It’s Professor Dave, let’s learn the substitution rule.
Hopefully we’ve gotten used to integration by now, and even feel that integrating polynomials
is very easy, as easy as taking a derivative.
This is because integrating a simple polynomial, much like taking a derivative, will follow
a simple algorithm that never changes.
But this will not always be the case, as integration doesn’t always follow such a rigid algorithm,
and in fact there will be times where it is not immediately obvious how to do it.
In such a case, there are a few tricks we can use, and the first one we will learn is
called the substitution rule.
The substitution rule allows us to take a complicated integrand and simplify it, by
representing some portion of it with the letter u.
Take this problem, the integral of two x times cosine (x squared plus one).
This looks like a mess.
The cosine is operating on a sum, and it is also part of a product, so nothing we have
learned so far will work here.
But let’s try something clever.
What if we make a substitution.
Let’s replace x squared plus one with the letter u.
We are still saying the same thing, since u equals x squared plus one, but we are simplifying
the integrand this way, because if cosine is operating on a lone variable rather than
an expression, we will be able to integrate it.
The only problem is that if we are changing things to u, this dx will be a problem.
We need things in terms of u now, so we need this to be du instead.
But we can fix that.
Let’s understand that d over dx of (x squared plus one), or the derivative of this function
with respect to x, is two x.
Now we can do something tricky.
Let’s bring the dx up to the other side to get two x dx.
This may seem strange, as dx is simply part of the notation d over dx, and it may not
look like something we can manipulate algebraically.
Later in the series we will justify why this is possible.
For now, let’s just do it, and observe how this is useful.
Now we can understand that this function on the left is what we are representing as u,
so let’s just change that to u, and we therefore see that du equals two x dx.
That’s quite convenient, because lo and behold, if we bring the two x inside the integrand
to the other side of this cosine term, we have two x dx right here, so according to
this equation, we can just rewrite that as du.
Now that we have a differential, du, that agrees with the rest of the integrand, we
are ready to integrate, and cosine u du can be integrated just like cosine x dx, or cosine
z dz, or cosine smiley face d smiley face.
The variable doesn’t matter.
The integral of cosine u du is sine u plus C. Now that integration is complete, let’s
return u back to its original form, and we have sine of x squared plus one, plus C, and
that’s the answer.
Just to be thorough, let’s differentiate this and make sure we get what we started with.
C will go away, so it’s just sine of (x squared plus one).
According to the chain rule, we do the outer function first, so we get cosine (x squared
plus one), times the derivative of what’s inside, which is two x.
That is indeed what we started with, so this approach worked like a charm.
In this way, we can think of the substitution rule as the chain rule in reverse, because
we symbolize some piece of the function with a single variable, allowing us to integrate
more simply, just the way that the chain rule ignores the inner function during the first
step of differentiation.
The only thing to remember is that while the chain rule will always work, the substitution
rule won’t always work.
It only worked here because our expression for du was indeed present in the integrand,
because the integrand was in the form: f of g of x times g prime of x.
If this wasn’t the case, we couldn’t do this, so we have to be able to look at an
integral and recognize whether or not the substitution rule can be applied.
If we do have an integrand in the form of f of g of x times g prime of x, dx, we can
represent g of x as u, g prime of x dx turns into du, and we end up with f of u du, which
we can then integrate.
Here’s another example.
X squared times the square root of x cubed plus one.
Is this in the form that we mentioned?
If the square root is f, and what’s inside is g, that does look like f of g of x.
But what about g prime?
Well g prime would be three x squared, and we pretty much have that, so let’s give
this a try and see what happens.
Let’s change x cubed plus one into u, so that we have root u.
Now du over dx, which means that we are differentiating the expression that u represents with respect
to x, will equal three x squared.
That means that du equals three x squared dx.
Now it’s true that we don’t have three x squared dx, but that’s no problem.
We have x squared dx, so let’s just bring the three over to the other side.
X squared dx will then equal du over three, or one third du, so let’s write one third
du in its place.
Now things are looking pretty good.
Let’s pull one third out of the integral, and then we can integrate root u du.
Remember, that’s u to the one half, so it will become u to the three halves over three
halves, or two thirds u to the three halves.
Times the one third we already had, and that will be two ninths u to the three halves.
Now we can return u to its original form, and we have two ninths times the quantity
x cubed plus one to the three halves power, and then plus C as we are used to.
Let’s do a couple with trig functions.
How about sine to the sixth x times cosine x dx.
The sine term has more going on with it, so why don’t we try representing sine x as u.
Remember, sine to the sixth x just means the quantity sine of x raised to the sixth power,
so this just becomes u to the sixth.
Then, du over dx equals cosine x, so du equals cosine x dx.
That is indeed present, so let’s just make the rest of this equal to du, and we are ready to go.
Integrating u to the sixth gives us u to the seventh over seven, we return u to its original
form, and we get sine to the seventh of x, over seven, plus C. Now how about tangent x dx.
This is an example where substitution might not jump out immediately as a possible strategy.
But if we remember that tangent is sine over cosine, and that the derivative of cosine
is negative sine, then we might see a plan unfold.
Let’s use cosine x as our u, and that means du will be negative sine x dx.
We have positive sine, so let’s make this negative du, and then pull the negative sign out.
Now du over u is the same as one over u times du, and we know that the integral of one over
u will be the natural log of u.
The only thing is that we have to put u in absolute value brackets, because this value
will sometimes be negative, as it’s a cosine function, and we know that the natural log
can’t take negative values.
So we get negative natural log of the absolute value of u plus C, and then change u back
to cosine, and that’s our answer.
So remember, any time we are integrating some function, and the function can be split up
into a smaller function followed by the derivative of that smaller function, or something very
close to it, the substitution rule is a good idea for something to try.
This won’t always work, but if it does, it can take a complicated integral and make
it very simple, so it is definitely a great start to our bag of tricks for integration.
Let’s check comprehension.