11} \, \right|_0^{\pi/2} - \int_0^{\pi/2} F(x)\, dx. $$ To find this, we need to find $F(x).$ \begin{align} F(x) & = \int \frac{\sin x\cos x}{\sin^4 x+ \cos^4 x}\, dx = \int \frac u {u^4 + (1-u^2)^2} \,du & & \text{where } u = \sin x, \\[10pt] & = \int \frac{dw/2}{w^2 + (1-w)^2} & & \text{where } w=u^2, \\[10pt] & = \frac 1 2 \int \frac{dw}{2w^2 - 2w+1}. \end{align}

And then: $$ 2w^2-2w+1 = \frac 1 2 \Big( (2w-1)^2+ 1 \Big) = \frac 1 2 (v^2 + 1), \qquad dw = dv/2 $$ So you get an arctangent. And $\arctan\left( 2\sin^2 x - 1\right) = -\arctan( \cos(2x)).$

And then you've got some more work to do.