If $g(x)=\int_0^x \cos ^4 t \mathrm{~d} t$, then $g(x+\pi)$ equals
If $g(x)=\int_0^x \cos ^4 t \mathrm{~d} t$, then $g(x+\pi)$ equals
$g(x)+g(\pi)$
$g(x)-g(\pi)$
$\frac{g(x)}{g(\pi)}$
$g(x) \cdot g(\pi)$
Solution
$\begin{aligned}
& g(x)=\int_0^x \cos ^4 t \mathrm{~d} t \\
& \Rightarrow g(x+\pi)=\int_0^{x+\pi} \cos ^4 t \mathrm{~d} t=\int_0^x \cos ^4 t \mathrm{~d} t+\int_x^{x+\pi} \cos ^4 t \mathrm{~d} t \\
& =g(x)+\int_x^{x+\pi} \cos ^4 t \mathrm{~d} t \\
& =g(x)+\int_0^\pi \cos ^4 t \mathrm{~d} t
\end{aligned}$
[as $\cos ^4 t$ is a periodic function with period $\left.\pi\right]=g(x)+g(\pi)$