Solution:
\(\displaystyle I=\int_{0}^{\pi}\frac{x^{2}\sin x\cos x}{\sin ^{4}x+\cos ^{4}x}dx \) \( \quad (1) \)
\(\displaystyle I=\int_{0}^{\pi}\frac{(\pi -x)^{2}\sin x\cos x}{\sin ^{4}x+\cos ^{4}x}dx \) \( \quad (2) \)
\(\displaystyle 2I=\int_{0}^{\pi}\frac{\sin x\cos x}{\sin ^{4}x+\cos ^{4}x}\left(x^{2}-(\pi -x)^{2}\right)dx=\int_{0}^{\pi}\frac{\sin x\cos x}{\sin ^{4}x+\cos ^{4}x}\left(-\pi ^{2}+2\pi x\right)dx \)
\(\displaystyle =-\pi ^{2}\int_{0}^{\pi}\frac{\sin x\cos x}{\sin ^{4}x+\cos ^{4}x}dx+2\pi \int_{0}^{\pi}\frac{x\sin x\cos x}{\sin ^{4}x+\cos ^{4}x}dx \)
\( 2I=2\pi I_{1}\)
\( I=\pi I_{1}\quad ...(1) \)
\(\displaystyle I_{1}=\int_{0}^{\pi}\frac{x\sin x\cos x}{\sin ^{4}x+\cos ^{4}x}dx \)
Let \(\displaystyle x=\dfrac{\pi }{2}-t =-\int_{\pi /2}^{-\pi /2}\dfrac{\left(\dfrac{\pi }{2}-t\right)\cos t\sin t\,dt}{\cos ^{4}t+\sin ^{4}t}\)
\(\displaystyle I_1=\int_{-\pi /2}^{\pi /2}\dfrac{\left(\dfrac{\pi }{2}-t\right)\sin t\cos t\,dt}{\cos ^{4}t+\sin ^{4}t} \)
\(\displaystyle \frac{\pi }{2}\int_{-\pi /2}^{\pi /2}\frac{\sin t\cos t}{\sin ^{4}t+\cos ^{4}t}dt+\int_{-\pi /2}^{\pi /2}\frac{t\sin t\cot t}{\sin ^{4}t+\cos ^{4}t}dt \)
\(\displaystyle I_{1}=2\int_{-0}^{\pi /2}\frac{t\sin +\cos t\ dt}{\sin ^{4}t+\cos ^{4}t}dt \quad ...(3) \)
\(\displaystyle I_1 = 2 \int_{0}^{\pi/2} \dfrac{\left(\dfrac{\pi}{2} - t\right) \sin t \cos t \, dt}{\sin^4 t + \cos^4 t} \quad ...(4) \)
\(\displaystyle 2I_{1}=2\int_{0}^{\pi /2}\frac{\dfrac{\pi }{2}\sin t\cos t \,dt }{\sin ^{4}t+\cos ^{4}t}=\pi \int_{0}^{\pi /2}\frac{\sin t\cos t}{\sin ^{4}t+\cos ^{4}t}dt \)
\( \sin ^{2}t=p\)
\( 2\sin t\cos t\,dt=dt \)
\(\displaystyle =\frac{\pi }{2}\int_{0}^{1}\frac{dt}{p^{2}+(1- p)^{2}}=\frac{\pi }{2}\int_{0}^{1}\frac{dt}{2p^{2}-2p+1}=\frac{\pi }{4}\int_{0}^{1}\frac{dt}{p^{2}-p^1\frac{1}{2}}=\frac{\pi }{4}\int_{0}^{1}\frac{dp}{\left(p-\dfrac{1}{2}\right)^{2}+\left(\dfrac{1}{2}\right)^{2}}=\frac{\pi }{4} \frac{1 }{4}\left[\tan ^{-1}\left(\frac{p-\dfrac{1}{2}}{\dfrac{1}{2}}\right)\right]_{0}^{1} \)
\(\displaystyle 2I_{1}=\frac{\pi }{4}\left(\frac{\pi }{4}+\frac{\pi }{4}\right)=\frac{\pi ^{2}}{8}\times 2=\frac{\pi ^{2}}{4} \)
Therefore, the answer is \( 15 \).

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