Answer
the volume of inhaled air in the lungs at time $t$ is.
$$
\begin{aligned}
V(t) &=\int_{0}^{t} f(u) d u\\
&=\int_{0}^{t} \frac{1}{2} \sin \left(\frac{2 \pi}{5} u\right) d u \\
&=\frac{5}{4 \pi}\left[1-\cos \left(\frac{2 \pi}{5} t\right)\right] \text { liters }
\end{aligned}$$
Work Step by Step
the volume of inhaled air in the lungs at time $t$ is.
$$
\begin{aligned}
V(t) &=\int_{0}^{t} f(u) d u\\
&=\int_{0}^{t} \frac{1}{2} \sin \left(\frac{2 \pi}{5} u\right) d u \\
&=\int_{0}^{2 \pi t / 5} \frac{1}{2} \sin v\left(\frac{5}{2 \pi} d v\right) \\
& \quad \quad \quad \quad\left[\text { substitute } v=\frac{2 \pi}{5} u, d v=\frac{2 \pi}{5} d u\right] \\
&=\frac{5}{4 \pi}[-\cos v]_{0}^{2 \pi t / 5} \\
&=\frac{5}{4 \pi}\left[-\cos \left(\frac{2 \pi}{5} t\right)+1\right] \\
&=\frac{5}{4 \pi}\left[1-\cos \left(\frac{2 \pi}{5} t\right)\right] \text { liters }
\end{aligned}$$
