You can use this function and integrate it to find the amount of volume of inhaled air inside the lungs.

Let's assume at t = 0, the volume of inhaled air is 0 L.

`f(t) = 2/5sin((2pit)/7)`

`V = int_0^tf(t)dt`

`V = int_0^t2/5sin((2pit)/7) dt`

`V = 2/5int_0^tsin((2pit)/7) dt`

`intsin((2pit)/7)dt =...

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You can use this function and integrate it to find the amount of volume of inhaled air inside the lungs.

Let's assume at t = 0, the volume of inhaled air is 0 L.

`f(t) = 2/5sin((2pit)/7)`

`V = int_0^tf(t)dt`

`V = int_0^t2/5sin((2pit)/7) dt`

`V = 2/5int_0^tsin((2pit)/7) dt`

`intsin((2pit)/7)dt = -cos((2pit)/7)/((2pi)/7)`

`intsin((2pit)/7)dt = -(7/(2pi))cos((2pit)/7)`

Therefore,

`V = 2/5int_0^tsin((2pit)/7) dt = -(7/(2pi))[cos((2pit)/7) - cos(0)]`

`V = -(7/(2pi))[cos((2pit)/7) - 1]`

`V = (7/(2pi))[1-cos((2pit)/7)]`