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How to Solve Integrals Using Substitution

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  • 0:06 Review of the Chain Rule
  • 2:14 Solving By U Substitution
  • 3:45 Examples of U Substitution
  • 5:55 Checking U Substitution
  • 9:46 Lesson Summary
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Lesson Transcript
Instructor: Kelly Sjol
Some integrals are as easy as riding a bike. But more often, integrals can look like deformed bikes from Mars in the year 3000. In this lesson, you will learn how to transform these scary-looking integrals into simpler ones that really are as easy as riding a bike.

When you're trying to find a definite or indefinite integral, things may not be particularly clear, and you can't just look it up in a table all the time. In order to find the integral, or the area under the curve, you need a few more tools.

The steps for using substitution to solve integrals
U Substitution Steps

Review of the Chain Rule

The first tool is the chain rule. Remember that if we have a derivative of a composite function like d/dx(f(g(x))), we sometimes write it as d/dx(f(u)), where u is g(x), so u is a function of x. Then its derivative becomes f`u * du/dx , which equals f `(u)u`. So we're taking the derivative of the outer function and multiplying it times the derivative of the inner function.

We also know that you use the chain rule for derivatives whenever you see parentheses, as well as for composite functions. A composite function is like when you take some meat and you put it into a function to make a patty. Then you put that patty into a function to make a burger. So you put the function for making a patty into the function for making a burger. You put meat in, and you get out a hamburger. How does this relate to integration?

Well, if by using the chain rule you write d/dx * f(u)= f`(u)u` and you integrate both sides, you end up with the integral of (f`(u)u`)dx = f(u). So this equals our original function. Hmm, what does this mean? This means you can solve by something called substitution.

Solving the first u substitution example problem
U Substitution Example 1

Solving by U Substitution

We're going to substitute something for the inner function in our composite function. Let me explain. You're going to take an integral f(g(x)), which is a composite function, times g`(x) (the derivative of the inside, the 'patty-forming,' function), times dx. That equals the anti-derivative of your function evaluated at g(x), or F(g(x)). And if this is an indefinite integral, you're going to add a constant of integration, C.

So how do you do this? First, you're going to make what's called a substitution, and you're going to say there's some function, u, that is equal to g(x). Then you're going to take the derivative of u - that's du - which is equal to g`(x)dx. Once you've made the substitution, you're going to find the anti-derivative of your function f. Finally, once you have the anti-derivative, you're going to substitute u back into your function and that should give you the integral of your original function. Make sure by checking it. So what does all of this mean? Let's do an example.

Examples of U Substitution

Let's say we have the integral of 3e^3x * dx. There are some implied parentheses here, around the 3x. So I'm going to substitute my variable u for 3x, u=3x. Then I'm going to take the derivative of u, differentiating this with respect to x, and write du=3dx. Then I'm going to use these two substitutions in my original integral. So I have the integral 3e^(3x) dx, which I can rewrite as the integral of e^(3x) 3dx. Well 3dx is equal to du, and e^(3x) is the same as e^u, because u=3x; that's how I defined it.

Checking the answer in the first u substitution example problem
Checking U Substitution Example 1

So my new function is e^u, and I'm integrating with respect to u now. Let's actually do that integral. The integral of e^u * du is just e^u + some constant C. We have our constant because this is an indefinite integral, and we have e^u because I know that the derivative of e^u is e^u and the integral of e^u is e^u. It's that function that never changes. Okay, so now I have everything in terms of u. Let's get rid of the u by substituting u=3x into this equation, and I get e^(3x) + C. By using what's called u substitution, I have solved the integral of 3e^(3x) dx as e^(3x) + C. But before I go ahead and say this is right, let's check the answer.

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