### Video Transcript

Integration by Parts

In this video, we will learn how to
use integration by parts to find the integral of a product of two functions. Weβll do this by reversing the
product rule for derivatives and by using the fact that integration gives us the
most general antiderivative of a function.

So, letβs start by recalling what
the product rule for differentiation tells us. It tells us if we have two
differentiable functions π of π₯ and π of π₯, then the derivative of π of π₯
multiplied by π of π₯ with respect to π₯ is equal to π prime of π₯ times π of π₯
plus π of π₯ multiplied by π prime of π₯. And this rule helps us
differentiate the product of two functions. We need to reverse this rule to get
an equivalent rule in terms of integration. This will help us integrate the
product of two functions.

We can do this by noting π of π₯
multiplied by π of π₯ is an antiderivative of the right-hand side of the equation:
π prime of π₯ times π of π₯ plus π of π₯ times π prime of π₯. And now, since integration gives us
the most general antiderivative of a function, we can rewrite this as an integral
result. We get the indefinite integral of
π prime of π₯ times π of π₯ plus π of π₯ times π prime of π₯ with respect to π₯
is π of π₯ multiplied by π of π₯.

However, we canβt yet use this
result to evaluate the integral of a product of two functions because our integrand
contains two terms. But letβs instead take the integral
of each term in the integrand separately. This gives us the indefinite
integral of π prime of π₯ times π of π₯ with respect to π₯ plus the indefinite
integral of π of π₯ times π prime of π₯ with respect to π₯ is equal to π of π₯
multiplied by π of π₯.

We can now find an equation for the
product of two functions by rearranging to make this the subject. This then gives us that the
indefinite integral of π of π₯ times π prime of π₯ with respect to π₯ is equal to
π of π₯ multiplied by π of π₯ minus the indefinite integral of π prime of π₯
times π of π₯ with respect to π₯. And this is called the integration
by parts method.

And at first, we might notice
something interesting. We want to use this method to
evaluate the integral of the product of two functions. However, in our formula, weβre
using the integration of the product of two functions. And so it might not seem that this
formula will be useful. However, we can see weβre
differentiating a different function inside of our integrand. And as weβll see, being able to
differentiate one of the factors of our integrand often makes the integral easier to
evaluate.

And before we move on to some
examples, itβs also worth noting the integration by parts formula is often written
in Leibnizβs notation. However, this time, weβll call our
original functions π’ of π₯ and π£ of π₯. This gives us the indefinite
integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the
indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯. Letβs now see an example of
applying integration by parts to evaluate the integral of a product of two
functions.

Use integration by parts to
evaluate the indefinite integral of π₯ times the sin of π₯ with respect to
π₯.

In this question, we are asked
to evaluate the indefinite integral of the product of two functions: an
algebraic function π₯ and a trigonometric function sin of π₯. And this should hint to us that
we could try using integration by parts to evaluate this integral, even if we
were not told to use this method in the question. So, letβs start by recalling
the formula for integration by parts.

It tells us the indefinite
integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus
the indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯. To apply this to evaluate our
integral, weβre going to need to choose which of the factors in our integrand
needs to be π’ and which one should be dπ£ by dπ₯. So, we need to determine how
weβre going to choose which factor will be π’ and which will be dπ£ by dπ₯.

To do this, we need to know
when weβre using this formula, the most difficult part of this expression to
evaluate is the integral. So, we want to choose our
functions to make this integral as easy to evaluate as possible. And usually, we want to do this
by choosing π’ which makes dπ’ by dπ₯ simpler. However, sometimes, we need to
choose our function π’ such that it cancels when multiplied by part of π£. In either case, letβs look at
our functions π₯ and sin of π₯.

We know when we differentiate
π₯, its degree will decrease. However, if we were to
differentiate the sin of π₯, we would just get another trigonometric
function. This would not be any
simpler. So, weβll set our function π’
to be equal to π₯ and dπ£ by dπ₯ to be equal to the sin of π₯. Now, to apply integration by
parts, weβre going to need to find an expression for dπ’ by dπ₯ and π£.

So, letβs start by finding dπ’
by dπ₯. Thatβs the derivative of π₯
with respect to π₯, which is just equal to one. We also want to find an
expression for π£ from dπ£ by dπ₯. And we can recall π£ will be an
antiderivative of this function. So, we know that π£ is going to
be equal to the indefinite integral of the sin of π₯ with respect to π₯, which
we can recall is negative the cos of π₯ plus the constant of integration. However, we donβt need to add a
constant of integration in this case.

To see this, consider what
would happen if we replace π£ by π£ plus πΆ in our integration by parts
formula. In the first term, we would get
π’ multiplied by the constant of integration πΆ. However, in our second term, we
get πΆ times dπ’ by dπ₯. Negative the integral of this
is negative π’ time πΆ, so these two terms cancel. So, itβs another case of
constants of integration canceling, so we donβt need to include this
constant.

Letβs now substitute our
expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯ into our integration by parts
formula. We get the indefinite integral
of π₯ times the sin of π₯ with respect to π₯ is equal to π₯ times negative the
cos of π₯ minus the indefinite integral of negative cos of π₯ multiplied by one
with respect to π₯. The first term simplifies to
give us negative π₯ times the cos of π₯. And in our second term, we take
out the factor of negative one. This means we add the
indefinite integral of the cos of π₯ with respect to π₯.

Now, all thatβs left to do is
evaluate this indefinite integral. We know that the sin of π₯ is
an antiderivative of cos of π₯. So, the indefinite integral of
cos of π₯ is sin of π₯ plus the constant of integration πΆ. Therefore, we were able to show
the indefinite integral of π₯ times the sin of π₯ with respect to π₯ is negative
π₯ cos of π₯ plus sin of π₯ plus πΆ. And it is worth noting we can
check our answer by differentiating it with respect to π₯ and checking we get π₯
multiplied by the sin of π₯.

In our next example, weβre going to
use integration by parts to evaluate the integral of the natural logarithm
function.

Integrate the natural logarithm
of π₯ with respect to π₯ by parts using π’ is equal to the natural logarithm of
π₯ and dπ£ is equal to dπ₯.

In this question, we want to
evaluate an indefinite integral by using integration by parts. We can do this by first
recalling integration by parts tells us the indefinite integral of π’ times dπ£
by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite integral
of π£ multiplied by dπ’ by dπ₯ with respect to π₯.

Usually when weβre using
integration by parts, we first need to determine which function we set as π’ and
which one we set as dπ£ by dπ₯. However, weβre already told
this information in the question. Weβre told to set π’ equal to
natural logarithm of π₯. And weβre told in terms of
differentials, dπ£ is equal to dπ₯. This means that dπ£ by dπ₯ is
equal to one. We can then see if we set π’
equal to the natural logarithm of π₯ and dπ£ by dπ₯ equal to one, then in our
integration by parts formula we get the indefinite integral of the natural
logarithm of π₯ with respect to π₯.

So to apply this, weβre going
to need to find expressions for π£ and dπ’ by dπ₯. Letβs start by differentiating
π’ with respect to π₯. Thatβs the derivative of the
natural logarithm with respect to π₯, which we know is the reciprocal function
dπ’ by dπ₯ is one over π₯. Since dπ£ by dπ₯ is equal to
one, the derivative of π£ with respect to π₯ is equal to one. In other words, π£ is an
antiderivative of one. And we know the derivative of
π₯ with respect to π₯ is one, so weβll set π£ equal to π₯.

Now, we just substitute all of
these expressions into our integration by parts formula. We get the indefinite integral
of the natural logarithm of π₯ with respect to π₯ is equal to π₯ times the
natural logarithm of π₯ minus the indefinite integral of π₯ multiplied by one
over π₯ with respect to π₯. And we can simplify this; π₯
multiplied by one over π₯ is just equal to one. So, weβre just left with π₯
times the natural logarithm of π₯ minus the indefinite integral of one with
respect to π₯. And we can evaluate this
integral. Itβs equal to π₯ plus the
constant of integration πΆ.

Therefore, we were able to show
the indefinite integral of the natural logarithm of π₯ with respect to π₯ is
equal to π₯ times the natural logarithm of π₯ minus π₯ plus πΆ. And integration by parts was
really useful for helping us evaluate this integral because the derivative of
the natural logarithm of π₯ is a much simpler expression than its integral.

Thus far, when weβve used
integration by parts, weβve either been given our function dπ’ and dπ£ by dπ₯ or
weβve needed to choose them ourselves. And itβs quite difficult to choose
these functions ourselves since itβs not always obvious which function to choose as
π’. But there is one method we can use
to help us choose our function π’; itβs called the LIATE method. And this tells us to choose our
function π’ based on which of five possible types of functions appears first in our
integrand.

The L stands for logarithmic, the I
stands for inverse trigonometric functions, the A stands for algebraic, the T stands
for trigonometric, and finally the E stands for exponential. We just choose our function π’
based on which of these five functions appears first in our integrand. And it is worth noting the LIATE
method does not always give us the best function π’. However, it does usually work. Letβs now see an example of
applying the LIATE method to evaluate an integral by parts.

Determine the indefinite
integral of three π₯ plus four all squared times π to the power of π₯ with
respect to π₯.

In this question, we are asked
to evaluate the indefinite integral of the product of two functions. So, weβre going to attempt to
do this by using integration by parts. And we can recall integration
by parts tells us the indefinite integral of π’ times dπ£ by dπ₯ with respect to
π₯ is equal to π’ times π£ minus the indefinite integral of π£ multiplied by dπ’
by dπ₯ with respect to π₯.

So now, we need to choose our
functions π’ and dπ£ by dπ₯. And weβre going to choose our
function π’ by using the LIATE method. We check our integrand for each
of the five types of functions in turn to determine our function π’. First, we start with
logarithmic functions. We can see thereβs no
logarithmic function in our integrand. So, we move on to the next type
of function, inverse trig. Once again, we can see thereβs
no inverse trigonometric functions in our integrand. So, weβll move on to the third
type of function, algebraic, and we can see that three π₯ plus four all squared
is a polynomial. So, itβs an algebraic
function. So, weβll set π’ to be three π₯
plus four all squared. And this means dπ£ by dπ₯ is
going to be equal to the remaining factor π to the π₯.

Now, to apply integration by
parts, weβre going to need to find expressions for dπ’ by dπ₯ and π£. Letβs start with dπ’ by
dπ₯. Weβre going to want to
differentiate π’ with respect to π₯. And thereβs a few different
ways of doing this. For example, π’ is the
composition of functions, so we could do this by using the chain rule. However, our outer exponent is
only two. So, itβs probably easier to
just distribute two over our parentheses. We get nine π₯ squared plus
24π₯ plus 16. This then allows us to
differentiate this term by term by using the power rule for differentiation. We get dπ’ by dπ₯ is equal to
18π₯ plus 24.

Letβs now find π£. We can see that dπ£ by dπ₯ is
equal to π the π₯. So, π£ is going to be an
antiderivative of π to the π₯. π£ is equal to the indefinite
integral of π to the π₯ with respect to π₯. And we know this is equal to π
to the power of π₯ plus the constant of integration. However, we donβt need the
constant of integration in this case, we can now substitute these expressions
into our integration by parts formula. We get the indefinite integral
of three π₯ plus four all squared times π to the power of π₯ with respect to π₯
is equal to three π₯ plus four squared times π to the power of π₯ minus the
indefinite integral of π to the power of π₯ multiplied by 18π₯ plus 24 with
respect to π₯.

And at this stage, we might be
worried weβve chosen the wrong function π’ since weβve ended up with the
integral of a product of two functions. However, we can notice
something interesting. If we differentiate 18π₯ plus
24 with respect to π₯, weβll get a constant. So, we can evaluate this
integral by applying integration by parts a second time. To do this, letβs start by
clearing some space and then choose our function π’. Weβll do this once again by
using the LIATE method. We can see thereβs no
logarithmic functions and no inverse trigonometric functions. However, there is an algebraic
function.

So, weβll set π’ to be 18π₯
plus 24 and dπ£ by dπ₯ to be π to the power of π₯. We now want to find dπ’ by dπ₯
and π£. Since π’ is a linear function,
its derivative is the coefficient of π₯. dπ’ by dπ₯ is 18. And π£ is an antiderivative of
π to the power of π₯. π£ is just π to the power of
π₯. We can now substitute these
values into our integration by parts formula to evaluate the second indefinite
integral. This then gives us three π₯
plus four squared times π to the power of π₯ minus 18π₯ plus 24 times π to the
power of π₯ minus the indefinite integral of π to the power of π₯ times 18 with
respect to π₯. And we can simplify this.

First, we can distribute the
negative over our parentheses, which gives us the following. Next, we can take 18 out of our
integral. Now, we can evaluate our final
indefinite integral. The indefinite integral of π
to the power of π₯ with respect to π₯ is π to the power of π₯ plus the constant
of integration, which then leaves us with the following expression.

Now we could leave our answer
like this. However, we can also notice all
three of these terms share a factor of π to the power of π₯. So, if we take out the shared
factor of π to the power of π₯, we get the following expression, and we can
simplify the algebraic expression inside of our parentheses. And if we were to do this, we
would get our final answer. The indefinite integral of
three π₯ plus four all squared times π to the power of π₯ with respect to π₯ is
equal to π to the power of π₯ times nine π₯ squared plus six π₯ plus 10 plus
the constant of integration πΆ.

Letβs now see an example of using
indefinite integration to evaluate the integral of the product of a trigonometric
function and an exponential function.

By setting π’ equal to π to
the power of π₯ and dπ£ equal to cos of π₯ dπ₯, evaluate the indefinite integral
of π to the power of π₯ multiplied by the cos of π₯ with respect to π₯ by
integrating by parts.

In this question, we are asked
to evaluate an indefinite integral by using integration by parts. And we can recall integration
by parts tells us the indefinite integral of π’ times dπ£ by dπ₯ with respect to
π₯ is equal to π’ times π£ minus the indefinite integral of π£ multiplied by dπ’
by dπ₯ with respect to π₯. And at this point, we usually
need to choose our functions π’ and dπ£ by dπ₯ to apply integration by
parts. However, weβre told what to
choose in the question.

Weβre told to choose π’ equal
to π to the power of π₯. And weβre told in terms of
differentials, dπ£ is equal to cos of π₯ dπ₯. This is equivalent to saying
that dπ£ by dπ₯ is equal to cos of π₯. So, π’ is π to the power of
π₯, and dπ£ by dπ₯ is cos of π₯. We need to use these to
determine dπ’ by dπ₯ and π£. First, dπ’ by dπ₯ is the
derivative of π to the power of π₯ with respect to π₯, which is just equal to
itself.

Next, π£ is the indefinite
integral of the cos of π₯ with respect to π₯. Thatβs the sin of π₯ plus the
constant of integration. However, in this case, we donβt
need the constant. We can now substitute these
expressions into our integration by parts formula. This then gives us π to the
power of π₯ multiplied by the sin of π₯ minus the indefinite integral of the sin
of π₯ times π to the power of π₯ with respect to π₯.

And at this point, we might be
worried. Weβve not made our indefinite
integral any easier to evaluate by using this integration by parts. Integrating sin of π₯ times π
to the power of π₯ is just as difficult as integrating π to the power of π₯
times the cos of π₯. However, we can notice
something interesting if we try and evaluate this integral by using integration
by parts. If we once again set our
function π’ to be the exponential function π to the power of π₯ and our
function dπ£ by dπ₯ to be the trigonometric function sin π₯, then π’ will be π
to the power of π₯ and π£ is going to be negative cos of π₯.

So, this indefinite integral
ends up being the indefinite integral of π to the power of π₯ times cos of π₯
with respect to π₯. This is the integral weβre
trying to evaluate, and weβll be able to use this to evaluate our integral. So, letβs start by setting π’
equal to π to the power of π₯ and dπ£ by dπ₯ equal to the sin of π₯. We differentiate π’ with
respect to π₯. We get dπ’ by dπ₯ is equal to
π to the power of π₯. And the indefinite integral of
the sin of π₯ with respect to π₯ is negative the cos of π₯ plus πΆ. So, π£ is negative cos of
π₯.

We can now substitute these
expressions into our integration by parts formula to evaluate the indefinite
integral. This then gives us π to the π₯
times sin π₯ minus π to the π₯ multiplied by negative cos of π₯ minus the
indefinite integral of negative cos of π₯ times π to the power of π₯ with
respect to π₯. And we can then simplify
this. Letβs start by distributing the
negative over our parentheses and rearranging. This gives us π to the π₯
times sin π₯ plus π to the π₯ times cos of π₯ minus the indefinite integral of
π to the π₯ multiplied by cos of π₯ with respect to π₯.

But this integral is exactly
the same as the integral we have on the left-hand side of the equation. So, letβs add the indefinite
integral of π to the power of π₯ times the cos of π₯ with respect to π₯ to both
sides of the equation. This gives us two times the
indefinite integral of π to the power of π₯ times cos of π₯ with respect to π₯
is equal to π to the power of π₯ times sin π₯ plus π to the power of π₯ times
cos of π₯ plus the constant of integration πΆ.

And it is worth reiterating we
do need to add a constant of integration here. We can now solve for the
integral by dividing both sides of the equation through by one-half. And of course, dividing our
constant of integration by two will still leave us with a constant. So, weβll just call this
πΆ. This then gives us our final
answer. The indefinite integral of π
to the power of π₯ multiplied by the cos of π₯ with respect to π₯ is equal to
one-half times π to the π₯ sin π₯ plus π to the π₯ cos of π₯ plus the constant
of integration πΆ.

Letβs now go over the key points of
this video. First, we showed that integration
by parts is the corresponding rule for the product rule. In other words, if we consider the
product rule in terms of indefinite integration, we get integration by parts. In particular, in terms of
Leibnizβs notation, integration by parts tells us the indefinite integral of π’
times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite
integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯.

And we saw that in general itβs a
good idea to choose our function π’ so that when we differentiate it, we make our
second integral as easy as possible to evaluate. And we also saw one method of
choosing our function π’ by using the LIATE acronym. We choose π’ to be the first
function in this list of five which appears in our integrand. However, it is worth noting the
LIATE acronym does not always give us the best function π’, so we should always be
careful of this.

Finally, itβs also worth noting we
saw we could integrate certain functions, such as the natural logarithm of π₯, by
using integration by parts. We just write it as one times the
natural logarithm of π₯ and then apply integration by parts. And finally, we saw that sometimes
we need to apply integration by parts multiple times to evaluate the indefinite
integral.