Explanation

(b)
$\because\sin^{4}{t + \cos^{4}t}$ is periodic with period $\frac{\pi}{2}$
Now, $f(x + \pi) = \int_{0}^{\pi + x}{(\sin^{4}{t + \cos^{4}{t)dt}}}$
$= \int_{0}^{x}{(\sin^{4}{t + \cos^{4}{t)dt}}} + \int_{0}^{x + \pi}{(\sin^{4}{t + \cos^{4}{t)dt}}}$
$= f(x) + \int_{0}^{\pi}{(\cos^{4}{t + \sin^{4}{t)dt}}} = f(x) + f(\pi)$
or
$= f(x) + 2\int_{0}^{\pi/2}{(\cos^{4}{t + \sin^{4}{t)dt}}}$
$= f(x) + 2f\left( \frac{\pi}{2} \right)$