U-substitution is a useful method for integrating composite functions. In this article, we’ll discuss the meaning of u-substitution, how to use u-substitution, common mistakes students make, and review some practice problems.
You’re probably familiar with the idea that integration is the reverse process of differentiation. U-substitution is an integration technique that specifically reverses the chain rule for differentiation. Because of this, it’s common to refer to u-substitution as the reverse chain rule. We may also refer to it as integration by substitution, or “change of variables” integration.
U-substitution integration allows us to find the antiderivative of composite functions.
Dr. Tim Chartier explains antiderivatives more:
You can think of u-substitution as the chain rule executed backward. To “undo” the chain rule, we rewrite the integral in terms of du and u.
Since u-substitution “undoes” the chain rule, we can use the chain rule formula to help determine which problems require u-substitution. If you can spot a function and its derivative in the same integrand, that indicates that u-substitution is likely the best integration method for that scenario.
How to Do U-Substitution
First, let’s review the chain rule below. The chain rule states:
An integrand that you can evaluate using u-substitution might look something like this. Do you see the chain rule formula?
Using u-substitution, we substitute u=g(x) and du=g’(x) into our integrand, which “undoes” the chain rule:
4 Steps to Follow
Here are 4 simple steps for u-substitution:
Pick your “u”. This expression is the "inside" part of the chain rule and is usually the term inside a radical, power, or denominator.
Differentiate u to find du. If your du does not match what’s left inside the integrand perfectly, you must rearrange your du so that it does match perfectly.
Substitute u and du into the integrand, and integrate using the key integration formulas that are already familiar to you.
Substitute the original value for u back into the equation after integrating. Remember to add the constant of integration to your final answer.
Let’s solve one problem together. We’ll evaluate ∫3(3x+1)5dx.
Examine this function carefully. First, notice that we have a composite of functions with 3x+1 raised to the fifth power. And, notice that we have a function and its derivative in the same integral. The derivative of 3x+1 is 3, and both 3x+1 and 3 are inside the integrand. This means that u-substitution is the way to go!
Let’s walk through our 4 steps.
Remember that we want our substitution integral to look like ∫f(u)du. Since 3x+1 is part of a composite of functions, and since the derivative of 3x+1 is present elsewhere in the integrand, we’ll let u=3x+1.
To find du, we can find dxdu and then multiply by dx. Since dxdu=3, then du=3dx.
Now we can evaluate our integral by substituting u=3x+1 and du=3dx into the integrand. Then, we can evaluate using the power rule for integrals.
4. Finally, we can substitute the original value for u, u=3x+1, back into our equation to get our final answer.
To help with Step 3, review the list of standard integral rules below. These formulas are essential for any integration method. Assume that f and g are continuous functions.
Sum and Difference Rule:
Constant Multiplier Rule:
∫kf(x)dx=k∫f(x)dx for some constant k
∫xndx=n+1xn+1+C for some real number n
∫adx=ax+C for some constant a
Exponential and Logarithmic Function Rules:
∫axdx=ln(x)ax+C, for any positive real number a
Trigonometric Function Rules:
For these rules, assume that x is in radians.
Two Common Mistakes Using U-Substitution
Choosing the wrong u or du
Many students struggle with determining which expression to designate as u and which expression to designate as du. Choosing the wrong u and du will result in an incorrect answer.
Remember, you’re looking for two functions within the integrand that fit the framework given by the chain rule. Make sure that u is equal to the “inside” function of the chain rule, or the inner part of the composite of functions. Double-check that you’ve differentiated u correctly to find du.
Forgetting to multiply/divide by a constant, when necessary.
Sometimes, your calculated du isn’t immediately visible inside the integrand. In this case, you can algebraically adjust your du substitution to match the expression that is present in the integrand. Usually, this means multiplying or dividing the integrand by a constant. For example, consider ∫xsin(x2+1)dx.
Solving for u and du, we find the u=x2+1 and du=2xdx. But it’s clear that du=2xdx isn’t part of our integrand, and can’t be immediately substituted.
This is easily fixed. While du=2xdx isn’t present, notice that xdx is. We can solve for xdx and substitute that value into our integrand instead. Dividing both sides of du=2xdx by 2, we find that 21du=xdx.
Now, we can substitute u=x2+1 and 21du=xdx into the integrand and solve normally.
Examples of U-Substitution
We’ll work through 3 more practice problems together.
Let u=3x+1. Then, du=3dx. Now, we can substitute u and du into the integrand.
Let u=5x+2. Then, du=5dx. But notice that 5dx is not immediately visible in our integrand. We can fix this by cleverly manipulating the integrand so that 5x becomes present. We’ll use the trick that 51⋅5=1. So, we have:
Now we can substitute u and du as normal, and use the constant multiplier rule to bring the constant 21 outside of the integral.
Let u=ln(x). Then, du=x1dx. Now, we can substitute u and du into the integrand.