### Video Transcript

In this lesson, weβll learn how to
use integration by substitution to evaluate definite integrals. At this stage, you should feel
comfortable finding the antiderivative for a variety of functions, including
polynomials, trigonometric, and logarithmic functions. In this lesson weβll look at how to
apply these rules to find the antiderivative for more complicated functions.

Because of the fundamental theorem
of calculus, itβs important to be able to find the antiderivative, but our formulae
donβt tell us how to evaluate integrals such as the definite integral between one
and two of π₯ squared times π₯ cubed minus three cubed. So, to find this integral, we use a
special strategy of introducing something extra, a new variable. This is called integration by
substitution. And itβs sometimes referred to as
the reverse chain rule.

The first step is always to get our
integral into this form. Itβs the integral of π of π of π₯
times π prime of π₯ with respect to π₯. Notice that we have π of π₯ as the
inside part of a composite function and then its derivative π prime of π₯. Then, once weβre sure that our
integral is in this form, we apply the substitution rule for definite integrals.

This says that if π prime is
continuous on the closed interval π to π, and π is continuous on the range of π’
which is π of π₯, then the definite integral between π and π of π of π of π₯
times π prime of π₯ with respect to π₯ is equal to the definite integral between π
of π and π of π of π of π’ with respect to π’. Itβs often best just to have a look
at an example of how this works.

Evaluate the definite integral
between one and two of π₯ squared times π₯ cubed minus three cubed with respect to
π₯.

This is not a polynomial thatβs
nice to integrate using our standard rules of finding the antiderivative. And we certainly donβt want to
distribute our parentheses and find the antiderivative for each term. Instead, we spot that our integral
is set up in this form. Itβs the definite integral between
some limits of π and π of some function of π of π₯ times the derivative of the
inner part of that composite function. Here, π of π₯, the inner part of
our composite function, is π₯ cubed minus three.

And then, we have a scalar multiple
of its derivative here. So, we use integration by
substitution to evaluate the integral. This says that if π prime, the
derivative of π, is continuous on some closed interval π to π, and π is
continuous on the range of π’, which is our function π of π₯. Then, then the definite integral is
equal to the definite integral between π of π and π of π of π of π’ with
respect to π’. So, we let π’ be equal to the
function that we defined as π of π₯. Itβs the function inside a function
whose derivative also appears.

So, weβre going to let π’ be equal
to π₯ cubed minus three. Now, this is great as when we
differentiate π’ with respect to π₯, we see that dπ’ by dπ₯ equals three π₯
squared. And in integration by substitution,
we think of dπ’ and dπ₯ as differentials. And we can alternatively write this
as dπ’ equals three π₯ squared dπ₯. Notice that whilst dπ’ by dπ₯ is
definitely not a fraction, we do treat it a little like one in this process. We divide both sides by three. And we see that a third dπ’ equals
π₯ squared dπ₯.

So, letβs look back to our original
integral. We now see that we can replace π₯
squared dπ₯ with a third dπ’. And we can replace π₯ cubed minus
three with π’. But what do we do with our limits
of one and two? Well, we need to replace them with
π of one and π of two. Well, we go back to our original
substitution. We said that π’ is equal to π₯
cubed minus three, and our lower limit is when π₯ is equal to one. So, thatβs when π’ is equal to one
cubed minus three, which is equal to negative two.

Our upper limit is when π₯ is equal
to two. So, at this stage, π’ is equal to
two cubed minus three, which is of course five. We can now replace each part of our
integral with the various substitution. And we see that weβre going to need
to work out the definite integral between negative two and five of a third π’ cubed
with respect to π’. Remember, we can take any constant
factors outside of the integral and focus on integrating π’ cubed.

And then, we recall that we can
integrate a polynomial term whose exponent is not equal to negative one by adding
one to that exponent and then dividing by that value. So, the integral of π’ cubed is π’
to the fourth power divided by four. And this means our integral is
equal to a third times π’ to the fourth power divided by four evaluated between
negative two and five. Weβll substitute π’ equals five and
π’ equals negative two, and find their difference.

In this case, thatβs simply a third
of five to the fourth power divided by four minus negative two to the fourth power
divided by four. That gives us a third of 609 over
four. And we can cancel through by
three. And we obtain our solution to be
equal to 203 over four, or 50.75. And so, the definite integral
between one and two of π₯ squared times π₯ cubed minus three cubed with respect to
π₯ is 50.75.

In this example, we saw that we
should try and choose π’ to be some factor of the integrand whose differential also
occurs, albeit some scalar multiple of it. If thatβs not possible though, we
try choosing π’ to be a more complicated part of the integrand. This might be the inner function in
a composite function or similar.

Letβs have a look at an example of
this form.

Find the definite integral between
negative one and four of π₯ times the square root of π₯ plus five with respect to π₯
to the nearest thousandth.

In this question, our integrand is
the product of two functions, one of which is itself a composite function. It certainly is a tricky one to
integrate. And really, we have two
options. We could try the substitution
method or integration by parts. Notice though that the inner part
of our composite function has a nice simple derivative. And that tells us we might be able
to use integration by substitution.

The substitution rule for definite
integrals says that if π prime is continuous on the closed interval π to π, and
π is continuous over the range of π’, which is equal to π of π₯. Then the definite integral between
π and π of π of π of π₯ times π prime of π₯ with respect to π₯ is equal to the
definite integral between π of π and π of π of π of π’ with respect to π’. So, with this method, we try to
choose π’, which is π of π₯, to be some factor of the integrand whose differential
also occurs, albeit a scalar multiple of it. Here though, itβs not instantly
obvious what that might be.

So, instead, we choose π’ to be
some more complicated part of the function, here, the inner part in a composite
function that has a nice derivative. Weβll try π’ equals π₯ plus
five. The derivative of π₯ plus five with
respect to π₯ is simply one. And whilst we know that dπ’ by dπ₯
isnβt a fraction, we do treat it a little like one. We treat dπ’ and dπ₯ as
differentials. And we can say that dπ’ equals
dπ₯. Now, this may not seem instantly
helpful, as if we replace dπ₯ with dπ’ and π₯ plus five with π’, we still have a
part of our function thatβs in terms of π₯.

However, if we look back to our
substitution, we can rearrange this and say that π₯ must be equal to π’ minus
five. So, we can now replace each part of
our integrand and dπ₯ with dπ’. But what about our limits? Well, here we use our substitution
to redefine these. Our lower limit is when π₯ is equal
to negative one. And since π’ was equal to π₯ plus
five, π’ is equal to negative one plus five, which is equal to four. Then, when π₯ is equal to four,
thatβs our upper limit, π’ is equal to four plus five, which is nine.

So, we rewrite our definite
integral as the definite integral between four and nine, those are our new limits,
of π’ minus five β remember, we said π₯ is equal to π’ minus five β times the square
root of π’ with respect to π’. And actually, letβs rewrite the
square root of π’ as π’ to the power of one-half. And then, we can distribute the
parentheses. When we multiply π’ by π’ to the
power of one-half, we add their exponents. And we end up with π’ to the power
of three over two. So, our integrand is π’ to the
power of three over two minus five π’ to the power of one-half.

When integrating simple polynomial
terms whose exponent is not equal to negative one, we add one to the exponent and
divide by this new value. So, when we integrate π’ to the
power of three over two, we get π’ to the power of five over two divided by five
over two. And thatβs the same as two-fifths
times π’ to the power of five over two. Similarly, when we integrate
negative five π’ to the power of one-half, we get negative five π’ to the power of
three over two divided by three over two. And that simplifies to negative 10
over three π’ to the power of three over two.

Now, of course, we mustnβt forget
that weβre going to need to evaluate this between the limits of four and nine. Thatβs two-fifths of nine to the
power five over two minus ten-thirds of nine to the power of three over two minus
two-fifths times four to the power of five over two minus ten-thirds times four to
the power of three over two. Thatβs 316 over 15, which correct
to the nearest thousandth is 21.067.

In our next example, weβll consider
how this process also works for fractional functions.

Determine the definite integral
between negative five and negative two of two over the square root of π₯ plus six
dπ₯.

It may not be instantly obvious how
weβre going to evaluate this integral. However, if we look carefully, we
can see that the numerator is a scalar multiple of the derivative of the inner
function on our denominator. In other words, the derivative of
π₯ plus six multiplied by two is equal to the numerator. Thatβs two. And thatβs a hint to us that weβre
going to need to use integration by substitution to evaluate our integral.

Remember, in integration by
substitution, we introduce a new function. That is π’, which we let π equal
to π of π₯. And we see that the definite
integral between π and π of π of π of π₯ times π prime of π₯ with respect to π₯
is equal to the definite integral between π of π and π of π of π of π’ with
respect to π’. Weβre going to let π’ be equal to
the inner part of our composite function, thatβs π₯ plus six, so that dπ’ by dπ₯ is
equal to one. And whilst dπ’ by dπ₯ isnβt a
fraction, we are allowed to treat it a little like one when weβre working with
integration by substitution. And we can say that dπ’ is equal to
dπ₯.

And so, we can replace dπ₯ with dπ’
and π₯ plus six with π’. But weβre going to need to do
something with our limits. We use our substitution to redefine
them. Our lower limit is when π₯ is equal
to negative five. So, π’ then is equal to π₯ plus
six, which is here negative five plus six, which is of course one. Then, our upper limit is when π₯ is
equal to negative two. So, π’ is equal to negative two
plus six, which is four. And so, our integral is equal to
the definite integral between one and four of two over the square root of π’ with
respect to π’.

Now, in fact, we can write one over
the square root of π’ as π’ to the power of negative one-half. And this next step isnβt entirely
necessary, but it can make the process simpler. We recall that we can take any
constant factors outside of the integral and focus on integrating the function in π’
itself. So, this is equal to two times the
definite integral between one and four of π’ to the power of negative one-half. Now, when we integrate π’ to the
power of negative one-half, we add one to the exponent and then divide by that new
number. So, π’ to the power of negative
half becomes π’ to the power of one-half divided by one-half, which is the same as
two times π’ to the power of one-half.

We then replace π’ with four and
one and find the difference. We get two times two lots of four
to the power of one-half minus two lots of one to the power of one-half. Well, four to the power of one-half
is two. And one to the power of one-half is
one. So, we have two times two times two
minus two times one, which is simply equal to four. And weβre done. Since we changed our limits, we
donβt need to do anything more. Our definite integral is equal to
four.

In the previous two examples, we
saw how we can perform integration by substitution, even if itβs not instantly
obvious what it might look like. Weβre now going to see how we can
use the process to integrate a more complicated trigonometric function.

Find the definite integral between
zero and π by four of negative nine tan π§ times sec squared π§ with respect to π§.

Firstly, letβs not worry that this
function is in terms of π§. Weβre integrating with respect to
π§, so we perform the process as normal. Then, we notice that sec squared π§
is the derivative of tan of π§. This tells us we can use
integration by substitution to evaluate it. Weβre going to let π’ be equal to
tan of π§. And we know that the first
derivative of tan of π§ is sec squared π§. And whilst dπ’ by dπ§ is not a
fraction, we treat it a little like one, and we say that dπ’ is equal to sec squared
π§ dπ§.

And we now see that we can replace
tan of π§ with π’ and we can replace sec squared π§ dπ§ with dπ’. Before we go any further though,
weβre going to need to work out what our new limits are. So, we use our substitution. We said π’ is equal to tan of π§
and our lower limit is when π§ is equal to zero, which is when π’ is equal to tan of
zero, which is zero. Our upper limit is π§ is equal to
π by four. So, π’ is equal to tan of π by
four, which is one. And thatβs great because we can now
rewrite our definite integral as the definite integral between zero and one of
negative nine π’ with respect to π’.

Now, we could, if we wanted, take
negative nine out as a constant factor. Of course, we donβt need to. And if we donβt, we find that the
integral of negative nine π’ is negative nine π’ squared divided by two. Weβre going to evaluate this
between zero and one. Thatβs negative nine times one
squared over two minus negative nine times zero squared over two, which is just
simply negative nine over two. Our definite integral is equal to
negative nine over two.

In our very final example, weβll
look at how to use integration by substitution to evaluate the integral of a
logarithmic function.

Determine the definite integral
between one and π of the natural log of π₯ over π₯ with respect to π₯.

In order to evaluate this integral,
we need to spot that the derivative of the natural log of π₯ is one over π₯ and that
a part of this function is indeed a scalar multiple of one over π₯. We, therefore, let π’ be equal to
the natural log of π₯. And when we differentiate π’ with
respect to π₯, we get one over π₯. Now, dπ’ by dπ₯ is not a fraction,
but we treat it a little like one. And we see that this is equivalent
to saying dπ’ equals one over π₯ dπ₯. And thatβs great because we can now
replace the natural log of π₯ with π’ and one over π₯ dπ₯ with dπ’.

We are going to need to change our
limits. So, we use our substitution. We said that π’ is equal to the
natural log of π₯. And when π₯ is equal to one, π’
must be equal to the natural log of one, which is zero. And when π₯ is equal to π, thatβs
the upper limit, π’ is equal to the natural log of π, which is one. And thatβs great because our
definite integral is now equal to the definite integral between zero and one of π’
with respect to π’.

The integral of π’ is π’ squared
over two. And when we evaluate this between
the limits of zero and one, we get one squared over two minus zero squared over two,
which is of course one-half. The definite integral between one
and π of the natural log of π₯ over π₯ with respect to π₯ is a half.

In this video, we saw that the
substitution rule for definite intervals says that if π prime, the first derivative
of π, is continuous on the closed interval of π to π, and π itself is continuous
on the range of π’, which is equal to π of π₯. Then the definite integral between
π and π of π of π of π₯ times π prime of π₯ with respect to π₯ is equal to the
definite integral between π of π and π of π of π of π’ with respect to π’.

We saw that we usually try to
choose π’ to be some factor of the integrand whose differential also occurs, albeit
some scalar multiple of it. If thatβs not instantly obvious
though, we try choosing π’ to be some more complicated part of the integrand. It might be the inner function in a
composite function. And we saw that itβs really
important that we use our substitution to change our limits before performing the
integral. And we saw this method can be used
to integrate functions involving roots, trigonometric functions, and logarithms.