If $g(x)=\int\limits_0^x \cos ^4 t d t$, then $g(x+\pi)=$

$g(x)+g(\pi)$ $g(x)-g(\pi)$ $g(x) g(\pi)$ $\frac{g(x)}{g(\pi)}$

$g(x)+g(\pi)$

$g(x+\pi)=\int\limits_0^{\pi+x} \cos ^4 t d t$ $=\int\limits_0^\pi \cos ^4 t d t+\int\limits_\pi^{\pi+x} \cos ^4 x d x$ (Put $t=\pi+\theta$ is second integral) $=\int\limits_0^\pi \cos ^4 t d t+\int\limits_0^x \cos ^4 t d t=g(\pi)+g(x)$ Hence (1) is the correct answer.