Trig substitution can be super tricky because there are just a lot of steps to follow which
means there's a lot of different opportunities to make mistakes on small details. But don't
worry because I've got a few tricks to help you stay organized that'll make trig sub problems
a whole lot easier. So if you're ready to finally understand what trig substitution
actually is, how it works and when you should use it, and how to easily solve any trig sub
problem that gets thrown at you, then you're definitely in the right place. So what is
trig substitution? Well it's just like these other rules for integration, like power rule,
u-substitution or just substitution, integration by parts and partial fractions. So we've already
learned all of these rules and these are different things that we use to help us evaluate integrals.
For example if we have a really simple integral like the integral of x^3+x^2, we can just
use power rule to evaluate that integral. We know how to do that. If we have something
a little more complicated, we might need to use substitution, or if we have the product
of two functions for our integrand then we might need to use integration by parts. Or
if we have a rational function then we might need to use partial fractions. All of these
methods just help us integrate different kinds of functions. And trig substitution is no
different. It's another method that we can use to find the value of an integral, and
it works for specific kinds of functions just like these work for specific kinds of functions.
So that kind of brings us to the question then, when do we want to use trig substitution?
Well there's two parts really to that answer. The first part is, you want to use trig substitution
when these other rules don't work. Now the reason that you look at these rules first
is because they're all generally simpler than trig substitution. So if I could use power
rule to evaluate an integral I would certainly rather do that rather than using trig substitution
because it's going to be easier and faster. And same thing goes for substitution, integration
by parts, and partial fractions. If I can use one of these methods, it's probably going
to be easier and quicker than using trig substitution, which is a little complicated. That means
that I want to go through a mental checklist of these other methods of integration before
I would use trig substitution. So I want to kind of look at my integral and think, can
I use power rule? No. Okay I can't use substitution, can't use integration by parts because it's
not the product of two functions, can't use partial fractions because it's not a rational
function. So at this point now I've kind of ruled these out. Maybe I want to think about
trig substitution and look at my integral and see if I might be able to apply this method.
So what kinds of integrals are you looking for? Well you're looking for things like this.
Here I've got three examples of integrals that you would apply trigonometric substitution
to in order to solve them. So what you want to be on the lookout for is a couple of things.
First of all we have to have one of these forms inside of our integral. We have to have
a^2-u^2, u^2+a^2 or a^2+u^2, or u^2-a^2. Now you might be like what the heck are you talking
about, a^2, u^2, I don't get it. All this means, think of a as a constant and u as a
variable. So for example if a were 4 then 4^2 is 16. So this a^2 term is always going
to be a constant number, meaning a constant has no variable attached to it. So like 16,
16, 7, those are constants. u^2 though is going to be a variable. So think of u like
the variable x. So u could be x, such that this is x^2, u could be 2x such that when
I square 2x I get 4x^2. u is just going to be something that involves x. So when you
say a^2-u^2, really think of this as a constant minus a variable. So here's a perfect example.
Inside of this first integral I have a constant 16, minus a variable x^2. This is a constant
term, this is a variable term, and I'm looking for a^2-u^2, constant minus variable, and
here I have constant minus variable. So that's what we mean when we say a^2-u^2. And of course
we can also have any of these forms. So u^2+a^2 would be like a variable plus a constant.
And remember because addition is commutative, u^2+a^2 is really the same thing as a^2+u^2.
They're just flipped around. It's no different than if I had 3+4 or 4+3 those are both 7,
it's just that the order is flipped around. So these two things mean the same thing, so
I could be looking for variable plus constant or constant plus variable. And then the third
form which would be the only other possibility would be variable term minus constant term,
u^2-a^2. So I'm looking for one of those four relationships somewhere in my integral. The
only time I can do trig substitution is if I have one of these four relationships or
I might be able to do a little manipulation with my integral to get it into one of these
forms so that I can then do trig substitution. So I'm looking for one of those forms and
for example here we have constant minus variable, 16 minus x^2, which matches this form of a^2-u^2
because it's also constant minus variable. Ideally I want both of these values to be
perfect squares. So for example 16 is the perfect square of 4, it's 4^2. And x^2 is
the perfect square of x. So this is x^2. So because these are both perfect squares, I
could match this up, 4^2-x^2, I could match that to a^2-u^2 and easily say that a has
got to be 4 and u has to be x. So that's kind of a perfect match for this a^2-u^2, so that's
a dead giveaway for trig substitution. Same thing here in this second integral, I have
x^4, a variable term, plus 16 the constant term. And that's just like u^2+a^2, variable
plus constant. And furthermore, if I look at these two values they're also perfect squares.
x^4 is a perfect square of x^2, so I would get (x^2)^2, and 16 is the perfect square
of 4, so this is 4^2. So I could easily say that u is x^2 and that a is 4 and this format
here matches u^2+a^2. So we're looking for those kinds of relationships inside of our
integral somewhere. Another dead giveaway is if you have a square root inside of your
integral like this here. We have this square root, we have this square root. And especially
when you have one of these relationships, a^2-u^2, one of these u^2+a^2 or a^2+u^2,
or a u^2-a^2 underneath your square root sign, that is a dead giveaway for trig substitution
and you should probably strongly consider using it. Now there's a couple of caveats
to that. First of all our integral doesn't have to have a square root sign in order to
be a trig substitution problem. For example this second integral is a trig substitution
problem but you'll notice, no square root. So it doesn't have to be there, it's just
an obvious sign that it might be a trig substitution problem. The second caveat is in these first
two examples we had perfect squares. Both of these values here worked out to perfect
squares: 4^2, x^2, (x^2)^2, 4^2. But like in this example, this is a trig substitution
problem, and we have this u^2-a^2 relationship. But these are not both perfect squares. This
turns out to be a perfect square x^2. But 7 is not a perfect square. So you might think,
oh this can't be a trig substitution problem because 7 is not a perfect square. Well that's
not true. This is a trig substitution problem. The way you get around your constant not being
perfect square, is you say well 7 actually is a perfect square of the square root of
7. So you say square root of 7 squared like this, and so then, if you are matching this
up to your u^2-a^2, then u would be equal to x and a would just be the square root of
7. So my point is that a trig substitution problem must have one of these relationships
inside of it between a and u. But a^2 might not always be a perfect square, like we saw
here, and especially if you have a square root in your integral, that's even more evidence
that this is probably a trig subproblem, but you don't necessarily have to have a square
root in order for it still to be trigonometric substitution. So before we continue on let's
take a look at a bunch of trig sub examples so that we can see what these problems typically
look like. So here are a bunch of examples. These are all trigonometric substitution problems.
On the left here these are all sine substitutions and we'll talk more about what this means
in a little bit. These in the middle here are all tangent substitutions, and these on
the right are all secant substitutions. So if you look here on the left you'll notice
this common theme of a^2-u^2. We have here an a^2-u^2, two perfect squares, the difference
of two perfect squares, where we have constant minus variable, we have constant minus variable,
two perfect squares. And both of those a^2-u^2 values are underneath square root signs. So
dead giveaways for a sine substitution. This one though, not so much. This is just a quadratic
under a square root so, the fact the square root is there, would maybe make you think
trig substitution, but you see no perfect square here. So here's the trick when you
see something like this. Trig substitution you often use when you have quadratics. This
is a quadratic where you have the -x^2+2x+48, the x^2 term, the x to the first term, and
the constant, so that's a quadratic. And when you have that, oftentimes you'll use trig
substitution. What you want to do is take this quadratic and you want to go through
the process of completing the square. You want to complete the square for this quadratic
because when you do that what you'll end up with is this form here of a^2-u^2. And so
you'll essentially turn this into a trig substitution problem because when you rewrite this quadratic
under the square root, and you change it into a perfect square, you'll have this a^2-u^2
format. If we look at these tangent substitutions notice here the pattern of either a^2+u^2
or u^2+a^2. So here we have a u^2+a^2, but notice no square root sign, just this value
inside parentheses and then we're squaring that whole quantity. Even though we don't
have a square root sign, this is still a trig sub problem. This one's perfect because we
have an a^2 plus a u^2, constant plus a variable, and we have the square root sign, so that
one's really obvious. But these two again, not quite as obvious. This one same thing
as before. This is a quadratic, the x^2+4x+5, that common form, it's a quadratic. We need
to go through the process of completing the square to turn this into a perfect square
such that you have a u^2+a^2 format and it becomes a trig sub problem. Even this problem
here, this one looks really crazy and I don't want to freak you out with it. But the point
here, what I'm trying to show you is that you can have all different kinds of functions
inside your integral and they could still be a trigonometric substitution problem. This
one, what you actually end up doing is, this is a u-substitution problem so you use u-substitution
first on this. Once you do that it turns into a partial fractions problem. And so then you
go through the process of partial fractions, you do your partial fractions decomposition,
and at the end of that what you end up with is this format here u^2+a^2, and it actually
at that point becomes a trigonometric substitution problem. And again you don't need to worry
about that, you rarely deal with problems this complicated. My point is only that you
may have to, like with these quadratics, or something like this, go through one of the
other methods and then apply trigonometric substitution. So if you're working through
a problem and you're doing another method and it's working and going well, and then
you get to a point midway through solving your integral, and you realize all of a sudden
you have a trig substitution problem. That's not a bad sign, it's just a sign that now
you need to transition to trigonometric substitution and start applying this process after some
work that you've already done. So it can pop up in the middle of a problem. And then this
last set of examples here, these are all secant substitutions in the form u^2-a^2. You see
the variable minus constant, both perfect squares underneath the square root sign, classic
classic trig sub. Or here u^2-a^2, variable minus constant. Yes it's cubed but it has
that format and it's underneath the square root perfect trig substitution problem. And
then here this example, similar to these other two that we talked about. This is a quadratic.
It needs to be factored using the process of completing the square, and once that's
done it will become a trig sub problem with a secant substitution because you'll have
the form u^2-a^2. That value will be underneath the square root. It'll be perfect for trig
sub. Now why does trig sub actually work? Well this is a little bit of a simplified
explanation but I want to give you a better intuition for why this works. So we're going
to go through this briefly and then we're going to talk about how to actually solve
trig sub problems. So if I start by drawing a right triangle, and I make this a right
triangle and I say that this is the angle theta here. Remember that any method of integration
whether, it's u-substitution, integration by parts, partial fractions, trigonometric
substitution, all these things that we use, all we're trying to do is rewrite the integrand,
we're trying to rewrite the function so that it turns into something we can actually integrate.
Because we're given all these integrals that we can't integrate directly and so we use
these methods to manipulate the functions, rewrite them, change them around into a different
form. Still the same value but a different form so that we can actually integrate it
with a simpler method like power rule. So trig substitution is no different, we're just
trying to simplify our integrals. And remember we had all these examples of integrals that
were perfect for trig sub that had these values inside them, the a^2-u^2, u^2-a^2, one of
these relationships inside of them. Well the reason that this works is because we look
for that relationship and then we want to replace that relationship with something simpler
and that's how we end up simplifying our integral. So why can we replace values like these? Well
it comes back to the Pythagorean theorem for right triangles. So remember that the Pythagorean
theorem says that if you have a side here, this is the adjacent side, this is the opposite
side of the angle theta. And then this is the hypotenuse c, that the relationship between
these sides is a^2+b^2=c^2. Now an interesting thing happens here. If I for example wanted
to solve for the length of side a, and I wanted to use the Pythagorean theorem, let's pretend
that I knew that the length of the hypotenuse was 4 and that the length of the opposite
side here was x. Well in that case, if I plug into my Pythagorean theorem I get a^2+, I
know b is x so I get x^2, and I know that c the hypotenuse is 4, so I get 4^2 or 16.
Now if I subtract x^2 from both sides because I'm trying to solve for a, I get 16-x^2. And
then if I take the square root of both sides, I get a is equal to the square root of 16-x^2.
Now here's the interesting thing. Remember all those integrals that we looked at where
I said like these are all perfect examples of trig sub problems? Well didn't they look
a lot like this value right here? The square root of 16-x^2? They did! Remember I was saying
the square root sign is a dead giveaway. We would have a relationship between values like
this, constant minus variable, this is a perfect 4^2-x^2, or we could call it a^2-u^2, right?
Which is this a^2-u^2 that we use for a sine substitution? So the point here is that trigonometric
substitution works, the reason why it works, is because we're given a value like this,
square root of 16-x^2 inside of our integral. Well if we can relate that back to a right
triangle, if we can sort of undo this Pythagorean theorem process, what we realize is that this
value is related to a few other values. It's related to the angle theta, it's related to
the length of the hypotenuse 4, and it's related to the length of the opposite side x. All
of which, if I think about theta, 4, and x, all of which are a lot simpler than this square
root of 16-x^2. So the point is that if I start out with something like square root
of 16-x^2, I can sort of work backwards, get to these values of theta, 4, and x, and I
can end up replacing this value inside of my integral with something that's maybe in
terms of theta, or with something involving 4 or x, a simpler value than what I originally
started with, which overall is going to make my integral a lot easier to solve. So that's
why trigonometric substitution actually works, because you're relating a value that you're
given in your integral to these associated components inside a right triangle and that
allows you to simplify that function that you're trying to integrate. So now that we
know how this actually works let's talk about how to solve a trig sub problem. So when we're
talking about solving a trigonometric substitution problem, the first thing we want to do is
go through the same set up process every single time. And the reason that we want to do this
is because we have lots of little values that we're going to need to use throughout our
problem so we want to get them all out in the open up front instead of having to pause
our work as we're going through the integral to find each one of these little pieces. We
want to get them all done first thing, that way we'll be prepared to just go through the
rest of our problem smoothly. So with that being said, remember before we had talked
about sine, tangent, and secant. Those are the substitutions we're going to be making.
So when you talk about trig substitution, you can make a sine substitution, a tangent
substitution, or a secant substitution. And the reason that we call it that is because
when you're doing a sine substitution, your substitution is built off of this u=asin(theta)
value, where u and a come from the a^2-u^2 that we find inside our integral. And a tangent
substitution comes from u=atan(theta), secant comes from u=asec(theta), so that's why we
call it a sine substitution or a tangent substitution. But let's quickly go through this setup process
so that you know what you're doing and you start getting comfortable setting up for a
trig sub problem. So let's pretend that you found this value inside your integral. The
square root of 1-x^2. Well the first thing you see right away is that you have constant
minus variable, 1-x^2, constant minus variable. That matches your a^2-u^2 format. So you want
to match those formats together and you want to say a^2 has to be equal to 1 and u^2 has
to be equal to x^2. Then you want to take the square root of both of those to get a
and u. So the square root of 1 is still 1 and the square root of x^2 is x. So we get
a=1 and u=x. Now we want to plug those into our sine substitution, u=asin(theta). So since
u=x and a=1 we get x=1sin(theta) or simply just x=sin(theta) so we say down here x=sin(theta).
And if this isn't already solved for x we want to go ahead and do that. But in this
case it is. So we have x=sin(theta), then we want to find dx, which is the derivative
of x. So we say the derivative of sin(theta) is cos(theta) so we get cos(theta) and then
we always want to remember here on the right side to put dtheta. Now for sin(theta) here,
what we're doing is we're solving this equation for just the trig function. In this case it
already is because a was equal to 1 so we just ended up with sin(theta) on the right
hand side. That won't always be the case and if it's not you want to make sure to solve
this for just the trig function sin(theta). So we get in this case sin(theta)=x. And then
what we want to do is we want to solve this equation for just theta. So the way we do
that is we take arcsin or the inverse sine function of both sides because arcsin of sin,
those things will cancel out leaving us with just theta on the left side. So we get theta=arcsin(x).
Now that was kind of a lot to do it once, but after we do two more examples you'll really
start to get the hang of it. So let's look at a tangent substitution here and pretend
that inside of our integral we found 4x^2+9. So when we look at that we see u^2+a^2 because
this was variable plus constant, so u^2+a^2. So if we compare that to u^2+a^2 then what
we can say is that u^2 has to be 4x^2 and that a^2 has to be 9. Then we want to take
the square root of both those values. So the square root of 9 is 3 and the square root
of 4x^2, square root of 4 is 2, the square root of x^2 is x, so we get 2x for u. Now
we want to plug these into our formula u=atan(theta), so in our case we get 2x=3tan(theta). And
in this case this equation is not solved for x and remember we said we always wanted it
to be solved for x so we divide both sides by 2 to get x=(3/2)tan(theta). Then we want
to take the derivative of that to get dx. So the derivative of tan(theta) is sec^2(theta),
so we get dx=(3/2)sec^2(theta)dtheta. Then we also want to solve x=(3/2)tan(theta) for
tan(theta) specifically, which we can do by multiplying both sides of this by 2/3. So
tan(theta)=2x/3. And then we take arctan of both sides to get theta=arctan(2x/3). And
then let's do this one more time. If we have in our integral the square root of x^2-25,
what we see is variable minus constant, and that matches u^2-a^2. So u^2 has to be x^2
and a^2 has to be 25. So we square root both of those and we get a is equal to square root
of 25 or 5, and u is equal to the square root of x^2 or x. Then we plug both of these into
u=asec(theta) and we get x=5sec(theta). We're already solved for x so we're done there,
and then we take the derivative to get dx. So the derivative of sec(theta) is secant
times tangent so we get 5sec(theta)tan(theta)dtheta, and then we want to solve this equation x=5sec(theta)
just for sec(theta), so we'll divide both sides by 5 to get sec(theta)=x/5. And then
we take arcsec of both sides to get theta=arcsec(x/5). So now that we understand this setup process
let's go through one trigonometric substitution so that we can talk about the steps involved
in solving a trig sub problem and how to go through this setup process again. So if we
have for example this problem. We have 1 divided by x^2 times the square root of 4-x^2 dx.
So we look at this and right away we can see that this looks like a trigonometric substitution
problem because we have constant minus variable and they're both perfect squares and they're
even underneath the square root sign. So remember that's a dead giveaway this is probably a
trig sub problem. So if we match this up, the 4-x^2, that's constant minus variable,
which means that's an a^2-u^2, and that is actually a sine substitution where we know
we're going to have u=asin(theta). So our first step in any trig sub problem is to number
one identify that it is actually a trig sub problem, and we did that. Number two is to
figure out which trig substitution to use. Is it a sine substitution, a tangent substitution,
or secant substitution? We found 4-x^2, constant minus variable, we matched that up to constant
minus variable, a^2-u^2. And we know because we memorized it that a^2-u^2 is a sine substitution
and therefore that our substitution will be u=asin(theta). Step three then is to go through
the setup process like we just did in those last three examples. So the setup process,
we're going to match all these values up to our 4-x^2. So 4 the constant has to be a^2,
and x^2 has to be u^2. Then we want to take the square root of both of those so u=x and
a=2. Then since we want u=asin(theta) and we know u is x and a is 2, we get x=2sin(theta)
so x=2sin(theta). This is already solved for x so we're good there. Then we want to take
the derivative of that and remember the derivative of sine is cosine, so we get 2cos(theta) and
we don't forget our dtheta. Then we want to solve x=2sin(theta) for just the trig function
so we divide both sides by 2 and we get sin(theta)=x/2, and then we want to solve this for theta by
taking arcsin of both sides to get theta=arcsin(x/2). That's how quick the set up process can be.
Now our fourth step is to take all this information that we found and actually make substitutions
into our integral. So we're going to take these values, we're going to plug into our
integral. And the goal here is to completely transform this integral. Right now it's in
terms of x everywhere, we have x^2, x^2, dx. We want to get rid of all the x's and instead
end up with only theta. So how do we do that? Well let's look at this here. If we have the
integral the 1 will stay in the numerator because we're only trying to replace the x's.
So we had that. x^2 here remember we found that the value of x was 2sin(theta) so we
can plug that in here for x. So we get 2sin(theta) and because we have x^2 we square that. And
then we have the square root of 4 minus, and then again we plug in for x, 2sin(theta) quantity
squared and that's all going to be underneath our square root. And then we have to replace
dx which we know is 2cos(theta)dtheta. So we can multiply here by 2cos(theta)dtheta.
Notice now that we have actually completely transformed the integral. Everything here
is in terms of theta. There are no x's remaining, which is perfect. That's exactly what we wanted
to do. So we're done with that step. The next step is just to simplify this integral as
much as we can down to a point where we can actually evaluate the integral. And as we
go through this process we may have to use some trigonometric identities to make this
simpler, we may have to use some other methods of integration, but we're trying to get to
a point where we have a function that's simple enough that we can integrate it directly.
So how will we do that? We can start by simplifying here in the denominator. So for example if
we look at just the square root here, we have the square root of 4 minus, we are squaring
2sin^2. So 2sin^2 quantity squared turns into 4sin^2(theta). That's all underneath our square
root so let's go ahead and cancel as we go. So we're replacing that. Then underneath the
square root we can factor out of 4. So this becomes 4 times 1 minus sin^2(theta) underneath
our square root. Now at this point what we recognize is that we have 1-sin^2(theta).
This is where the first trigonometric identity comes in. So we want to remember the trig
identity sin^2(theta)+cos^2(theta)=1. If we subtract sine from both sides we get cos^2(theta)
is equal to 1-sin^2(theta). And 1-sin^2(theta) is exactly what we have. Which means that
we can replace it with cos^2(theta). So this becomes 4 times cos^2(theta) underneath our
square root sign. Now of course the square root of 4cos^2(theta), the square root of
4 is 2 and the square root of cos^2 is just cosine, so this becomes then 2cos(theta) and
we get rid of the entire square root and everything underneath it. Now if we go ahead and move
this up underneath here what we see then is that we can cancel a 2cos(theta) from the
numerator and denominator. So this is going to cancel with this. So this integral becomes
1 over 4 sin^2(theta) dtheta. This is where our next trig identity comes in and it's a
reciprocal identity so remember that cosecant is the same as 1 over sine because sine and
cosecant are reciprocals of one another. So whenever we have 1 over sine that's the same
thing as cosecant. So here we have 1 over sin^2 which means we can change that to csc^2.
So this becomes the integral of 1/4 csc^2(theta) dtheta. And at this point we're lucky because
we actually already know the integral of csc^2(theta). It's a common trig integral. The integral
of this value is -cot(theta) so what we can say is that this is going to be equal to -1/4
cot(theta)+C. And at this point we've actually managed to evaluate the integral. Our last
step at this point is to get this value back in terms of x. Remember we started with an
integral that was in terms of x but this value still in terms of theta. We need to put it
back so that it's in terms of x. So the way that we always do that is we draw a reference
triangle, back to the right triangle like we did before. So let's go ahead and do that.
We want to draw a right triangle for reference we always put theta here the right angle here.
So now what we need to remember is that building this reference triangle, we already have back
at the beginning of the problem sin(theta)=x/2. This is the last piece that we need to remember
from our trigonometric identities. And it's that old phrase you might remember SOH-CAH-TOA,
which reminds you that sine is equal to opposite over hypotenuse, that cosine is equal to adjacent
over hypotenuse, and that tangent is equal to opposite over adjacent. So in this case
we have sine is equal to opposite over hypotenuse. We have sin(theta)=x/2. So if we equate that
to opposite over hypotenuse then when we look at our angle theta we say that the opposite
side has to be x, opposite has to be x, hypotenuse has to be 2. So the opposite side has to be
x, the hypotenuse has to be 2. And then we can solve for the length of this third side.
We do that using Pythagorean theorem again. So if we call this side a and we get with
Pythagorean theorem a^2+x^2=2^2 or 4 and then we solve this for a. So a^2 is 4=x^2 and then
a is the square root of 4-x^2. So the third side then is square root of 4-x^2. Now why
did we build that reference triangle? Well the reason is because we're trying to get
-(1/4)cot(theta)+C back in terms of x and in order to find the value of cot(theta),
we need our reference triangle. So what we need to remember is that tangent is equal
to sine over cosine always. And cotangent is the reciprocal of tangent which means that
if tangent is sine over cosine then cotangent is the reciprocal. It's cosine over sine.
So when we transform this value we want -(1/4) and then we want cosine over sine. Well cosine
of the angle theta going back to SOH-CAH-TOA, cosine is adjacent over hypotenuse. So cosine
is the adjacent side here over the hypotenuse 2. So we're going to multiply this by square
root 4-x^2 over 2. But then for cotangent we're dividing that by sine, so we're dividing
this whole thing by sine. Sine is opposite over hypotenuse so opposite over hypotenuse,
x over 2, which remember we already had here. Sine of theta is x over 2. So we get x over
2, and then plus C. And then finally we simplify this. Instead of dividing by the fraction
x over 2, we can multiply by its reciprocal, multiply by 2 over x instead of x over 2.
Our 2's cancel, 2 and 2 here, which means that our final answer is negative square root
of 4-x^2, all divided by 4x and then plus C. I hope that video helped you, and if it
did hit that like button, make sure to subscribe, and I'll see you in the next video.
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