Explanation

(a)
$g(x) = \int_{0}^{x}{\cos^{4}{t\ dt}}$ (given)
$g(x + \pi) = \int_{0}^{\pi + x}{\cos^{4}{t\ dt}}$
$= \int_{0}^{\pi}{\cos^{4}{t\ dt + \int_{\pi}^{\pi + x}{\cos^{4}{t\ dt = I_{1} + I_{2}}}}}$
$\Rightarrow \ I_{1} = g(\pi)$
$I_{2} = \int_{\pi}^{\pi + x}{\cos^{4}{t\ dt}}$
Put $t = \pi + y\ \Rightarrow dt = dy$
$I_{2} = \int_{0}^{x}{\cos^{4}{(y + \pi)dy =}}\int_{0}^{x}{\left\lbrack \cos(\pi + y) \right\rbrack^{4}dy}$
$= \int_{0}^{x}{( - {\cos{y)}}^{4}dy = \int_{0}^{x}{\cos^{4}{y\ dy = g(x)}}}$
$\therefore\ \ g(x + \pi) = g(x) + g(\pi)$