Evaluate $\int_0^{π/2} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx$ 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?
 A: 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*}
A: 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.
