\(\int_{0}^{\pi} \frac{\cos ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x\) is equal to

Let,
\(I=\int_{0}^{\pi} \frac{\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x\)
\(I=2 \int_{0}^{\pi / 2} \frac{\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x...(i)\)
\(\left[\because \int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x\right.\), if \(\left.f(2 a-x)=f(x)\right]\)
\(\begin{aligned}
&\Rightarrow I=2 \int_{0}^{\pi / 2} \frac{\cos ^{4}\left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x \\
&\Rightarrow I=2 \int_{0}^{\pi / 2} \frac{\sin ^{4} x}{\cos ^{4} x+\sin ^{4} x} d x...(ii)
\end{aligned}\)
On adding Eqs. (i) and (ii), we get
\(\begin{aligned} 2 l &=2 \int_{0}^{\pi / 2} \frac{\sin ^{4} x+\cos ^{4} x}{\sin ^{4} x+\cos ^{4} x} d x \\ &=2 \int_{0}^{\pi / 2} d x=2[x]_{0}^{\pi r}=2\left(\frac{\pi}{2}-0\right) \\ \Rightarrow \quad 2 l &=\pi \Rightarrow l=\frac{\pi}{2} \end{aligned}\)