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# 𝘶-substitution

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 ${x}^{2}$ is $2x$, so $\int 2x\phantom{\rule{0.167em}{0ex}}dx={x}^{2}+C$. We can use this straightforward reasoning with other basic functions, like $\mathrm{sin}\left(x\right)$, ${e}^{x}$, $\frac{1}{x}$, etc.
Other cases, however, are not that simple. For example, what is $\int \mathrm{cos}\left(3x+5\right)\phantom{\rule{0.167em}{0ex}}dx\phantom{\rule{0.167em}{0ex}}$? Hint: it's not $\mathrm{sin}\left(3x+5\right)+C$. 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 $u$‍ -substitution with indefinite integrals

Imagine we are asked to find $\int 2x\mathrm{cos}\left({x}^{2}\right)\phantom{\rule{0.167em}{0ex}}dx$. Notice that $2x$ is the derivative of ${x}^{2}$, which is the "inner" function in the composite function $\mathrm{cos}\left({x}^{2}\right)$. In other words, letting $u\left(x\right)={x}^{2}$ and $w\left(x\right)=\mathrm{cos}\left(x\right)$, we have:
$\stackrel{{u}^{\prime }}{\stackrel{⏞}{2x}}\underset{w}{\underset{⏟}{\mathrm{cos}\left(\stackrel{u}{\stackrel{⏞}{{x}^{2}}}\right)}}={u}^{\prime }\left(x\right)w\left(u\left(x\right)\right)$
This suggests that $u$-substitution is called for. Let's see how it's done.
First, we differentiate the equation $u={x}^{2}$ according to $x$, while treating $u$ as an implicit function of $x$.
$\begin{array}{rl}u& ={x}^{2}\\ \\ \frac{d}{dx}\left[u\right]& =\frac{d}{dx}\left[{x}^{2}\right]\\ \\ \frac{du}{dx}& =2x\\ \\ du& =2x\phantom{\rule{0.167em}{0ex}}dx\end{array}$
In that last row we multiplied the equation by $dx$ so $du$ is isolated. That's somewhat unorthodox, but useful for our next step. So we have $u={x}^{2}$ and $du=2x\phantom{\rule{0.167em}{0ex}}dx$. Now we can perform a substitution in the integral:
$\begin{array}{rl}& \phantom{=}\int 2x\mathrm{cos}\left({x}^{2}\right)\phantom{\rule{0.167em}{0ex}}dx\\ \\ & =\int \mathrm{cos}\left(\underset{u}{\underset{⏟}{{x}^{2}}}\right)\underset{du}{\underset{⏟}{2x\phantom{\rule{0.167em}{0ex}}dx}}& \text{Rearrange.}\\ \\ & =\int \mathrm{cos}\left(u\right)\phantom{\rule{0.167em}{0ex}}du& \text{Substitute.}\end{array}$
After the substitution we are left with an expression for the antiderivative of $\mathrm{cos}\left(u\right)$ in terms of $u$. How convenient! $\mathrm{cos}\left(u\right)$ 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 $x$:
$\begin{array}{rl}& \phantom{=}\int \mathrm{cos}\left(u\right)\phantom{\rule{0.167em}{0ex}}du\\ \\ & =\mathrm{sin}\left(u\right)+C\\ \\ & =\mathrm{sin}\left({x}^{2}\right)+C\end{array}$
In conclusion, $\int 2x\mathrm{cos}\left({x}^{2}\right)\phantom{\rule{0.167em}{0ex}}dx$ is $\mathrm{sin}\left({x}^{2}\right)+C$. You can differentiate $\mathrm{sin}\left({x}^{2}\right)+C$ 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 $w\left(u\left(x\right)\right)$ is ${w}^{\prime }\left(u\left(x\right)\right)\cdot {u}^{\prime }\left(x\right)$.
• In $u$-substitution, we take an expression of the form ${w}^{\prime }\left(u\left(x\right)\right)\cdot {u}^{\prime }\left(x\right)$ and find its antiderivative $w\left(u\left(x\right)\right)$.
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.
$\int \left(6{x}^{2}\right)\left(2{x}^{3}+5{\right)}^{6}\phantom{\rule{0.167em}{0ex}}dx=?$
How should we define $u$?

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

Choosing the wrong expression for $u$ will result in a wrong answer. For example, in Problem set 1, $u$ must be defined as $2{x}^{3}+5$. Letting $u$ be $6{x}^{2}$ or $\left(2{x}^{3}+5{\right)}^{6}$ will never work.
Remember: For $u$-substitution to apply, we must be able to write the integrand as $w\left(u\left(x\right)\right)\cdot {u}^{\prime }\left(x\right)$. Then, $u$ must be defined as the inner function of the composite factor.
Another crucial step in this process is finding $du$. Make sure you are differentiating $u$ correctly, because a wrong expression for $du$ will also result in a wrong answer.
Problem 2
Tim was asked to find $\int \mathrm{cos}\left(5x-7\right)dx$. This is his work:
$\int \mathrm{cos}\left(5x-7\right)dx=\mathrm{sin}\left(5x-7\right)+C$
Is Tim's work correct? If not, what is his mistake?

### Common mistake: not realizing $u$‍ -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:
$\int w\left(u\left(x\right)\right)\phantom{\rule{0.167em}{0ex}}dx\ne W\left(u\left(x\right)\right)+C$

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

Imagine you're trying to find $\int {x}^{2}\mathrm{cos}\left(2x\right)\phantom{\rule{0.167em}{0ex}}dx$. You might say "since $2x$ is the derivative of ${x}^{2}$, we can use $u$-substitution." Actually, since $u$-substitution requires taking the derivative of the inner function, ${x}^{2}$ must be the derivative of $2x$ 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 $\int \mathrm{sin}\left(3x+5\right)\phantom{\rule{0.167em}{0ex}}dx$. Notice that while we have a composite function $\mathrm{sin}\left(3x+5\right)$, it is not multiplied by anything. That might seem weird at first, but let's proceed and see what happens.
We let $u=3x+5$, then $du=3\phantom{\rule{0.167em}{0ex}}dx$. Now we substitute $u$ into the integral, not before we perform this clever manipulation:
$\int \mathrm{sin}\left(3x+5\right)\phantom{\rule{0.167em}{0ex}}dx=\frac{1}{3}\int \mathrm{sin}\left(3x+5\right)3\phantom{\rule{0.167em}{0ex}}dx$
See what we did there? In order to have $3\phantom{\rule{0.167em}{0ex}}dx$ in the integrand, we multiplied the entire integral by $\frac{1}{3}$. That way we allowed for $u$-substitution while keeping the value of the integral the same.
Let's continue with the substitution:
$\begin{array}{rl}& \phantom{=}\frac{1}{3}\int \mathrm{sin}\left(\underset{u}{\underset{⏟}{3x+5}}\right)\underset{du}{\underset{⏟}{3\phantom{\rule{0.167em}{0ex}}dx}}\\ \\ & =\frac{1}{3}\int \mathrm{sin}\left(u\right)\phantom{\rule{0.167em}{0ex}}du\\ \\ & =-\frac{1}{3}\mathrm{cos}\left(u\right)+C\\ \\ & =-\frac{1}{3}\mathrm{cos}\left(3x+5\right)+C\end{array}$
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
$\int \left(2x+7{\right)}^{3}\phantom{\rule{0.167em}{0ex}}dx=?$

Want more practice? Try this exercise.

## Want to join the conversation?

• who's confused cos I'm definitely baffled o_0
• At least you're introducing trigonometry into your communication, cos that's not easy.
• 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"
(1 vote)
• 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.
• if trigonometry enters this field, am I gone?
• Pretty vague question. Trigonometry is extensively integrated into calculus, so you'll eventually encounter it. That said, if you know your trig identities, it's really easy. Trust me, going forward, the easiest thing about integrals is doing the integral itself, be it trig, logs or anything else. Once you learn about double and triple integrals (yeah they exist), you'll see that the integration is the easiest part of it.
• 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.
• 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.
• can we multiply/divide by a variable?
• As long as you do it to both sides of the equation, yes, you can (with the exception of the variable being = 0)
• 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!
• So I have this question that has me baffled. It is: find the integral of 2sec^2⋅(x)tan(x)*

Now to solve it, I need u substitution. Well, when I saw the question, I found that I have two options for u, and both will work. Here's what I did:

*u = tan(x)*
du = sec^2(x) dx
dx = du/sec^2(x)

Putting this in the question, I get:
int(2sec^2(x)⋅u du/sec^2(x)) = int(2u du)
= 2int(u du) = 2×1/2 u^2 = u^2 = tan^2(x) + C

So here we got the integral is *tan^2(x) + C
.

Now the second option

u = sec(x)*
du = sec(x)tan(x) dx
dx = du/sec(x)tan(x)

Putting this in the question, *but only substituting for one sec(x)*, I get:
int(2u⋅sec(x)tan(x) du/sec(x)tan(x)) = int(2u du) = u^2 (as done above) = sec^2(x) + C

Here, I get *sec^2(x) + C
!!

How's that possible??