Using properties $of$ the integral, $we$ know the following.

${\displaystyle \underset{0}{\overset{12}{}}}(41+10sin(\frac{t}{12}))dt={\displaystyle \underset{0}{\overset{12}{}}}41dt+10{\displaystyle \underset{0}{\overset{12}{}}}sin(\frac{t}{12})dt$

$By$ evaluating the first integral, $we$ get

${\displaystyle \underset{0}{\overset{12}{}}}41dt=[x{]}_0^{12}$

$=$