Integration by Parts

Integration by Parts Video

Hey guys! Welcome to this video on integration by parts! At this point, you should know how to integrate basic functions, including some polynomials and other problems that can be solved with \(u\)-substitution. However, not all functions fall into these categories, and today we will discuss another method of integration that will enable us to solve more integration problems. This method is called integration by parts.

To begin, let’s take a look at a problem we should already be able to solve.

\(\int \left [ x+\sin \left ( 2x \right )\right ]dx\)


First, we can break this into two separate integrals by using the addition rule. So the addition rule says that we can break it up where there’s a plus sign, so we’ll have:

\(\int x~dx+\int \sin\left ( 2x \right )dx\)

The first integral, \(\int x~dx\), is simple to solve. Just increase the power by 1, and divide by the new exponent. So this part is equal to, if you increase \(x^{1}\) by 1, you’ll get \(x^{2}\), and then divide by 2, is the same as multiplying by \(\frac{1}{2}\), plus \(c\), our integration constant.

\(\int x~ dx=\frac{1}{2}x^{2}+c\)


For the second integral, notice that the derivative of \(2x\) is 2, which is a scalar multiple of the implied 1 in front of the sine function. Because of this relationship, we can apply \(u\)-substitution to solve the second integral.

\(\int \sin\left ( 2x \right )dx\rightarrow \int \frac{1}{2}\sin\left ( u \right )du\)


So, if we have \(\int \sin(2x)dx\), and we call \(2x\) our \(u\), then \(du=2~dx\). So this will be equal to \(\int \sin(u)du\).



\(\int \sin\left ( 2x \right )dx=\int \sin\left ( u \right )du\)


But we’re actually going to have to use \(\frac{1}{2}du\) to get it to be equal to \(dx\). We just divided by 2 on both sides.



And since we have a \(dx\) here, that’s what we’re going to want to do. So we’re going to write \(\frac{1}{2}\) right here:

\(\int \sin\left ( 2x \right )dx=\int \frac{1}{2}\sin\left ( u \right )du\)


Now we integrate this, and we’ll get:

\(-\frac{1}{2}\cos\left ( u \right )+c\)


Because \(-\cos(u)\) is the integral of \(\sin(u)\). Now all we have to do is plug in our \(u\), which is \(2x\), and we get:

\(-\frac{1}{2}\cos\left ( 2x \right )+c\)


Now, all we have to do is put our addition terms back together. So this is equal to our first part, \(\frac{1}{2}x^{2}\). We’re going to leave the \(c\) alone for a second. And then \(-\frac{1}{2}\cos(2x)+c\).

\(\int x~ dx+\int \sin\left ( 2x \right )dx=\frac{1}{2}x^{2}-\frac{1}{2}\cos\left ( 2x \right )+c\)


Don’t be concerned by the fact that one of our \(c\)’s went away. When we add together our two integration constants from the first and second terms, they combine to form another constant, so we are safe to lump them together using just one letter.

Solving that problem wasn’t so bad, but consider this next problem:

\(\int x\sin\left ( 2x \right )dx\)


\(\int x\sin(2x)~dx\) is almost identical to the one we just solved, except now we are dealing with multiplication where before we were simply adding the two terms. Unfortunately, we can’t just break this integral into two separate integrals like we can with the addition rule. What can we do, then?

Think back to what we know about taking derivatives of products. We had a special rule, the product rule, to help us with those types of problems. Similarly, with integration, we have a defined process for handling products, and it was actually derived by mathematicians from the product rule for derivatives. This process, as you might have guessed, is called integration by parts, and it goes like this.

If you have an integral that can be written in the form \(\int u~dv\), then its solution is equal to \(uv-\int v~ du\).

\(\int u~ dv=uv-\int v~ du\)


At first glance, this may seem a bit confusing or unhelpful, but let’s break it down and then apply it to our example.

To begin, there are two functions, \(u\) and  \(dv\), whose product we are interested in taking the integral of. If we know how to take the integral of  \(dv\), and we know how to take the derivative of \(u\), then we can use a little bit of algebra to apply the integration by parts formula and determine our answer.

Consider again our example: \(\int x\sin\left ( 2x \right )dx\)

The two functions here are \(x\) and \(\sin2x~dx\), so we will assign them to be our \(u\) and  \(dv\), respectively.

\( dv=\sin\left ( 2x \right )dx\)


Now, notice that I kept the \(dx\) with the \(dv\). This is important because we’re going to integrate this part later on.

In order to determine our solution using the integration by parts formula, we need to calculate \(uv\) and \(\int v~du\). We know that \(u=x\), but what is \(v\)? To find \(v\), we just need to take the antiderivative of \(dv\).

\(v=\int \sin\left ( 2x \right )dx\)

Now, remember, we found this earlier, so this was equal to \(-\frac{1}{2}\cos(2x)\).

\(v=\int \sin(2x)dx=-\frac{1}{2}\cos(2x)\)


And we’re not going to write the \(+c\) yet because we’ll do that at the end. Since we’re integrating a whole indefinite integral, we’ll do that at the very end. We don’t have to worry about it in our little steps in between.

So now we can calculate \(uv\).

\(uv=x\left [ -\frac{1}{2}\cos(2x) \right ]=-\frac{1}{2}x\cos(2x)\)


To find the last piece, \(\int v~ du\), we need to compute \(du\) and then integrate. The derivative of \(u\) is the derivative of \(x\), which is 1, so \(du\) is equal to  \(dx\).

\(du=  dx\)


Putting this together:

\(\int v~ du=\int -\frac{1}{2}\cos(2x)dx\)


Let’s put together what we have so far. Remember, the formula tells us that

\(\int u~ dv=uv-\int v~ du\)


We assigned \(u=x\), and \(dv=\sin(2x) dx\), which made \(du=dx\), and \(v=-\frac{1}{2}\cos(2x)\).

\(u = x\) \(du =  dx\)
\(v=-\frac{1}{2}\cos(2x)\) \(dv=\sin(2x) dx\)


Putting these pieces together into the formula, we now have that

\(uv-\int v~ du=-\frac{1}{2}x\cos(2x)-\int \left [ -\frac{1}{2}\cos(2x) \right ]dx\)


All we need to do from here is compute the second integral and simplify the entire expression. Let’s move that \(-\frac{1}{2}\) outside of the integral, which will cancel out the negative sign.

So this first part’s going to stay the same And we can pull this \(-\frac{1}{2}\) out, so we’ll have minus a negative, which becomes plus.

\(-\frac{1}{2}x\cos(2x)+\frac{1}{2}\int \cos(2x)dx\)


Now, the antiderivative of cosine is sine, and we pull out a \(\frac{1}{2}\) because of this \(2x\) in here. So that will look like this. Again, this first part stays the same, and \(\int \cos(2x)dx=\frac{1}{2}\sin{2x}\).



And this is where we’re going to add our \(+c\), because we’re at the very end and we just need to simplify.

So now, all we have to do is simplify, and we do that by, again, keeping this first part the same. There’s nothing we can simplify in this part. And then, multiplying \(\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\).

\(-\frac{1}{2}x\cos(2x)+\frac{1}{4}\sin\left ( 2x \right )+c\)


So our final answer is:

\(\int x\sin(2x)dx=-\frac{1}{2}x\cos(2x)+\frac{1}{4}\sin(2x)+c\)


This process may seem long, but with practice it will come along without much difficulty.

At the beginning of this problem, we decided to let \(u=x\), and to let \(dv=\sin(2x) dx\). This enabled us to solve the integral using the integration by parts formula. However, if we would have tried to solve the integral with these switched (that is, \(u=\sin(2x)\), and \(dv=x~ dx\)), we would have run into some problems. The new integral, \(\int v~ du\), would have been \(\int x^{2}\cos(2x)dx\). Like the problem we started with, we cannot easily solve this new integral with previous methods. By choosing incorrectly, we effectively go in circles.

Clearly, it is important to choose correctly which functions will be \(u\) and which will be \(dv\) when we integrate by parts. But how can we know the correct choice, apart from trial and error? Some mathematicians have developed a method for making this choice, called “ILATE.” ILATE is an acronym to help us choose our u, based on an order of priority.

When approaching a problem for which you will need to use integration by parts, first check to see if there is any…

I, inverse trig function. If so, select this to be \(u\). If not, then look for an…
L, logarithmic function. If there is one, then it will be what you choose for \(u\). Otherwise, check for an…
A, algebraic expression, such as \(x\) or \(x^2\). If there is none, see if there is any…
T, trigonometric functions, like \(\sin(x)\) or \(\cos(x)\). If none of the previous expressions are in the problem, then finally, check for…
E, exponential functions, such as \(e^x\).

Whichever of these you find in order of ILATE, that will be your \(u\). The part of the integral that is not chosen to be \(u\) will be your \(dv\).

Let’s look back to our problem from earlier one more time to apply ILATE.

\(\int x\sin(2x)dx\)


First, are there any (I) inverse trig functions? No, there are not. Next is L. Are there any log functions present? Again, no. Are there any (A) algebraic expressions? Yes! By using ILATE, we see that \(x\) is the best choice for our \(u\), because it is an algebraic expression. Notice that the sine function is a (T) trig function, but we don’t choose that to be u because the T comes after A in ILATE.

Let’s work one more problem.

\(\int x^{2}\ln(x)dx\)


We can’t use simple antiderivatives or \(u\)-substitution to solve this integral, and because we have a product of two functions, \(x^{2}\) and \(\ln(x)\), we’re going to use integration by parts. Which of the two will be our \(u\), and which will be our \(dv\)? We will use ILATE to decide.

There are no (I) inverse trig functions. There is, however, an (L) log function. We assign \(u=\ln(x)\), and we have no need to continue with A, T, and E. The leftover portion of this integral, \(x^{2}~dx\), is our \(dv\).

\(dv=x^{2} dx\)


To use the integration by parts formula, let’s go ahead and determine what \(du\) and \(v\) are, based on our choice. So \(du\) is the derivative of \(\ln(x)\), which is equal to \(\frac{1}{x} dx\), and \(v=\int x^{2}~dx\), which is equal to \(\frac{1}{3}x^{3}\). Remember, we don’t have to put the \(+c\) yet. We’ll put that at the very end.



Because the formula for integration by parts is:

\(\int u~ dv=uv-\int v~ du\)


We plug in our substitutions and get this. So \(uv=\ln(x)\frac{1}{3}x^{3}\), so I’m going to write the \(\frac{1}{3}x^{3}\) in front (that’s just the more formal way to write it), then \(-\int v~ du\).

\(\frac{1}{3}x^{3}ln(x)-\int v~ du\)


So, \(\int \frac{1}{3}x^{3}\times du\), which is \(du=\frac{1}{x} dx\).

\(\frac{1}{3}x^{3}\ln(x)-\int \frac{1}{3}x^{3}(\frac{1}{x})dx\)


Now we’re going to go ahead and simplify. So this is equal to:

\(\frac{1}{3}x^{3}ln(x)-\int \frac{1}{3}x^{2}~dx\)


The next step is that we’ll pull out our \(\frac{1}{3}\). So again, this first part stays the same, and we can subtract \(\frac{1}{3}\int x^{2}~dx\).

\(\frac{1}{3}x^{3}\ln(x)-\frac{1}{3}\int x^{2}~dx\)


And then we can integrate \(x^{2}\) to get \(\frac{1}{3}x^{3}\). And this is where we’ll add our \(+c\).

\(\frac{1}{3}x^{3}\ln(x)-\frac{1}{3}\left ( \frac{1}{3x}x^{3} \right )+c\)


And now all we have to do is simplify this last part right here. So we have



And that’s our final answer!

Let’s briefly run back through everything we’ve talked about. First, not all integrals can be solved using basic antiderivatives and u-substitution. Integration by parts allows us to work out more integrals, specifically those containing products of functions, using the formula

\(\int u~ dv=uv-\int v~ du\)


To make the best choice for which function should be \(u\), we use the acronym ILATE: inverse trig functions, log functions, algebraic expressions, trig functions, and finally, exponential functions. The part of the original integrand that we do not select to be \(u\) will be \(dv\). Before plugging everything into the formula, we need to calculate \(du\), the derivative of \(u\), and \(v\), the antiderivative of \(dv\). Once we have all four of those pieces, we simply substitute them all into the formula and simplify.

With a little practice, you’ll be fluent with integration by parts in no time!

Thanks for watching, and happy studying!


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by Mometrix Test Preparation | This Page Last Updated: December 20, 2023