in this video we're presented with a fairly simple trig integral you can see that inside
the square root of this thing we can use a pythagorean identity to simplify it
but there's a really subtle sign issue that arises and it involves the absolute value function so
what we're going to do in this video is just churn through it and ignore the absolute value and see
what goes wrong and then we'll go back and think carefully about how things have to be adjusted
and we should be able to get a sensible result out of the integral so to begin with we're going to
apply the identity sine squared x equals 1 minus cosine squared x and this means the inside of the
square root reduces to the square of the sine function now what i mean by proceeding naively
through the integral is that we could just say the square root undoes the squaring function
and again i'm working through it this way to illustrate how it gets us into trouble to be
that casual about the square root of a square and there's a couple ways to argue what happens
next i could guess the anti-derivative and evaluate across the endpoints or i could say
look the sine function spins half of its time above the x-axis and half of its time below
the x-axis on each period and i'm integrating on exactly one period so i know this has to
come out to zero if you want a little more formal argument i guess the anti-derivative
evaluate across the endpoints and when i plug in the upper limit i get a negative one
then i subtract what i get when i plug in the lower limit and i get zero and then i have to ask
myself does it make any sense that this integral came out to zero if i go back to my original
integral that square root instead of the integral is always non-negative and usually it's positive
when the square of the cosine is less than one so how can i possibly integrate a thing that's
usually positive and never negative and come out with zero this is impossible so something's gone
wrong and where it's gone wrong is back here where i said the square root of the square of the sine
function is just equal to the sine function that's not quite right because the square of the sine
function is always positive and then when i square root it i'm guaranteed to get a positive result
the square root of a square of a thing is just the absolute
value of it in other words it erases minus signs just like the absolute value does so really what
i'm looking at here is the integral from 0 to 2 pi of the absolute value of the sine of x
dx and so the absolute value is going to leave the sine function alone
provided the sine function is non-negative and that happens on the interval 0 to pi
so we're going to split up the integral and then the absolute value function multiplies the
sine function by negative 1 in order to make it positive whenever it's negative and that happens
on pi to 2 pi now we can guess anti-derivatives and evaluate across the limits of integration
and when i plug the upper limit into that first term i get negative negative 1 that's
positive 1 and then i subtract negative cosine 0 so i'm adding cosine zero and that's a one
in my second half i get the cosine of two pi which is one and then i subtract
the cosine of pi which is negative one okay so this turns into one plus one plus one plus one
and that is four and that makes a lot more sense than saying this thing is equal to zero so just to
wrap things up here i popped in a picture real quick of the square root of one minus cosine
squared x and i shaded in the area represented by the integral that we've just computed now clearly
this thing is not zero and if i check to see how reasonable four is as an answer i put grid
spaces on this graph and you can see that in the first bump i have about two total spaces
in the second bump i have about two total spaces and so it looks like four is a reasonable answer
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