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# 𝘶-substitution

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
𝘶-Substitution essentially reverses the chain rule for derivatives. In other words, it helps us integrate composite functions.
When finding antiderivatives, we are basically performing "reverse differentiation." Some cases are pretty straightforward. For example, we know the derivative of start color #1fab54, x, squared, end color #1fab54 is start color #7854ab, 2, x, end color #7854ab, so integral, start color #7854ab, 2, x, end color #7854ab, d, x, equals, start color #1fab54, x, squared, end color #1fab54, plus, C. We can use this straightforward reasoning with other basic functions, like sine, left parenthesis, x, right parenthesis, e, start superscript, x, end superscript, start fraction, 1, divided by, x, end fraction, etc.
Other cases, however, are not that simple. For example, what is integral, cosine, left parenthesis, 3, x, plus, 5, right parenthesis, d, x? Hint: it's not sine, left parenthesis, 3, x, plus, 5, right parenthesis, plus, C. Try differentiating that and you will see why.
One method that can be very useful is u-substitution, which basically reverses the chain rule.

## Using $u$u-substitution with indefinite integrals

Imagine we are asked to find integral, start color #7854ab, 2, x, end color #7854ab, start color #e07d10, cosine, left parenthesis, end color #e07d10, start color #1fab54, x, squared, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, start color #7854ab, d, x, end color #7854ab. Notice that start color #7854ab, 2, x, end color #7854ab is the derivative of start color #1fab54, x, squared, end color #1fab54, which is the "inner" function in the composite function start color #e07d10, cosine, left parenthesis, end color #e07d10, start color #1fab54, x, squared, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10. In other words, letting start color #1fab54, u, left parenthesis, x, right parenthesis, equals, x, squared, end color #1fab54 and start color #e07d10, w, left parenthesis, x, right parenthesis, equals, cosine, left parenthesis, x, right parenthesis, end color #e07d10, we have:
start color #7854ab, start overbrace, 2, x, end overbrace, start superscript, u, prime, end superscript, end color #7854ab, start color #e07d10, start underbrace, cosine, left parenthesis, start color #1fab54, start overbrace, x, squared, end overbrace, start superscript, u, end superscript, end color #1fab54, right parenthesis, end underbrace, start subscript, w, end subscript, end color #e07d10, equals, start color #7854ab, u, prime, left parenthesis, x, right parenthesis, end color #7854ab, start color #e07d10, w, left parenthesis, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, right parenthesis, end color #e07d10
This suggests that u-substitution is called for. Let's see how it's done.
First, we differentiate the equation start color #1fab54, u, equals, x, squared, end color #1fab54 according to x, while treating u as an implicit function of x.
\begin{aligned} u&=x^2 \\\\ \dfrac{d}{dx}[u]&=\dfrac{d}{dx}[x^2] \\\\ \dfrac{du}{dx}&=2x \\\\ \purpleD{du}&\purpleD{=2x\,dx} \end{aligned}
In that last row we multiplied the equation by d, x so d, u is isolated. That's somewhat unorthodox, but useful for our next step. So we have start color #1fab54, u, equals, x, squared, end color #1fab54 and start color #7854ab, d, u, equals, 2, x, d, x, end color #7854ab. Now we can perform a substitution in the integral:
\begin{aligned} &\phantom{=}\displaystyle\int \purpleD{2x}\goldD{\cos(}\greenD{x^2}\goldD )\,\purpleD{dx} \\\\ &=\displaystyle\int \goldD{\cos(\greenD{\underbrace{x^2}_{u}})}\purpleD{\underbrace{2x\,dx}_{du}}&\gray{\text{Rearrange.}} \\\\ &=\displaystyle\int \goldD{\cos(}\greenD{u}\goldD )\purpleD{\,du}&\gray{\text{Substitute.}} \end{aligned}
After the substitution we are left with an expression for the antiderivative of start color #e07d10, cosine, left parenthesis, end color #e07d10, start color #1fab54, u, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10 in terms of u. How convenient! cosine, left parenthesis, u, right parenthesis is a basic function so we can find its antiderivative in a straightforward way. The only thing left to do is return the function to be in terms of x:
\begin{aligned} &\phantom{=}\displaystyle\int \goldD{\cos(}\greenD{u}\goldD )\,du \\\\ &=\sin(\greenD u)+C \\\\ &=\sin(\greenD{x^2})+C \end{aligned}
In conclusion, integral, 2, x, cosine, left parenthesis, x, squared, right parenthesis, d, x is sine, left parenthesis, x, squared, right parenthesis, plus, C. You can differentiate sine, left parenthesis, x, squared, right parenthesis, plus, C to verify that this is true.
Key takeaway #1: u-substitution is really all about reversing the chain rule:
• According to the chain rule, the derivative of start color #e07d10, w, left parenthesis, end color #e07d10, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10 is start color #e07d10, w, prime, left parenthesis, end color #e07d10, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, dot, start color #7854ab, u, prime, left parenthesis, x, right parenthesis, end color #7854ab.
• In u-substitution, we take an expression of the form start color #e07d10, w, prime, left parenthesis, end color #e07d10, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, dot, start color #7854ab, u, prime, left parenthesis, x, right parenthesis, end color #7854ab and find its antiderivative start color #e07d10, w, left parenthesis, end color #e07d10, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10.
Key takeaway #2: u-substitution helps us take a messy expression and simplify it by making the "inner" function the variable.
Problem 1.A
Problem set 1 will walk you through all the steps of finding the following integral using u-substitution.
integral, left parenthesis, 6, x, squared, right parenthesis, left parenthesis, 2, x, cubed, plus, 5, right parenthesis, start superscript, 6, end superscript, d, x, equals, question mark
How should we define u?

### Common mistake: getting incorrect expressions for $u$u or $du$d, u

Choosing the wrong expression for u will result in a wrong answer. For example, in Problem set 1, u must be defined as 2, x, cubed, plus, 5. Letting u be 6, x, squared or left parenthesis, 2, x, cubed, plus, 5, right parenthesis, start superscript, 6, end superscript will never work.
Remember: For u-substitution to apply, we must be able to write the integrand as start color #e07d10, w, left parenthesis, end color #e07d10, start color #1fab54, u, left parenthesis, x, right parenthesis, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, dot, start color #7854ab, u, prime, left parenthesis, x, right parenthesis, end color #7854ab. Then, u must be defined as the inner function of the composite factor.
Another crucial step in this process is finding d, u. Make sure you are differentiating u correctly, because a wrong expression for d, u will also result in a wrong answer.
Problem 2
Tim was asked to find integral, cosine, left parenthesis, 5, x, minus, 7, right parenthesis, d, x. This is his work:
integral, cosine, left parenthesis, 5, x, minus, 7, right parenthesis, d, x, equals, sine, left parenthesis, 5, x, minus, 7, right parenthesis, plus, C
Is Tim's work correct? If not, what is his mistake?

### Common mistake: not realizing $u$u-substitution is called for

Remember: When integrating a composite function, we can't simply take the antiderivative of the outer function. We need to use u-substitution.
Letting W be an antiderivative of w, this point can be expressed mathematically as follows:
integral, w, left parenthesis, u, left parenthesis, x, right parenthesis, right parenthesis, d, x, does not equal, W, left parenthesis, u, left parenthesis, x, right parenthesis, right parenthesis, plus, C

### Another common mistake: confusing the inner function and its derivative

Imagine you're trying to find integral, x, squared, cosine, left parenthesis, 2, x, right parenthesis, d, x. You might say "since 2, x is the derivative of x, squared, we can use u-substitution." Actually, since u-substitution requires taking the derivative of the inner function, x, squared must be the derivative of 2, x for u-substitution to work. Since that's not the case, u-substitution doesn't apply here.

## Sometimes we need to multiply/divide the integral by a constant.

Imagine we are asked to find integral, start color #e07d10, sine, left parenthesis, end color #e07d10, start color #1fab54, 3, x, plus, 5, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, start color #7854ab, d, x, end color #7854ab. Notice that while we have a composite function start color #e07d10, sine, left parenthesis, end color #e07d10, start color #1fab54, 3, x, plus, 5, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, it is not multiplied by anything. That might seem weird at first, but let's proceed and see what happens.
We let start color #1fab54, u, equals, 3, x, plus, 5, end color #1fab54, then start color #7854ab, d, u, equals, 3, d, x, end color #7854ab. Now we substitute u into the integral, not before we perform this clever manipulation:
integral, start color #e07d10, sine, left parenthesis, end color #e07d10, start color #1fab54, 3, x, plus, 5, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, start color #7854ab, d, x, end color #7854ab, equals, start fraction, 1, divided by, 3, end fraction, integral, start color #e07d10, sine, left parenthesis, end color #e07d10, start color #1fab54, 3, x, plus, 5, end color #1fab54, start color #e07d10, right parenthesis, end color #e07d10, start color #7854ab, 3, d, x, end color #7854ab
See what we did there? In order to have start color #7854ab, 3, d, x, end color #7854ab in the integrand, we multiplied the entire integral by start fraction, 1, divided by, 3, end fraction. That way we allowed for u-substitution while keeping the value of the integral the same.
Let's continue with the substitution:
\begin{aligned} &\phantom{=}\dfrac13\displaystyle\int\goldD{\sin(\greenD{\underbrace{3x+5}_u})}\purpleD{\underbrace{3\,dx}_{du}} \\\\ &=\dfrac13\displaystyle\int\goldD{\sin(}\greenD{u}\goldD )\purpleD{\,du} \\\\ &=-\dfrac13\cos(\greenD{u})+C \\\\ &=-\dfrac13\cos(\greenD{3x+5})+C \end{aligned}
Key takeaway: Sometimes we need to multiply or divide the entire integral by a constant, so we can achieve the appropriate form for u-substitution without changing the value of the integral.
Problem 3
integral, left parenthesis, 2, x, plus, 7, right parenthesis, cubed, d, x, equals, question mark

Want more practice? Try this exercise.

## Want to join the conversation?

• who's confused cos I'm definitely baffled o_0
• wait a sec hold on
(1 vote)
• If u substitution does not apply in ∫x^2 cos(2x)dx which approach should we use?
• The product of a polynomial and a trig function is a classic example of integration by parts.
• Repeatedly the idea of multiplying the dx out is being referred to as unorthodox. Is there a more formal or mainstream way of doing or thinking about this? It makes intuitive sense to me, but I'm wondering if there's something I should know about the "proper" method.
• try googling "integration by parts"
• Can I use u-substitution to find anti-derivative of (x^2 + 1)^2? u = (x^2 + 1) and du/dx = 2x
• That won't work. Replacing x²+1 by u leaves you with u²dx, but the du=2xdx part doesn't help you.

However, this is just a polynomial. You can expand it as x⁴+2x²+1, then use the power rule.
• Is there a video somewhere that goes over just u, du, dx and how they are chosen or derived?
• u is just the variable that was chosen to represent what you replace.

du and dx are just parts of a derivative, where of course u is substituted part fo the function. u will always be some function of x, so you take the derivative of u with respect to x, or in other words du/dx.

There should be videos on this playlist such as "u-substitution intro" and a few others I would watch, but you would also want to watch chain rule videos as well, though here is my explanation. am assuming familiarity with the chain rule.

The chain rule starts with a composite function f(g(x)). Such as sin(x^2), where one function is the sine operation and the other is the squared operation. For the sake of clarity g(f(x)) would be sin^2(x).

Anyway, the chain rule says if you take the derivative with respect to x of f(g(x)) you get f'(g(x))*g'(x). That means if you have a function in THAT form, you can take the integral of it to look like f(g(x)). The process of doing this is traditionally u substitution.

So you start with f'(g(x))*g'(x). the first step is to make u=g(x) that way, when you take the derivative of u with respect to x (in other words du/dx) this gets you g'(x) So now you know what g(x) is in f(g(x)). since you start with f'(g(x))*g'(x) you ust have to take the integral of f'(g(x)) to get f(g(x)), though it's easier if g(x) is just a single variable, so we substitute in u for g(x). of course at the end you need to re-substitute g(x) for it.

For notation's sake, and to ensure everything works out logically du/dx = g'(x) is changed to du = g'(x) dx because when you take the integral of something you add dx (as long as it is with respect to x.) so the integral of f'(g(x))*g'(x) dx gets g'(x) dx replaced with du because f'(g(x)) becomes f'(u). so now you have the integral of f'(u) du which of course becomes f(u), then you replace u with g(x) to get f(g(x)) effectively undoing the chain rule.

Let me know if this did not help. And of course, I should mention with indefinite integrals you always put a + C at the end of your answer.
• Hi, would you please help me solve (2x^3)/(1+x^2) using you-substitution? I don't quite understand why the answer I am looking at online calls for a subtraction of two integrals. Thanks!
• Why is it that we are only manipulating dx to make our u substitution work but all other times we seem to ignore it's even there?
• Shouldn't it be 1/3(sin3x+5)+C instead of 3(sin3x+5)+C for the first example?
• I think you meant to w(x)=cos (x^2) in the initial example of using u substitution. If so u = x^2 and w = cos u.