If $\mathrm{f}(x)=\frac{\sin ^2 \pi x}{1+\pi^x}$, then $\int(\mathrm{f}(x)+\mathrm{f}(-x)) \mathrm{d} x$ is equal to

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$\begin{aligned} & \int(\mathrm{f}(x)+\mathrm{f}(-x)) \mathrm{d} x \ & =\int\left[\frac{\sin ^2 \pi x}{1+\pi^x}+\frac{\sin ^2(-\pi x)}{1+\pi^{-x}}ight] \mathrm{d} x \ & =\int\left(\frac{\sin ^2 \pi x}{1+\pi^x}+\frac{\pi^x \sin ^2 \pi x}{\pi^x+1}ight) \mathrm{d} x\end{aligned}$ $\begin{aligned} & =\int \sin ^2 \pi x\left(\frac{1+\pi^x}{1+\pi^x}ight) d x \ & =\int \sin ^2 \pi x \mathrm{~d} x \ & =\int\left(\frac{1-\cos 2 \pi x}{2}ight) \mathrm{d} x \ & =\frac{x}{2}-\frac{1}{2} \cdot \frac{\sin 2 \pi x}{2 \pi}+c=\frac{x}{2}-\frac{\sin 2 \pi x}{4 \pi}+c\end{aligned}$

If $\mathrm{f}(x)=\frac{\sin ^2 \pi x}{1+\pi^x}$, then $\int(\mathrm{f}(x)+\mathrm{f}(-x)) \mathrm{d} x$ is…

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