in this video we're looking at an interesting integral that popped up for me recently in the
context of a larger problem it's the integral from 0 to 2 pi square root 1 minus cosine t
dt now the first solution method that occurred to me was to use what i'll call the conjugate
method on this integral and something really interesting happens when you take this approach
to the integral if you aren't extremely careful about what you're doing with plus and minus signs
when you evaluate this integral you get zero and i know that's impossible because the integrand of
this thing is always non-negative it's got to come out to a positive answer so i'll discuss sort of
what goes wrong and how to fix it then we're going to look at a different approach to the
integral that i'll call the trig identity method it's just a different trig identity that cleverly
bypasses the sine issues from the first method so here's the basic idea of the conjugate method
i want to build a 1 minus cosine squared inside of a square root because i know there's an identity
that can reduce that to sine squared so i'm going to multiply top and bottom by the conjugate
then i go ahead and multiply the interiors of those two square roots in the numerator
and that's a difference of two squares by design and i'm going to get a 1 minus cosine squared
inside that square root then we simplify the 1 minus cosine squared to a sine squared and i have
the square root of sine squared now we're used to in so many problems just simplifying that to the
sine of t but i'll give you a warning that this is where our tricky sine issue is hiding today
first let's see what goes wrong if we don't notice that so i'm going to manipulate the integral
a little more just to highlight how the chain rule is working backwards here
i'm putting a minus sign on the sine t because that's the derivative of the interior of the
square root piece and then i'm compensated out in front with another minus sign all right so
i have 1 plus cosine of t raised to a power right next to the derivative of that interior
and i'm ready to guess the antiderivative so i add one to the exponent divide by the resulting number
so this gives me a one plus cosine t to the one half divided by one half which is
the same as multiplying by two and i've got my antiderivative and now i see a serious problem
if i plug in the upper limit the cosine of 2 pi is 1 and i end up with negative 2 root 2
and then i subtract what i get when i plug in the lower limit and again the cosine is 1. the cosine
is periodic with a period of 2 pi so i've got to get the same number for the upper and lower limits
and these are going to add to zero well this is impossible because i have the integral of
something that's always non-negative i know i need to get a positive result so as i already said the
gremlin that's lurking in this problem is a really tricky sine issue that occurs right at this step
if i look at the square root of the square of the sine function i know that that has got to always
be a positive number because i'm squaring the sine function which gives me guaranteed
a positive number before i square root it in fact the square root of the square of something
is really just another way of saying the absolute value so it takes whatever value is there
and erases the minus sign so really at this step i should have said it's the absolute value
of the sine function i'm guaranteed to get a positive result
and on the interval 0 to pi there's no difference between the absolute value of sine t and sine
t because the sine function is positive once you cross over to the interval from pi to 2 pi
then the sine function becomes negative and i have to write the absolute value as the negative
of the sine function so the way to proceed with this integral is to actually split the integration
interval at pi and do two separate integrals so we'll go ahead and erase our work from before
and we're going to split the integration interval so i go from zero to pi
sine t over square root 1 plus cosine t
dt plus the integral from pi to 2 pi
of negative sine t again the sine function is negative on that interval so negative sine t is
positive now the actual symbolic manipulations to get the antiderivative we've already done it once
before so on the first integral i throw in a minus sign to get ready to guess the anti-derivative
put a minus sign out in front and now i have the derivative of the interior of this thing
and i just use the power rule and i get negative 2 square root 1 plus cosine t evaluated from 0 to pi
and my second integral already comes with the minus sign that i needed inside the integrand
so i can go ahead and guess the anti-derivative so in the first piece when i plug in the upper limit
of pi the interior of the square root vanishes because the cosine of pi is negative 1. and when i
plug in the lower limit i need to subtract that so i'm going to get a plus 2 and then the square root
of 1 plus cosine 0 cosine 0 is 1 so i have 2 root 2. then in my second piece
i plug in the 2 pi for cosine in the upper limit
again the cosine of 2 pi is just 1 so i end up with 2 root 2 and then in my lower limit
i have a 1 plus cosine pi which is 0 inside that square root so in the end i end up with
4 square root of 2 for the value of this integral all right so let's look at a method
that bypasses all of these issues entirely and the idea is to just find a different trig identity
to simplify the interior of that square root right in the first step so i want to remind
you of a really useful trig identity when you have the square of a sine or a cosine
these identities can be immensely useful so when i have the square of a sine
function in particular that's one half times the quantity one minus cosine of twice the angle and
what i notice in this identity is the one minus cosine of something is related to a square of
a sine function directly well this means i might be able to change the interior of my square root
to just the square of a sine function so i'll go ahead and multiply both sides by
two and now i'll just replace theta with t over two and i end up with two sine squared t over two
is equal to one minus cosine of t so i'm going to replace 1 minus cosine t in the
integral with 2 sine squared t over 2. and the square root undoes the squaring inside of that
i'm also going to pull out a factor of square root of 2 out in front as i'm going and at this
point i want to highlight that same sign issue can we really just say the square root undoes
the squaring without worrying about these absolute value bars like we did in the last example
and the fortunate thing about this approach to the integral if i look at the argument of
the sine function here that's t over 2 and if i go to my upper limit of integration i'm only
getting to the sine of pi in other words we're staying on that first half of the first period
of the sine function where it's always positive and that's why we get away with just ignoring the
sine issue this time so to get the integral done i just i set up to make the chain rule backwards
easier to think about so i need a one half in here and i'll compensate with a 2 out in front
that one half is just the derivative of the argument of the sine function it makes it easier
to see the chain rule backwards so i have a 2 root 2 out in front and then the thing i differentiate
to get 1 half sine t over 2 is negative cosine t over two the derivative would undo that minus sign
and the chain rule would produce that one half so when i plug in the upper limit i get negative
cosine pi so negative negative one then when i plug in the lower limit and i'm subtracting
that so i've already undone my minus sign here i get the cosine of zero which is one
and one plus one of course is two and two times two is four and i end up with the same
value 4 root 2. okay so this comes out to 4 root 2 and for those of you who are curious
this integral popped up when i was computing the arc length of the arch of a cycloid function
and i'll post a link real quick to that arc length video
if you're curious to see how this integral came up in the context of a larger problem
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