in this video we're looking at a particularly challenging trigonometric integral it's the
integral of square root 1 plus sine X on the interval 0 to Pi and I actually wrote this
problem during the pandemic because I had too many students copying solutions from simbolab and this
was intentionally designed to break simbo Labs so this integral is weird enough where a lot of
computer algebra systems will fail to compute the integral and I wasn't too terribly mean with this
problem I actually gave a lot of hints on how to proceed with the problem because it's so exotic so
as you can see simbolab has given up and it says it's just evaluating using a numerical calculation
method but there are actually two different ways of evaluating this integral symbolically
and getting an exact answer of 4 and one of them is to use an exotic combination of trigonometric
identities to manipulate this integrand until it simplifies but the first one that occurred
to me was to use a conjugate so we're going to multiply the numerator and denominator by the
conjugate of that interior and the point of that is to intentionally build a difference
of two squares which then opens up a trig identity approach so that conjugate is the square root of 1
minus sine X again the whole point of this is to intentionally build a difference of two squares
so when we multiply the Interiors of these two square roots we get one for the first term the
cross terms cancel and then we get Negative sine squared X for the second term and this is where
we get the payoff of the conjugate strategy 1 minus sine squared is cosine squared and this
is probably the hardest point about this whole approach to the integral you can't just say that
the square root of cosine squared x is just cosine of x and the reason why is that the square root of
the square of the thing is always going to be positive because squaring it makes it positive
and then you square root it and you get the same magnitude as whatever you're squaring in there
we get the same magnitude as cosine X but it's guaranteed to be positive in other words we're
talking about an absolute value and this has to be treated very carefully so if that's not totally
clear then just let me give a simple example really quick if I take the square root of the
square of negative 2 for example that gives me the square root of 4 and that's positive 2. so I plug
negative 2 into this thing and positive 2 pops out this is just another way of saying the absolute
value function so how do you deal with an absolute value of cosine of x inside of an integral
well you have to understand where the cosine function is positive and where it's negative
so here's a picture of the cosine function on 0 to Pi and it crosses the x-axis at pi over two and I
can see the cosine function is positive from zero to pi over two but negative from pi over two to Pi
so we're going to have to split this integral into two pieces one for each of those signs where the
cosine function is positive the absolute value does nothing and I get the integral of cosine X
over the square root of 1 minus sine X DX and where the cosine function is negative the absolute
value multiplies it by negative one so on this interval pi over 2 to Pi the absolute value of
cosine X is negative cosine of x guaranteeing that we get a positive output now we can finally get to
the work of finding the antiderivative of each of these integrands and for each of these I notice
that I have a 1 over a square root 1 minus sine X and then I have the derivative of the Interior
function so the interior function here is 1 minus sine X the derivative of that is negative cosine
X so I'm so close to having that tacked onto my integrands in other words I'm recognizing the
results of the chain rule so in my first integral I'm going to supply that minus sign to the cosine
of x in the numerator and compensate for it out in front in my second integral I already have
the minus sign so I don't need to worry about supplying an extra factor of negative one now
just to make it exceptionally clear how to get the antiderivative I'm going to rewrite the one
over square root 1 minus sine X as 1 minus sine X to the negative one-half now when I look at each
of these integrands I'm seeing 1 minus sine X raised to a power and then the derivative
of that interior function tacked on right next to it so I can recognize the chain rule backwards if
you prefer you could do an explicit U substitution and what you would do there is let U equal 1 minus
sine X and then d u would be negative cosine X DX so you'd be ready to go ahead and use the
power rule I prefer the approach of just directly using the chain rule backwards here so we apply
the power rule and we end up with 1 minus sine X to the one-half power we're adding 1 to the
exponent there and then dividing by the resulting exponent well dividing by one half is the same as
multiplying by two so I get negative 2 times the quantity 1 minus sine X to the one half evaluated
from 0 to pi over 2. we do the same thing with the second integral and it's really just the
same integrand so we already know what the answer is I end up with a 2 times the quantity one minus
sine X to the positive one-half evaluated from pi over 2 to Pi now we can sub in our limits of
integration and simplify the result so I notice when I sub in the upper limit on the first one I
plug in pi over 2 for x and the sine of pi over 2 is 1 and that gives me 1 minus 1 which is zero
so that's gone I then subtract what I get when I sub in the lower limit well the sine of zero is
zero so I just end up with a 1 to the one half in that piece and 2 times that is just two all right
we're subtracting that lower limit which gives us a positive two as a result in my second term
if I sub in pi for X the sine of Pi is 0 and I end up with a positive 2 for that upper limit
then I subtract what I get when I sub in the lower limit which is that same thing pi over 2.
if I replace X by pi over 2 I end up with 1 minus 1 and that's 0. so we end up with a 2 plus
2 or 4 for the numerical value of this integral there's one other point I want to mention for the
particularly rigorous among you and that is if I go back to this step where I've split the integral
into two pieces I notice that subbing in pi over 2 for X creates a zero denominator in other words
this is actually an improper integral and the second integral is as well if a sub and pi over
two I get a zero denominator that means these are actually improper integrals and to handle
these in a technically correct way you would want to replace the problematic limiting point with a
variable we usually use a t for that and then take the limit as T goes to pi over 2 from the left for
the first integral take the limit as T goes to pi over 2 from the right in the second integral
so you just replace the problematic limit with a t you do the antiderivative exactly the same
way we did you take the limit as T goes to pi over 2 and you're going to find the same
answer there's nothing problematic in these limits if you enjoyed this video or at least
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