Video: Integration by Parts

In this video, we will learn how to use integration by parts to find the integral of a product of functions.

16:11

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 π‘₯.

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 π‘₯. This is sometimes alternatively written using functions 𝑒 and 𝑣 and Leibniz notation. Such that the integral of 𝑒 times d𝑣 by dπ‘₯ evaluated with respect to π‘₯ is equal to 𝑒𝑣 minus the integral of 𝑣 times d𝑒 by dπ‘₯ evaluated with respect to π‘₯. We’ll begin by considering a fairly straightforward example of the application of this formula.

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

The function we’re looking to integrate is the product of two functions. That’s π‘₯ and sin π‘₯. This indicates that we might need to use integration by parts to evaluate our integral. Remember the formula here says the integral of 𝑒 times d𝑣 by dπ‘₯ is equal to 𝑒𝑣 minus the integral of 𝑣 times d𝑒 by dπ‘₯. If we compare this formula to our integrand, we see that we’re going to need to decide which function is 𝑒. And which function is d𝑣 by dπ‘₯. 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 𝑒 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 π‘₯ and sin π‘₯, the function that becomes simpler when differentiated is π‘₯. So we let 𝑒 be equal to π‘₯ and d𝑣 by dπ‘₯ be equal to sin π‘₯. The derivative of 𝑒 with respect to π‘₯ is one. And what do we do with d𝑣 by dπ‘₯? Well, we’re going to need to find 𝑣. So we find the antiderivative of sin of π‘₯, which is of course negative cos of π‘₯.

Let’s substitute what we have into our formula. We see that our integral is equal to π‘₯ times negative cos of π‘₯ minus the integral of negative cos of π‘₯ times one dπ‘₯. This simplifies to negative π‘₯ cos of π‘₯. And then we take the factor of negative one outside of our integral. And we see that we add the integral of cos of π‘₯ evaluated with respect to π‘₯. The antiderivative of cos of π‘₯ is sin of π‘₯. 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 π‘₯ cos of π‘₯ plus sin of π‘₯ plus 𝑐. Now it’s useful to remember that we can check our answer by differentiating. If we do, we indeed obtain π‘₯ times sin π‘₯ as required. We’re now going to look at a common example. That’s the integral of the natural log of π‘₯.

Integrate the natural log of π‘₯ dπ‘₯ by parts using 𝑒 equals the natural log of π‘₯ and d𝑣 equals dπ‘₯.

Well, this is great. We’ve been told to use integration by parts. And we’ve also been told to let 𝑒 be equal to the natural log of π‘₯ and d𝑣 be equal to dπ‘₯. So we recall the formula for integration by parts. The integral of 𝑒 times d𝑣 by dπ‘₯ is equal to 𝑒𝑣 minus the integral of 𝑣 times d𝑒 by dπ‘₯. Now we can rewrite d𝑣 equals dπ‘₯ slightly. And we can say that if d𝑣 is equal to dπ‘₯, then d𝑣 by dπ‘₯ must be equal to one. Our job is going to be to find the missing parts of the formula. d𝑒 by dπ‘₯ is fairly straightforward. The derivative of the natural log of π‘₯ is one over π‘₯. And if we integrate both sides of the equation d𝑣 equals dπ‘₯, we obtain 𝑣 as being equal to π‘₯. Substituting everything into the formula, and we get π‘₯ times the natural log of π‘₯ minus the integral of π‘₯ times one over π‘₯ dπ‘₯.

Well, this integral simplifies really nicely. We’re actually going to be integrating one with respect to π‘₯. The integral of one is, of course, simply π‘₯. 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 π‘₯ dπ‘₯ as being π‘₯ times the natural log of π‘₯ minus π‘₯ plus 𝑐. And integration by parts was really effective for integrating the natural log of π‘₯. Because the derivative of the natural log of π‘₯ is much simpler than the original function the natural log of π‘₯. 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 π‘₯ times π‘₯ over three times π‘₯ plus one all squared, evaluated with respect to π‘₯.

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 𝑒 times d𝑣 by dπ‘₯ is equal to 𝑒𝑣 minus the integral of 𝑣 times d𝑒 by dπ‘₯. Now, if we compare this formula to our integrand, we see that we need to decide which function is 𝑒. And which function is d𝑣 by dπ‘₯. 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 𝑒 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 𝑒 to be. So a little trial and error might be in order.

Let’s rewrite our integrand as two-thirds π‘₯ times 𝑒 to the power of π‘₯ times one over π‘₯ plus one squared. And in fact, let’s take the constant factor of two-thirds outside of the integral. Let’s try 𝑒 equals π‘₯ times 𝑒 to the power of π‘₯ and d𝑣 by dπ‘₯ as being one over π‘₯ plus one all squared, which I’ve written as π‘₯ plus one to the power of negative two. To find d𝑒 by dπ‘₯, we’re going to need to use the product rule. If we do, we see that d𝑒 by dπ‘₯ is equal to π‘₯ times the derivative of 𝑒 to the power of π‘₯ plus 𝑒 to the power of π‘₯ times the derivative of π‘₯. Well, the derivative of 𝑒 to the power of π‘₯ is 𝑒 to the power of π‘₯. And the derivative of π‘₯ is one. So we’ve obtained d𝑒 by dπ‘₯ to be equal to π‘₯𝑒 to the power of π‘₯ plus 𝑒 to the power of π‘₯. We can use the reverse of the chain rule to work out the antiderivative of π‘₯ plus one to the power of negative two. It’s negative π‘₯ plus one to the negative one. Rewriting π‘₯ plus one to the power of negative one as one over π‘₯ 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 π‘₯. And then we can see we can divide through by π‘₯ plus one. And our second integrand becomes negative 𝑒 to the π‘₯. We take out that factor of negative one. And we know that the integral of 𝑒 of the power of π‘₯ is simply 𝑒 to the power of π‘₯. And so we have two-thirds of π‘₯ times 𝑒 to the power of π‘₯ over negative π‘₯ plus one plus 𝑒 to the power of π‘₯ plus 𝑐. We rewrite this first fraction slightly and then multiply the numerator and the denominator of 𝑒 to the power of π‘₯ by π‘₯ plus one. So that we can add these fractions. And we see that the sum of these fractions is negative π‘₯ times 𝑒 to the power of π‘₯ plus π‘₯ times 𝑒 to the power of π‘₯, which is zero, plus 𝑒 to the power of π‘₯ over π‘₯ plus one.

Our final step is to distribute the parentheses. And we get two 𝑒 to the power of π‘₯ over three times π‘₯ plus one. And I’ve changed our constant of integration to capital 𝐢. 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 𝑒 to be. But there is a bit of a trick that can help us identify how to choose the function for 𝑒. It’s the letters L-I-A-T-E or LIATE. Essentially, you set 𝑒 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 π‘₯ squared times 𝑒 to the power of π‘₯ dπ‘₯.

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 𝑒 times d𝑣 by dπ‘₯ is equal to 𝑒𝑣 minus the integral of 𝑣 times d𝑒 by dπ‘₯. So how do we decide what we’re going to let 𝑒 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 𝑒 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 𝑒 be equal to 𝑒 to the power of π‘₯. Since the derivative of 𝑒 to the power of π‘₯ is just 𝑒 to the power of π‘₯. So, instead, we’re going to let 𝑒 be equal to π‘₯ squared. And d𝑣 by dπ‘₯ is therefore 𝑒 to the power of π‘₯. d𝑒 by dπ‘₯ is then two π‘₯. And the antiderivative of 𝑒 to the power of π‘₯ is 𝑒 to the power of π‘₯. So 𝑣 is 𝑒 to the power of π‘₯. And we obtain the integral to be equal to π‘₯ squared 𝑒 to the power of π‘₯ minus the integral of two π‘₯𝑒 to the power of π‘₯.

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 π‘₯ times 𝑒 to the power of π‘₯? Well, we’re going to need to use integration by parts again. For much the same reason, we once again choose d𝑣 by dπ‘₯ to be equal to 𝑒 to the power of π‘₯. And then 𝑒 is equal to two π‘₯. So we see that d𝑒 by dπ‘₯ is equal to two. And 𝑣 is equal to 𝑒 to the power of π‘₯.

So let’s just quickly evaluate the integral of two π‘₯𝑒 to the power of π‘₯. It’s two π‘₯𝑒 to the power of π‘₯ minus the integral of two times 𝑒 to the power of π‘₯. Well, the integral of two 𝑒 to the power of π‘₯ is just two 𝑒 to the power of π‘₯. 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 π‘₯𝑒 to the power of π‘₯ with two π‘₯𝑒 to the power of π‘₯ minus two 𝑒 to the π‘₯. And we obtain the integral of π‘₯ squared 𝑒 to the power of π‘₯ to be equal to π‘₯ squared 𝑒 to the power of π‘₯ minus two π‘₯𝑒 to the power of π‘₯ plus two 𝑒 to the power of π‘₯.

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 π‘₯ squared times 𝑒 to the power of π‘₯ 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 π‘₯ with respect to π‘₯.

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 π‘₯. We then let 𝑒 be equal to the inverse tan of π‘₯. Remember, we know how to find the derivative of this. And we let d𝑣 by dπ‘₯ be equal to one. The derivative of the inverse tangent function is one over one plus π‘₯ squared. So that’s d𝑒 by dπ‘₯. And the antiderivative of one is simply π‘₯. So 𝑣 is equal to π‘₯. Great! Substituting everything we have into the formula for integration by substitution. And we see that the integral of the inverse tan of π‘₯ is π‘₯ times the inverse tan of π‘₯ minus the integral of π‘₯ over one plus π‘₯ squared.

Now we’re actually going to need to use integration by substitution here to evaluate the integral of π‘₯ over one plus π‘₯ squared. Now, usually, when performing integration by substitution, we introduce a new variable 𝑒. 𝑒 already has a meaning in this example, though. So we’re going to let 𝑑 be equal to one plus π‘₯ squared, which means that d𝑑 by dπ‘₯ is equal to two π‘₯. Now, d𝑑 by dπ‘₯ 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 d𝑑 equals π‘₯ dπ‘₯. So actually, we’re going to be working out the integral of one over two 𝑑 or taking the constant factor of a half outside the integral. So we can just actually integrate one over 𝑑. But we’re going to need to do something with these limits. We’re going to use the substitution 𝑑 equals one plus π‘₯ squared. And the first limit, that we’re interested in, is one. So we substitute π‘₯ equals one. And we get one plus one squared, which is two.

Down here, we’re substituting π‘₯ equals zero. That’s one plus zero squared, which is equal to one. So we’re going to evaluate the integral of one over 𝑑 between the limits of one and two. The integral of one over 𝑑 is simply the natural log of 𝑑. So we can evaluate the natural log of 𝑑 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 π‘₯ times the inverse tan of π‘₯ between the limits of zero and one. Well, that’s one times the inverse tan of one minus zero, which is equal to πœ‹ by four. And we’re done. The integral of the inverse tan of π‘₯ between the limits of zero and one is πœ‹ 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 𝑒 times d𝑣 by dπ‘₯ equals 𝑒𝑣 minus the integral of 𝑣 times d𝑒 by dπ‘₯. We saw that, generally, we choose our function 𝑒, 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 𝑒. 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.

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