Question

If $g\left(x\right)={\int }_0^x\cos 4tdt,$ then $g(x+\pi )$ equals

Answer 1

$g\left(x\right)={\int }_0^x\cos 4tdt,$
$={\left[\frac{\sin 4t}{4}\right]}_0^x$
$g\left(x\right)=\frac{\sin 4x}{4}-0$
$g\left(x\right)=\frac{\sin 4x}{4}$
$g(x+\pi )=\frac{\sin 4(x+\pi )}{4}$
$=\frac{\sin (4\pi +4x)}{4}$
$=\frac{\sin 4x}{4}$