Video Transcript
In this video, we’ll learn how to
use integration by parts to find the integral of a product of functions. Because of the fundamental theorem
of calculus, we can integrate a function if we know its antiderivative. And by this stage, you should feel
confidence in evaluating the integral of polynomial functions as well as
trigonometric and exponential functions. It’s not entirely necessary to be
confident in applying integration by substitution to access this video. Though it would be beneficial.
Every differentiation rule has a
corresponding integration rule. For example, integration by
substitution is the rule that corresponds to the chain rule for differentiation. We’re now going to be looking at
integration by parts. This rule for integration
corresponds to the product rule for differentiation. The product rule states that, for
two differentiable functions 𝑓 and 𝑔, the derivative of their product is 𝑓 times
𝑔 prime plus 𝑔 times 𝑓 prime. That is, 𝑓 times the derivative of
𝑔 plus 𝑔 times the derivative of 𝑓. By integrating both sides of this
equation, we obtain 𝑓 times 𝑔 to be equal to the integral of 𝑓 times 𝑔 prime
plus 𝑓 prime times 𝑔, evaluated with respect to 𝑥.
Now we know that the integral of
the sum of two functions is equal to the sum of the integral of each function. And we can rewrite 𝑓 times 𝑔 as
shown. We rearrange. And we obtain the formula for
integration by parts. We see that the indefinite integral
of 𝑓 times 𝑔 prime is equal to 𝑓 times 𝑔 minus the indefinite integral of 𝑓
prime times 𝑔, evaluated with respect to 𝑥. This is sometimes alternatively
written using functions 𝑢 and 𝑣 and Leibniz notation. Such that the integral of 𝑢 times
d𝑣 by d𝑥 evaluated with respect to 𝑥 is equal to 𝑢𝑣 minus the integral of 𝑣
times d𝑢 by d𝑥 evaluated with respect to 𝑥. We’ll begin by considering a fairly
straightforward example of the application of this formula.
Use integration by parts to
evaluate the integral of 𝑥 times sin 𝑥 with respect to 𝑥.
The function we’re looking to
integrate is the product of two functions. That’s 𝑥 and sin 𝑥. This indicates that we might need
to use integration by parts to evaluate our integral. Remember the formula here says the
integral of 𝑢 times d𝑣 by d𝑥 is equal to 𝑢𝑣 minus the integral of 𝑣 times d𝑢
by d𝑥. If we compare this formula to our
integrand, we see that we’re going to need to decide which function is 𝑢. And which function is d𝑣 by
d𝑥. So how do we decide this? Well, our aim is going to be to
ensure that the second integral we get over here is a little simpler. We therefore want 𝑢 to be a
function that either becomes simpler when differentiated or helps to simplify the
integrand when multiplied by the function 𝑣. It should be quite clear that, out
of 𝑥 and sin 𝑥, the function that becomes simpler when differentiated is 𝑥. So we let 𝑢 be equal to 𝑥 and d𝑣
by d𝑥 be equal to sin 𝑥. The derivative of 𝑢 with respect
to 𝑥 is one. And what do we do with d𝑣 by
d𝑥? Well, we’re going to need to find
𝑣. So we find the antiderivative of
sin of 𝑥, which is of course negative cos of 𝑥.
Let’s substitute what we have into
our formula. We see that our integral is equal
to 𝑥 times negative cos of 𝑥 minus the integral of negative cos of 𝑥 times one
d𝑥. This simplifies to negative 𝑥 cos
of 𝑥. And then we take the factor of
negative one outside of our integral. And we see that we add the integral
of cos of 𝑥 evaluated with respect to 𝑥. The antiderivative of cos of 𝑥 is
sin of 𝑥. And of course, because this is an
indefinite integral, we must add that constant of integration 𝑐. And we see that the solution is
negative 𝑥 cos of 𝑥 plus sin of 𝑥 plus 𝑐. Now it’s useful to remember that we
can check our answer by differentiating. If we do, we indeed obtain 𝑥 times
sin 𝑥 as required.
We’re now going to look at a common
example. That’s the integral of the natural
log of 𝑥.
Integrate the natural log of 𝑥 d𝑥
by parts using 𝑢 equals the natural log of 𝑥 and d𝑣 equals d𝑥.
Well, this is great. We’ve been told to use integration
by parts. And we’ve also been told to let 𝑢
be equal to the natural log of 𝑥 and d𝑣 be equal to d𝑥. So we recall the formula for
integration by parts. The integral of 𝑢 times d𝑣 by d𝑥
is equal to 𝑢𝑣 minus the integral of 𝑣 times d𝑢 by d𝑥. Now we can rewrite d𝑣 equals d𝑥
slightly. And we can say that if d𝑣 is equal
to d𝑥, then d𝑣 by d𝑥 must be equal to one. Our job is going to be to find the
missing parts of the formula. d𝑢 by d𝑥 is fairly straightforward. The derivative of the natural log
of 𝑥 is one over 𝑥. And if we integrate both sides of
the equation d𝑣 equals d𝑥, we obtain 𝑣 as being equal to 𝑥. Substituting everything into the
formula, and we get 𝑥 times the natural log of 𝑥 minus the integral of 𝑥 times
one over 𝑥 d𝑥.
Well, this integral simplifies
really nicely. We’re actually going to be
integrating one with respect to 𝑥. The integral of one is, of course,
simply 𝑥. And of course, since this is an
indefinite integral, we must remember to add that constant of integration 𝑐. So we obtain the integral of the
natural log of 𝑥 d𝑥 as being 𝑥 times the natural log of 𝑥 minus 𝑥 plus 𝑐. And integration by parts was really
effective for integrating the natural log of 𝑥. Because the derivative of the
natural log of 𝑥 is much simpler than the original function the natural log of
𝑥.
In our next example, we’re going to
consider how integration by parts can be used to find the integral of a
quotient.
Determine the indefinite integral
of two 𝑒 to the power of 𝑥 times 𝑥 over three times 𝑥 plus one all squared,
evaluated with respect to 𝑥.
It might not be instantly obvious
how we are going to evaluate this. However, there’s no clear
substitution we can make. And it’s certainly not something we
can evaluate in our heads. So that tells us we might need to
use integration by parts. Remember, this says the integral of
𝑢 times d𝑣 by d𝑥 is equal to 𝑢𝑣 minus the integral of 𝑣 times d𝑢 by d𝑥. Now, if we compare this formula to
our integrand, we see that we need to decide which function is 𝑢. And which function is d𝑣 by
d𝑥. So how do we decide this? Well, our aim is to make sure that
the second integral we get over here is a little simpler. We therefore want 𝑢 to be a
function that either becomes simpler when differentiated or helps us to simplify the
integrand when multiplied by this 𝑣. Now it isn’t hugely obvious what we
should choose 𝑢 to be. So a little trial and error might
be in order.
Let’s rewrite our integrand as
two-thirds 𝑥 times 𝑒 to the power of 𝑥 times one over 𝑥 plus one squared. And in fact, let’s take the
constant factor of two-thirds outside of the integral. Let’s try 𝑢 equals 𝑥 times 𝑒 to
the power of 𝑥 and d𝑣 by d𝑥 as being one over 𝑥 plus one all squared, which I’ve
written as 𝑥 plus one to the power of negative two. To find d𝑢 by d𝑥, we’re going to
need to use the product rule. If we do, we see that d𝑢 by d𝑥 is
equal to 𝑥 times the derivative of 𝑒 to the power of 𝑥 plus 𝑒 to the power of 𝑥
times the derivative of 𝑥. Well, the derivative of 𝑒 to the
power of 𝑥 is 𝑒 to the power of 𝑥. And the derivative of 𝑥 is
one. So we’ve obtained d𝑢 by d𝑥 to be
equal to 𝑥𝑒 to the power of 𝑥 plus 𝑒 to the power of 𝑥. We can use the reverse of the chain
rule to work out the antiderivative of 𝑥 plus one to the power of negative two. It’s negative 𝑥 plus one to the
negative one. Rewriting 𝑥 plus one to the power
of negative one as one over 𝑥 plus one. And we can substitute everything we
have into our formula for integration by parts.
Now over here, this next step is
important. We factor by 𝑒 to the power of
𝑥. And then we can see we can divide
through by 𝑥 plus one. And our second integrand becomes
negative 𝑒 to the 𝑥. We take out that factor of negative
one. And we know that the integral of 𝑒
of the power of 𝑥 is simply 𝑒 to the power of 𝑥. And so we have two-thirds of 𝑥
times 𝑒 to the power of 𝑥 over negative 𝑥 plus one plus 𝑒 to the power of 𝑥
plus 𝑐. We rewrite this first fraction
slightly and then multiply the numerator and the denominator of 𝑒 to the power of
𝑥 by 𝑥 plus one. So that we can add these
fractions. And we see that the sum of these
fractions is negative 𝑥 times 𝑒 to the power of 𝑥 plus 𝑥 times 𝑒 to the power
of 𝑥, which is zero, plus 𝑒 to the power of 𝑥 over 𝑥 plus one.
Our final step is to distribute the
parentheses. And we get two 𝑒 to the power of
𝑥 over three times 𝑥 plus one. And I’ve changed our constant of
integration to capital 𝐶. Since our original constant of
integration was multiplied by two-thirds.
Now, in this example, it was a
little bit tricky to figure out what we needed 𝑢 to be. But there is a bit of a trick that
can help us identify how to choose the function for 𝑢. It’s the letters L-I-A-T-E or
LIATE. Essentially, you set 𝑢 to the
first term you see in the list. L is logarithm. I is inverse trig. A stands for algebraic. T stands for trigonometric. And finally, you look for an
exponential function. Now this rule doesn’t cover
everything. No rule can. But it does work remarkably well
and can be a good starting point. In our next example, we’ll see how
sometimes integration by parts might be required twice.
Evaluate the definite integral
between the limits of zero and one of 𝑥 squared times 𝑒 to the power of 𝑥
d𝑥.
In this example, we have the
product of two functions. That’s a hint that we might need to
use integration by parts. This says that the integral of 𝑢
times d𝑣 by d𝑥 is equal to 𝑢𝑣 minus the integral of 𝑣 times d𝑢 by d𝑥. So how do we decide what we’re
going to let 𝑢 be equal to? Well, remember, we want to ensure
that this second integral over here is a little simpler. Therefore, we’re going to want 𝑢
to be a function that either becomes simpler when differentiated or helps to
simplify the integrand when multiplied by this 𝑣. It makes no sense for us to let 𝑢
be equal to 𝑒 to the power of 𝑥. Since the derivative of 𝑒 to the
power of 𝑥 is just 𝑒 to the power of 𝑥. So, instead, we’re going to let 𝑢
be equal to 𝑥 squared. And d𝑣 by d𝑥 is therefore 𝑒 to
the power of 𝑥. d𝑢 by d𝑥 is then two 𝑥. And the antiderivative of 𝑒 to the
power of 𝑥 is 𝑒 to the power of 𝑥. So 𝑣 is 𝑒 to the power of 𝑥. And we obtain the integral to be
equal to 𝑥 squared 𝑒 to the power of 𝑥 minus the integral of two 𝑥𝑒 to the
power of 𝑥.
Now notice, I’ve used this kind of
funny half bracket here. This is just a way of reminding
ourselves that we’re dealing with a definite integral. And we’re going to need to evaluate
both parts of it between the limits of zero and one. But how do we evaluate the integral
of two 𝑥 times 𝑒 to the power of 𝑥? Well, we’re going to need to use
integration by parts again. For much the same reason, we once
again choose d𝑣 by d𝑥 to be equal to 𝑒 to the power of 𝑥. And then 𝑢 is equal to two 𝑥. So we see that d𝑢 by d𝑥 is equal
to two. And 𝑣 is equal to 𝑒 to the power
of 𝑥.
So let’s just quickly evaluate the
integral of two 𝑥𝑒 to the power of 𝑥. It’s two 𝑥𝑒 to the power of 𝑥
minus the integral of two times 𝑒 to the power of 𝑥. Well, the integral of two 𝑒 to the
power of 𝑥 is just two 𝑒 to the power of 𝑥. And I’ve put plus 𝑐 in brackets
because the integral we’ve just done is an indefinite integral. But actually the one we’re really
going to be doing is between the limits of zero and one. Replacing the integral of two 𝑥𝑒
to the power of 𝑥 with two 𝑥𝑒 to the power of 𝑥 minus two 𝑒 to the 𝑥. And we obtain the integral of 𝑥
squared 𝑒 to the power of 𝑥 to be equal to 𝑥 squared 𝑒 to the power of 𝑥 minus
two 𝑥𝑒 to the power of 𝑥 plus two 𝑒 to the power of 𝑥.
Now, we’re going to need to
evaluate this between the limits of zero and one. Substituting zero and one and
finding their difference. And we have 𝑒 to the power of one
minus two 𝑒 to the power of one plus two 𝑒 to the power of one. All these disappear and minus
two. And we’re done. The integral evaluated between zero
and one of 𝑥 squared times 𝑒 to the power of 𝑥 is 𝑒 minus two.
In our final example, we’re going
to look at how integration by parts can help us to evaluate the integral of an
inverse trigonometric function.
Calculate the definite integral
evaluated between zero and one of the inverse tan of 𝑥 with respect to 𝑥.
When integrating inverse
trigonometric functions, we plump for using integration by parts. And the formula for that is as
shown. And we’re going to do something a
little bit strange. We’re going to rewrite our
integrand as one times the inverse trigonometric function. That’s one times the inverse tan of
𝑥. We then let 𝑢 be equal to the
inverse tan of 𝑥. Remember, we know how to find the
derivative of this. And we let d𝑣 by d𝑥 be equal to
one. The derivative of the inverse
tangent function is one over one plus 𝑥 squared. So that’s d𝑢 by d𝑥. And the antiderivative of one is
simply 𝑥. So 𝑣 is equal to 𝑥. Great! Substituting everything we have
into the formula for integration by substitution. And we see that the integral of the
inverse tan of 𝑥 is 𝑥 times the inverse tan of 𝑥 minus the integral of 𝑥 over
one plus 𝑥 squared.
Now we’re actually going to need to
use integration by substitution here to evaluate the integral of 𝑥 over one plus 𝑥
squared. Now, usually, when performing
integration by substitution, we introduce a new variable 𝑢. 𝑢 already has a meaning in this
example, though. So we’re going to let 𝑡 be equal
to one plus 𝑥 squared, which means that d𝑡 by d𝑥 is equal to two 𝑥. Now, d𝑡 by d𝑥 is, of course, not
a fraction. But we treat it a little like one
when working with integration by substitution. And we see that we can equivalently
say that a half d𝑡 equals 𝑥 d𝑥. So actually, we’re going to be
working out the integral of one over two 𝑡 or taking the constant factor of a half
outside the integral. So we can just actually integrate
one over 𝑡. But we’re going to need to do
something with these limits. We’re going to use the substitution
𝑡 equals one plus 𝑥 squared. And the first limit, that we’re
interested in, is one. So we substitute 𝑥 equals one. And we get one plus one squared,
which is two.
Down here, we’re substituting 𝑥
equals zero. That’s one plus zero squared, which
is equal to one. So we’re going to evaluate the
integral of one over 𝑡 between the limits of one and two. The integral of one over 𝑡 is
simply the natural log of 𝑡. So we can evaluate the natural log
of 𝑡 between the limits of one and two by substituting them in and finding their
difference. That’s the natural log of two minus
the natural log of one. And since the natural log of one is
zero, we found this integral to be equal to a half times the natural log of two. We can pop this back into our
original integral. And we see that we’re going to need
to evaluate 𝑥 times the inverse tan of 𝑥 between the limits of zero and one. Well, that’s one times the inverse
tan of one minus zero, which is equal to 𝜋 by four. And we’re done. The integral of the inverse tan of
𝑥 between the limits of zero and one is 𝜋 by four minus the natural log of two
over two.
In this video, we’ve seen that
integration by parts is the corresponding rule for integration to the product rule
for differentiation. We saw that using Leibniz notation,
the formula is the integral of 𝑢 times d𝑣 by d𝑥 equals 𝑢𝑣 minus the integral of
𝑣 times d𝑢 by d𝑥. We saw that, generally, we choose
our function 𝑢, with an aim of ensuring that the second integral we get is a little
simpler. But we also saw that the acronym
LIATE can help us decide which function is going to be 𝑢. We finally saw that we can use this
to evaluate the integrals of the product of functions, quotients, and reciprocal
functions. And sometimes it might be applied
more than once.
