# Basic U-Substitutions

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Lesson Priority: VIP Knowledge

Objectives
• Learn the integration method of U Substitutions, and when it does and does not work
• Specific to U Substitutions, learn the Mr. Math method to write scratch work to avoid calculation errors
• Practice integrating indefinitely using the U Substitution method
• Learn the minute yet important differences between indefinite and definite integrals when it comes to the U Substitution approach
Lesson Description

The method of U Substitutions is a very common technique in integration. This lesson shows you how to do it, and when the method is and is not applicable. We will learn the mechanics to apply this method for both indefinite and definite integrals.

Practice Problems

Practice problems and worksheet coming soon!

## A New Trick For U

Integration via $u$ substitution, often called the "change of variable" method, is like a backward Chain Rule ». It is a very common integration method, so make sure you're a certified expert on both how to use it and when to use it.As an overview, we will rename part of the expression we are trying to integrate, and use its derivative to make substitutions that result in a new but equivalent integral, in a simpler form that we already know how to integrate. Don't worry if that sounds strange right now - this is very much a concept that requires demonstration to understand. Focus on the what, not the why, for this particular method, at least to start.To use the method, we will call part of the expression you are trying to integrate by a new name: $u$. It is our job to choose which part of the expression to name $u$, using the idea that when you take your $u=$ expression and find $du/dx$, the result will allow us to rewrite the original integral wholly in terms of $u$. Moreover, the new integral will be feasible to execute using already-familiar integration rules (usually the Power Rule »).
Pro Tip
When first learning this method, it can be a bit frustrating to learn which part of the expression to choose to call $u$. This takes a little practice to master but rest assured that a little practice goes a long way.
The mechanics of executing a $u$ substitution are particular. Let's look at an illustrative example to see how it works.

Example 1$$\int 2x \sqrt{x^2+5} \, dx$$$\blacktriangleright$ First, recognize and understand why we cannot use the power rule for this integral. That only works for monomial terms.To implement the $u$ sub method, let's call the expression under the root $u$, and proceed with finding its derivative, $du/dx$.$$u=x^2+5$$$$\frac{du}{dx} = 2x$$Now comes the peculiar mechanics that $u$ substitution requires. We are going to treat the derivative notation as a true fraction, and multiply both sides by $dx$.$$\longrightarrow du = 2x \, dx$$Turn your attention back to the original integral - we are going to replace $x^2 + 5$ with $u$, and replace $2x \, dx$ with $du$:$$\int \left(\sqrt{x^2 +5} \right) 2x \, dx$$$$\int \sqrt{u} \, du$$This is the payoff - the remaining integral is a single term that can be done with the power rule.$$\int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{2u^{3/2}}{3} + C$$Finally, we will replace $u$ with its $x$ expression equivalent.$$\frac{2u^{3/2}}{3} + C \longrightarrow \frac{2\sqrt{\left(x^2 +5 \right)^3}}{3} + C$$

It's a little awkward but the steps are very repeatable, and as noted earlier, the process is very commonly required (and I mean very).
Pro Tip
While choosing the right $u$ sometimes seems like an art form, use the following advice as a guideline: when choosing between two expressions to call $u$, if one of the expressions is the derivative of the other expression, pick the other expression.In general, $u$ subs work because the derivative of the $u$ expression appears elsewhere in the problem, so that we can perform the replacement of the $dx$ term.

## Mismatched Coefficients

You may have noticed how convenient it was to have a $2$ coefficient in Example 1. As we're about to see, the $u$ sub method would have worked whether or not that coefficient was there, but the substitution step is much easier to digest when it is. More often than not, the coefficient will not match exactly to our $u$ derivative. In these cases we should multiply by a form of $1$ to make it easier for ourselves.

Example 2$$\int \frac{x^2}{\left( x^3 - 8 \right)^6} \,\, dx$$$\blacktriangleright$ Let's call $u$ the expression trapped in the denominator parenthesis:$$u=x^3 -8$$$$\frac{du}{dx} = 3x^2$$$$\Rightarrow du = 3x^2 dx$$When we go back to our problem to perform substitution, we see that we do not have a $3$ coefficient in front of the $x^2$ term. That's ok - the best way to adjust for this problem properly is to multiply the original problem by $3/3$.$$\int \frac{x^2}{\left( x^3 - 8 \right)^6} \,\, dx$$$$\longrightarrow \int \frac{3}{3} \cdot \frac{x^2}{\left( x^3 - 8 \right)^6} \,\, dx$$$$\longrightarrow \frac{1}{3} \, \int \frac{3x^2}{\left( x^3 - 8 \right)^6} \,\, dx$$Now we have a $3x^2 \, dx$ term that we can replace with $du$:$$\frac{1}{3} \, \int \frac{1}{u^6} \; du$$Integrate this problem using the Power Rule.$$\frac{1}{3} \, \int u^{6} \, dx$$$$= \frac{1}{3} \, \Bigg[ \frac{u^{7}}{7} \Bigg] + C$$$$= \frac{u^{7}}{21} + C$$Finally, as with the first example, we must replace the $u$ variable with the $x$ variable expression that it represents.$$\longrightarrow = \frac{\left( x^3 -8 \right)^7}{21} + C$$
Remember!
When calculating an indefinite integral (one with no integration limits), you must always remember to substitute back the $x$ expression into your final answer. We started the problem looking for the integral of an $x$ expression, so we seek an answer that is also in terms of $x$.

## Definite Integral u Subs

Definite integrals that require $u$ substitutions are mechanically very similar. There is one extra step in the beginning but one fewer step at the end.The extra step comes from the requirement that our limits match our integration variable. For example, when you look at the integral$$\int_{2}^{5} 4x \, dx$$we know the $2$ and $5$ limits tell us to integrate from $x=2$ to $x=5$. In other words, it is implied that the integration limits are defined using whatever variable you are integrating with, which is usually obvious but technically indicated by the $dx$ at the end.In $u$ sub problems that have limits of integration, we will need to change the limits to be defined in terms of $u$, once we translate the original $x$ problem into a $u$ problem. Fortunately this extra step is super quick - we just have to remember to do it.Let's illustrate how this works with an example, and toward the end of the problem, we'll see which step we won't have to worry about for definite $u$ subs relative to indefinite ones.

Example 3$$\int_{2}^{5} \frac{3x^2}{x^3-4} \; dx$$$\blacktriangleright$ We'll start off the same by recognizing the need for a $u$ substitution and selecting the correct choice of $u$. Here we can see that the derivative of the denominator is present in the numerator, so we should select $u$ to be $x^3 -4$.$$u = x^3 - 4$$$$\frac{du}{dx} = 3x^2$$$$du = 3x^2 \, dx$$Now for the extra step. In this definite integral, we must also translate the limits to be measured in terms of $u$.The original problem integrates from $x=2$ to $x=5$. Using the relationship $u = x^3 -4$, we can find that when $x=2$, $u=4$, and when $x=5$, $u=121$.Now let's plug all that information in.$$\int_{2}^{5} \frac{1}{x^3-4} \cdot 3x^2 \, dx$$becomes$$\int_{4}^{121} \frac{1}{u} \; du$$Now we'll finish the problem completely with $u$ as the variable.
Warning!
The single largest source of lost test points for $u$ sub problems with limits is forgetting to translate the limits. If you keep the integration limits as $2$ and $5$, you'll get the wrong answer! Don't forget this important step!
Integrating this problem using common antiderivatives laws, $$\int_{4}^{121} \frac{1}{u} \, du$$$$=\bigg[ \ln(u) \Bigg\rvert_{4}^{121}$$We don't need to convert this problem back to $x$ like we did for indefinite problems, because we properly translated our limits to the basis of our new variable. We can plug in our limits as we usually do for definite integrals.$$=\bigg[ \ln(u) \Bigg\rvert_{4}^{121}$$$$=\ln(121) - \ln(4)$$$$=\ln \left( \frac{121}{4} \right)$$
You Should Know
It's very rare but I've seen some teachers not change the limits, favoring to write something like$$\int_{x=0}^{x=4} u^2 \, du$$and proceeding to work the problem identically to indefinite $u$ sub problems by "changing back to $x$" at the end. I and many others think this is less appropriate and more confusing, so the only time you should ever do it that way is if your teacher demands it.

## Put It To The Test

As noted, this method is very commonly needed, so make sure you review and practice until there's absolutely no mystery about how this works. For each of the following practice problems, integrate each definite or indefinite expression, making sure your final answer follows the conventions described above.

Example 4$$\int 2\sec^{2}(x) \cdot \tan(x) \, dx$$
Show solution
$\blacktriangleright$ With trig integrals, we can identify whether they are good candidates for the $u$ substitution method by asking the same question we typically ask for $u$ subs - does the derivative of one of the functions appear elsewhere?In this case, the derivative of $\tan(x)$ is $\sec^2(x)$, so we're in business. Let's let $u$ be $\tan(x)$.$$u = \tan(x)$$$$\frac{du}{dx} = \sec^2(x)$$$$du = \sec^2(x) \, dx$$We can now make our substitutions and proceed with integration (which will only require the power rule).$$\int 2 \tan(x) \cdot \sec^{2}(x) \, dx$$becomes$$\int 2 u \, du$$$$= u^2 + C$$Because this is an indefinite integral, we need to replace the $u$ variable in our final answer with the original $x$ expression.$$\longrightarrow \tan^2(x) + C$$
You Should Know
This is a somewhat unusual problem in that it has an alternate solution. You can choose to set $u$ equal to $\sec(x)$, whose derivative is $\sec(x)\tan(x)$ which will lead to the answer to this problem being $\sec^2(x) + C$. The reason that both solutions can be correct the $+C$ term. Recall that $\sec^2(x) = 1 + \tan^2(x)$, so plus or minus a constant, the values of $\tan^2(x)$ and $\sec^2(x)$ can be equal when an arbitrary constant is present in both cases.

Example 5$$\int x^2 \, \left( 3x^3 + 1 \right)^{-4} \, dx$$
Show solution
$\blacktriangleright$ As will often be the case with polynomial terms, we'll want to pick the higher power term to be $u$ so that its derivative will help with the $dx$ substitution.$$u = 3x^3 + 1$$$$\frac{du}{dx} = 9x^2$$$$\longrightarrow du = 9x^2 \, dx$$In the original problem, we have $x^2$, not $9x^2$, so let's multiply our problem by $9/9$.$$\frac{1}{9} \, \int 9x^2 \, \left( 3x^3 + 1 \right)^{-4} \, dx$$becomes$$\frac{1}{9} \, \int u^{-4} \, du$$Integrate with the power rule and polish up the solution.$$=\frac{1}{9} \cdot \frac{u^{-3}}{-3} + C$$$$=-\frac{1}{27u^3} + C$$$$=-\frac{1}{27\left( 3x^3 + 1 \right)^3} + C$$

Example 6$$\int^{3}_{1} \frac{x^2}{5x^3 - 2} \, dx$$
Show solution
$\blacktriangleright$ Similar to the previous exercise, we'll want to pick $u$ to be the larger powered polynomial term. This time, we'll also change the integration limits to be in terms of $u$, since it is a definite integral.$$u = 5x^3 - 2$$$$\frac{du}{dx} = 15x^2$$$$du = 15x^2 \, dx$$Additionally, when $x=0$, $u = 5(1)^3 - 2 = 3$, and when $x=3$, $u = 5(3)^3 - 2 = 133$.We also need to multiply the original problem by $15/15$ to make the $dx$ substitution manageable.$$\frac{1}{15} \, \int^{3}_{1} 15x^2 \frac{1}{5x^3 - 2} \, dx$$becomes$$\frac{1}{15} \, \int^{133}_{3} \frac{1}{u} \, du$$$$=\frac{1}{15} \bigg[ \ln(u) \Bigg\rvert^{133}_{3}$$$$=\frac{1}{15} \left(\ln(133) - \ln(3) \right)$$$$=\frac{1}{15} \cdot \ln\left(\frac{133}{3}\right)$$As with any definite integral, our result is completely numeric (remembering for $u$ subs that we won't need to do any switch-back to $x$ at the end).

Example 7$$\int \sin(4x+3) \, dx$$
Show solution
$\blacktriangleright$ While you may have learned a rule for $\sin(ax+b)$ and/or $\cos(ax+b)$, it's fairly common that we're expected to show how these types of integrals work via a proper $u$ substitution derivation. Let's let $u = 4x+3$.$$u = 4x+3$$$$\frac{du}{dx} = 4$$$$du = 4 \, dx$$Multiply and divide the original problem by $4$ before substituting:$$\frac{1}{4} \int \sin(4x+3) \cdot 4 \, dx$$becomes$$\frac{1}{4} \int u \, du$$$$\longrightarrow \frac{1}{4} \cdot \frac{u^2}{2} + C$$$$\longrightarrow \frac{\sin^2(4x+3)}{8} + C$$

Example 8$$\int^{16}_{1} \frac{1}{\left( 1 + \sqrt{x} \right)^2 \sqrt{x}} \, dx$$
Show solution
$\blacktriangleright$ We often are looking in the numerator for the derivative, but when root or fraction expressions are in play, their derivatives are fraction results. Typically you should trust your intuition - which hopefully tells you to let $u$ equal $1 + \sqrt{x}$.$$u = 1 + \sqrt{x}$$$$\frac{du}{dx} = \frac{1}{2\sqrt{x}} \, dx$$Also, because this is a definite integral, we need to change our integration limits to be in terms of $x$, using the relationship $u = 1 + \sqrt{x}$.$$x = 1 \longrightarrow u = 1 + \sqrt{1} = 2$$$$x = 16 \longrightarrow u = 1 + \sqrt{16} = 5$$Now let's re-write our integral and see what happens.$$\int^{16}_{1} \frac{1}{\left( 1 + \sqrt{x} \right)^2 \sqrt{x}} \, dx$$becomes$$\int^{5}_{2} \frac{1}{u^2} \cdot \frac{1}{\sqrt{x}} \, dx$$To aide in the substitution of the $du$ expression, let's multiply this integral by $2/2$:$$\frac{2}{2} \int^{2}_{5} \frac{1}{u^2} \cdot \frac{1}{\sqrt{x}} \, dx$$$$2 \int^{5}_{2} \frac{1}{u^2} \cdot \frac{1}{2 \sqrt{x}} \, dx$$Now it's cleaner to see the verbatim replacement of $1/2\sqrt{x}$ $dx$ with $du$, and the integration can be completed using the power rule.$$2 \int^{5}_{2} \frac{1}{u^2} \, du$$$$= 2 \bigg[\frac{u^{-1}}{-1} \Bigg\rvert^{5}_{2}$$$$=-2 \bigg[5^{-1} - 2^{-1}\bigg]$$$$= \frac{3}{5}$$As we've seen, we don't have any need to replace $u$ back with $x$ because a definite integral gets converted along the way, and a number result is obtained.As an FYI, the challenge of problems like this with all the action in the denominator comes later once you learn other methods beside $u$ substitution, and at that point need to decide which method to use.

Example 9$$\int 2x \sin(9x^2 - 1) \cos^{4}(9x^2 - 1) \, dx$$
Show solution
$\blacktriangleright$ While many times we can get away with choosing either $\sin$ or $\cos$ for our choice of $u$, hopefully your inclination is to pick $u$ to be $\cos(9x^2 - 1)$. After seeing other $u$ sub problems, the pattern you notice tends to be that we should pick the one that "more stuff is happening to". Here, the cosine function is being raised to the fourth power, while the sine function doesn't have anything being done to it.$$u = \cos(9x^2 - 1)$$$$\frac{du}{dx} = -18x \sin(9x^2 - 1)$$$$\longrightarrow du = -18x \sin(9x^2 - 1) \, dx$$To aide in the substitution step, we should multiply by $-9/-9$.$$-\frac{1}{9} \int -18x \sin(9x^2 - 1) \cos^{4}(9x^2 - 1) \, dx$$becomes$$-\frac{1}{9} \int u^4 \, du$$$$-\frac{1}{9} \cdot \frac{u^5}{5} + C$$$$-\frac{1}{9} \cdot \frac{\cos(9x^2-1)}{5} + C$$$$\frac{-\cos(9x^2-1)}{45} + C$$

Example 10$$\int^{\ln(5)}_{0} \frac{e^{x} - e^{-x}}{e^x + e^{-x}} \, dx$$
Show solution
$\blacktriangleright$ Here, hopefully it isn't too far of a stretch to see that the derivative of the denominator is equal to the numerator, which is a strong indication that we should let $u$ be equal to the denominator.$$u = e^x + e^{-x}$$$$\frac{du}{dx} = e^x - e^{-x}$$$$du = \left[ e^x - e^{-x} \right] \, dx$$Additionally, since this is a definite integral, we'll need to convert our limits to $u$ using the $u$ and $x$ relationship. Fortunately, this isn't a nightmare because of the limits we happen to have been given (a common occurence for these problems):$$x = 0 \longrightarrow u = 1 + 1 = 2$$$$x = \ln(5) \longrightarrow u = 5 + (1/5) = 26/5$$$$\int^{\ln(5)}_{0} \frac{e^{x} - e^{-x}}{e^x + e^{-x}} \, dx$$becomes$$\int^{26/5}_{2} \frac{1}{u} \; du$$$$=\ln(u) \Bigg\rvert^{26/5}_{2}$$$$=\ln\left(\frac{26}{5}\right) - \ln(2)$$$$=\ln\left(\frac{13}{5}\right)$$

Lesson Takeaways
• Learn how the method of $u$ substitution works and why we need it
• Become an expert at choosing which part of a problem to label $u$
• Know the specific mechanics of indefinte integration with $u$ substitution
• Know what's similar and different about evaluating definite integrals with $u$ substitution
• Become so well-practiced at these problems that it's as well-understood as any common antiderivative process you knew up to this point - this is a very, very commonly needed method!
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Remember! - Remember notes need to be in your head at the peril of losing points on tests.

You Should Know - Somewhat elective information that may give you a broader understanding.

Warning! - Something you should be careful about.