We will use these Results :
# (R_1) : int_0^af(x)dx=int_0^af(a-x)dx#.
# (R_2)(i) : int_0^(2a)f(x)dx=0, if f(2a-x)=-f(x), and, #
# (R_2)(ii) : int_0^(2a)f(x)dx=2int_0^af(x)dx, if f(2a-x)=f(x)#.
Let, #I=int_0^pi (xsin^4x)/(sin^4x+cos^4x)dx#.
#:. I=int_0^pi{(pi-x)sin^4(pi-x)}/{sin^4(pi-x)+cos^4(pi-x)}dx,...[(R_1)]#
#=int_0^pi {(pi-x)sin^4x}/(sin^4x+cos^4x)dx#,
#=piint_0^pi sin^4x/(sin^4x+cos^4x)dx-int_0^pi(xsin^4x)/(sin^4x+cos^4x)dx#,
#i.e., I=piint_0^pi sin^4x/(sin^4x+cos^4x)dx-I#,
#or, 2I=piint_0^pi sin^4x/(sin^4x+cos^4x)dx#.
Now, before applying #(R_2),# let us work out
#f(2a-x)=f(pi-x)" for "f(x)=sin^4x/(sin^4x+cos^4x)#.
From above, we see that, #f(pi-x)=f(x)#.
Hence, by #(R_2)(ii)#,
#(2I)/pi=2int_0^(pi/2)sin^4x/(sin^4x+cos^4x)dx#,
#or, I/pi=int_0^(pi/2)sin^4x/(sin^4x+cos^4x)dx...(1)#.
But for #f(x)=sin^4x/(sin^4x+cos^4x)#,
#f(pi/2-x)=sin^4(pi/2-x)/{sin^4(pi/2-x)+cos^4(pi/2-z)}dx#,
#=cos^4x/(cos^4x+sin^4x)#.
Therefore, by applying #(R_1)#, on #I/pi#,
#I/pi=int_0^(pi/2)cos^4x/(cos^4x+sin^4x)dx......(2)#.
Finally, we add #(1) and (2)# to get,
#(2I)/pi=int_0^(pi/2){(sin^4x+cos^4x)}/{(cos^4x+sin^4x)}dx#,
#=int_0^(pi/2)1dx#,
#=[x]_0^(pi/2)#,
#:. (2I)/pi=pi/2," giving, "#
#I=pi^2/4#.
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