Evaluate: ∫π/2 0 xsinxcosx/(sin4x+cos4x)dx

I=∫π/2 0 xsinxcosx/(sin4x+cos4x)dx— (1)
I=∫π/2 0 ((π/2−x)sinxcosx)/(sin4x+cos4x)dx—(2)
Adding (1) and (2), we get
2I=π/2∫π/2 0 sinxcosx/(sin4x+cos4x)dx
Putsin2x=t⇒2sinxcosxdx=dt
Therefore, sinxcosxdx=dt/2
⇒2I=π/2×1/2∫10 dt/(t^2+(1-t)^2)
I=π/8∫10 dt/(2t^2𕒶t+1)
⇒I=π/16∫10 dt/(t󔼔)^2+1/4
=π/16[1/2tan𕒵⎛⎜⎜⎜⎝t�⎞⎟⎟⎟⎠∣∣∣∣∣10
=π/32