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17Calculus - Integration By Substitution

Single Variable Calculus
Multi-Variable Calculus

Integration By Substitution

Integration by substitution is the first major integration technique that you will probably learn and it is the one you will use most of the time. In fact, as you learn more advanced techniques, you will still probably use this one also, in addition to the more advanced techniques, on the same problem. So, this is a critically important technique to learn.

Integration by substitution is sometimes called u-substitution since the letter \(u\) is often used as the substitution variable.


Topics You Need To Understand For This Page

[basics of integrals] - [differentials] - [substitution from precalculus]

If you want a complete lecture on integration by substitution, we recommend this video.

Prof Leonard - Calculus 1 Lecture 4.2: Integration by Substitution [1hr-33mins-57secs]

video by Prof Leonard

Main Idea of The Substitution Technique

The main idea of this technique is to reduce the complexity of an integral by introducing a new variable to replace a part of the integrand.   Then we convert the entire integrand in terms of the new variable.   This should reduce the integral to one that we can evaluate using a more basic and simpler technique.

Where Integration by Substitution Comes From

The idea of integration of substitution comes from something you already now, the derivative chain rule. Remember, the chain rule for \(y=f(g(x))\) looks like \(y'=f'(g(x))g'(x)\). Integration by substitution reverses this by first giving you \(f'(g(x))g'(x)\) and expecting you to come up with \(f(g(x))\). This is easier than you might think and it becomes easier as you get some experience. The key is to know how to choose \(g(x)\). First, let's see how the chain rule becomes an integral.

As we said above, the chain rule for \(y=f(g(x))\) is \(y'=f'(g(x))g'(x)\). This can also be written \(dy/dx=f'(g(x))g'(x)\) or in differential form \(dy=f'(g(x))g'(x)~dx\). Now when we integrate both sides we have \(y=\int{ f'(g(x))g'(x)dx }\). The key is to let \(u=g(x)\) and then \(du=g'(x)dx\). The integral can now be written \(y=\int{ f'(u)~du } = f(u)+C = f(g(x))+C\).

So the key to this technique is knowing what to choose for \(g(x)\). As mentioned above, the letter \(u\) is often used as the substitution variable, letting \(u=g(x)\).

How To Choose \( u \)

Let's watch a video clip with a VERY quick example. In this video she gives some really good advice on what to choose for \(u\).

Krista King Math - Solving Integrals [1min-39secs]

video by Krista King Math

Whew! That video clip went through the example pretty quickly without much explanation. Let's take it a bit slower and explain what is going on. In the video, the presenter evaluated the integral \(\int{x \sqrt{x^2+1} ~dx}\). Using her ideas, she chose \(u=x^2+1\). She actually lays out the solution very well but doesn't allow you much time to digest it. So, go back to the video and pause it at the moment she shows her complete solution and make sure you can follow each step.

Knowing what to choose for \(u\) is a skill that you get better at with practice.   Fortunately, there are some specific guidelines to get you started.   We discuss each guideline in detail on separate pages and give you a chance to practice each technique individually.   Here is the list of guidelines with links to each page.   If you are just learning this technique, we recommend that you go through these guidelines in this order and work the practice problems on each page before moving on.

Once you have gone through each guideline page, we recommend that you work some additional practice problems.   We have set up a special page that contains practice problems that require only the basic integral techniques and integration by substitution.   We can't emphasize enough how important this technique is to your ability to evaluate integrals.   So the more practice problems you work, the better.

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