We will use these Results :
(R_1) : int_0^af(x)dx=int_0^af(a-x)dx(R1):∫a0f(x)dx=∫a0f(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.
Feel and Spread the Joy of Maths.!
