Answer
The volume of inhaled air in the lungs at time $~t~$ is $~\frac{5}{4\pi}~ (1-cos~\frac{2\pi~t}{5})~liters$
Work Step by Step
We can find an expression for the volume of inhaled air in the lungs at time $t$:
$\int_{0}^{t}f(t)~dt$
$= \int_{0}^{t}\frac{1}{2}sin(\frac{2\pi~t}{5})~dt$
Let $u = \frac{2\pi~t}{5}$
$\frac{du}{dt} = \frac{2\pi}{5}$
$dt = \frac{5~du}{2\pi}$
$\int_{0}^{2\pi~t/5}\frac{1}{2}sin~u~\frac{5~du}{2\pi}$
$=\int_{0}^{2\pi~t/5}\frac{5}{4\pi}sin~u~du$
$=\frac{5}{4\pi}(-cos~u)~\vert_{0}^{2\pi~t/5}$
$=\frac{5}{4\pi}\cdot [(-cos~\frac{2\pi~t}{5})-(-cos~0)]$
$=\frac{5}{4\pi}~ (1-cos~\frac{2\pi~t}{5})$
The volume of inhaled air in the lungs at time $~t~$ is $~\frac{5}{4\pi}~ (1-cos~\frac{2\pi~t}{5})~liters$
