Integration by parts is a "fancy" technique for solving integrals. It is usually the last resort. Actually, it is quite simple. It is just the application of the product rule to solve integrals.
First of all, let's remember the product rule:
This formula is easy to memorize.
Now, what we have is an equality of two functions. If two functions are equal, their integrals must also be equal.
So, let's find the antiderivatives of both sides:
As integration reverses differentiation, we have:
Now, arranging this equality:
Look at that. Using just the product rule we obtained a formula for integration. This formula is called integration by parts.
Many people use the letters u and v instead of f and g. In the videos I'll use these letters. You can start with this one:
When we use this formula, we "divide the integral in parts". We must find another integral, but this one is hopefully simpler.
Integration by parts is useful when the function we want to integrate can be written as:
That is, the product of a function f and the derivative of another function g.
Let's do an example.
Let's find the integral:
According to our formula:
So, here we have all set:
We can make:
So, we have:
Easy enough. Now, you might ask a very important question: how did we choose our f(x) and g'(x)?
As a rule of thumb, you must choose f(x) such that its derivative is simpler than f(x).
Don't worry, the ability to choose f(x) correctly will be second nature to you after doing many exercises. In the previous example we had f(x) = x. Its derivative is one. That is much simpler!
In a later section I'll show you a very easy to remember rule for choosing f(x), called the LIATE rule.
You can watch the video for this section:
We have our formula:
So, we choose:
And our integral becomes:
And the antiderivative of -cos(x) is -sin(x). So, our integral is:
As always we must add the constant of integration.
Integration by parts is very "tricky" by nature. Here I'll show you one special trick. In the formula:
We can consider g'(x) = 1. This is useful because that function can always be written in an integral.
It is surprising we don't know this integral yet! We can write it like this:
So we have:
And our integral is:
And then, simplifying the x's in the integral:
And this is an easy integral:
As an exercise you might want to find the integral:
This second trick could be considered more a method than a trick. Integration by parts is usually the last resort for solving an integral.
When you apply it you may still get an integral that you don't know how to solve. What do we do in those cases? Integrate by parts again! An example:
Here we'll choose:
And we have:
Taking the 2 out of the integral sign:
In example 1, we've found that:
We can replace this result to get:
If you find this integral in an exam, for example, you'll need to use the formula twice.
When you first learn this technique of integration, the difficulty is in choosing f(x) and g'(x) correctly. What happens if you don't choose well? Your integral won't simplify.
Here's a rule of thumb I learned when studying integration, and I've been using it since then.
It was proposed by Herbert Kasube of Bradley University. It is called the LIATE rule. LIATE stands for:
L: Logarithmic functions
I: Inverse trigonometric functions
A: Algebraic functions: x squared, etc.
T: Trigonometric functions
E: Exponential functions
How do we use it? Whatever function from that list comes first should be our f(x).
For example, let's consider again the integral:
Thee we have the product of two functions: one algebraic (x) and one trigonometric.
According to LIATE, A (algebraic) comes before T (trigonometric). So, the algebraic function (x in this case) should be our f(x).
If you check all the examples we did you'll see tat we followed LIATE.
The next time you need to apply integration by parts try to use the LIATE rule, I'm sure you'll be surprised at how effective it is.
Here we'll solve one of the few integrals that don't follow the LIATE rule. This one is a trick question in many exams:
Here we won't apply LIATE and make:
You'll have to trust me on this one. We have:
Here we'll apply a trick we've already learned and use integration by parts again:
This time we call our functions u(x) and v'(x):
Our formula says:
Applying it in our integral:
And now we have something interesting. Look at this:
Here we have an equation. So, we can still apply the rules of algebra. We can add the same integral to both sides and get:
Dividing both sides by two:
And that's the integral we were looking for!
This is a very tricky integral. As an exercise, you might want to solve this similar integral:
Intuitive Calculus Insights is a free e-zine that provides you with unique ideas and tips for succeeding in calculus. I send it every two weeks, and you'll only receive valuable and unique information.
Subscribe now, and you'll also receive Free and Unlimited access to the two resources only available to subscribers:
If you have just a general doubt about a concept, I'll try to help you. If you have a problem, or set of problems you can't solve, please send me your attempt of a solution along with your question. These will appear on a new page on the site, along with my answer, so everyone can benefit from it.
Click below to see contributions from other visitors to this page...
Sign up below to get access to the two resources only available to subscribers:
Plus: Intuitive Calculus Insights. A free subscription to my newsletter, which I send once every two weeks. You'll receive study tips and other valuable things that'll help you succeed in calculus. Learn more here...