Trigonometric Integral U-Substitution Homework

U Substitution

Written by tutor Michael B.


By now, you have seen one or more of the basic rules of integration. These rules are so important and commonly used that many calculus books have these formulas listed on their inside front and/or back covers. Here are a few of them (you may not have learned all of these yet):

These rules are so commonly published, and yet there is a subtle and misleading point about them that very few books seem to discuss. And it is unfortunate because this issue could be so easily clarified if only the books would publish the same formulas in the ever-so-slightly different form shown below.

See the difference? Well, the second set of equations uses the variable 'u' instead of 'x'.

"But," I hear you say, "there isn't any real difference. The choice of a variable in any equation is arbitrary. It shouldn't matter that I write 'u' instead of 'x', right?" And yes, you would be right. However, the fact that these are supposed to be general rules conflicts with the fact that most functions that we write already use the variable 'x'. It's not that the typically published general equations are wrong, it's just that they aren't conducive to passing along one of the most critical rules that you must follow when performing integration problems, and that is the fact that you are ONLY allowed to integrate according to the KNOWN rules of integration.

"But," you further protest, "I know for a fact it's possible to integrate some expressions even though they aren't written as rules." Again, you would be correct. But what most often is happening in those situations is that some steps are not being displayed, because after a certain amount of experience and practice, those steps are considered trivial, and it is often assumed that the reader understands how the rules were applied to perform the integration. Those hidden steps make all the difference when you are learning how to integrate. So let me restate the most important integration rule of all: you are ONLY allowed to integrate expressions that EXACTLY match the KNOWN rules of integration, and no others. In other words, until you gain experience, if what you are trying to integrate doesn't exactly match a known rule of integration, you cannot integrate it.

So, what does all this have to do with "U-Substitution"? Well, U-Substitution is one of the most common methods that can be used to transform a problem that doesn't exactly match a known rule of integration to one that does, without changing the essence of the problem. Then, because there is an exact match, you can perform the integration.

This concept shouldn't be all that strange, actually. U-substitution is only one transformation technique, and in fact, you probably have already used other techniques of transformation to solve integration problems that haven't required the U-Substitution method. Here's an example:

So this doesn't look too hard. But before we integrate it, let's revisit the most important rule. Have you ever seen a published rule that describes how to integrate exactly "(x+1)²"? Probably not! Therefore, according to the most important rule, this expression cannot be integrated. But the most important rule doesn't say that we cannot transform the expression into an equivalent form that can be integrated. We can do this quite easily by FOILing the (x+1)(x+1)dx expression, which results in (x² + 2x + 1)dx. But wait a moment, there's no published rule that specifies how to integrate (x² + 2x + 1)dx either! But now we can utilize the properties of integrals to split this up into three integrals, extract constants, and then each integral does exactly match a rule - the Power Rule of integrals specifically:


The astute reader will point out a difficulty with this approach, however. What if we weren't just squaring (x+1), but raising it to the 215th power? Surely, there must be an easier way to integrate (x+1)215 than expanding the polynomial! And yes, there is -- this is where U-substitution comes in. To put it succinctly, U-Substitution allows you, in some cases, to make the integration problem at hand look like one of the known integration rules. Just as FOILing (x+1)² doesn't change the expression, neither does U-substitution, from a naive standpoint. U-Substitution can be a very powerful method of transformations, and it isn't hard, but it does have some quirks that we must be careful to handle properly.

Performing U-substitution

I started this article by spurning the traditional method presenting integration formulas with x's instead of u's. The reason for this, although it is somewhat contrived, is beause it makes it difficult to understand why we need to learn u-substitution. The ultimate goal of the U-Substitution technique is to write an expression for u as a function of x that simplifies the integral, resulting in an expression that exactly maps to one of the known rules of integration, while still accounting for the details that calculus imposes on such an action. This technique is also often called the "change of variable" technique, because we are essentially replacing expressions of 'x' with expressions of 'u'. But it's not as simple as changing x's into u's - to do so is considered a trivial change, and ultimately doesn't transform the problem at all.

Let's return to the example above, and see how we can perform the same integration using the U-substitution technique. If we look at the Power Rule for integration, by saying "something" instead of "u" or "x", it might be read aloud as "the integral of something raised to a power n is equal to the something raised to the power n+1 all divided by n+1." Well, the (x+1) is a "something" isn't it?

So since we are free to assign anything we want to a variable, we could assign u=x+1. Then, we might naively write:

The reason that the second integral is identified as bad is because you cannot (in the context of this discussion of U-Substitution) integrate a variable 'u' with a 'dx'. An integral with "u's" must be matched up with "du's". So how do we get a "du" instead of a "dx"? This is accomplished by taking the derivative of u with respect to x:

Because du=dx (or equivalently, dx=du), we can now properly write the integral in terms of u and du, complete the integration, and then we re-substitute the expression we chose for u back into the equation to complete the solution:

Note that this answer is exactly the same as the answer we previously obtained by FOILing (x+1)² although it may not immediately appear so. If we expand the (x+1)3 that we obtained with the U-Substitution method, we would obtain x3+3x²+3x+1, which when we divide all of those terms by 3, we obtain (1/3)x3+x²+x+(1/3). It might appear that the answer we obtained by FOILing is missing the constant 1/3, but in fact this is OK because both of the integrals contain the "unknown constant C". The fact that the 1/3 term is missing simply means that it is accounted for in the C from the FOILed version of the integral. In other words, the two C's are different, and the difference exactly accounts for the missing constant.

Note: Quite often, choosing the correct u is more an art than science. Sometimes you might choose a u only to find that you can't make the substitution look like a known integral. Perhaps you need to choose a different u, or perhaps you need to do something a bit different.

A more interesting example #1

The example given above was a bit trivial, because of the fact that we determined that dx=du. Let us now examine a more interesting example in which this is not the case. In this process, we will now see that the choice of 'u' must be carefully chosen so that our du exactly accounts for all of the other variable components of the integral.

As with all U-Substitution problems, our objective is to pick an expression of x that we can assign to u, and more specifically, one for which the derivative of u covers any remaining variable portions of the integral that are not covered by u. When dealing with polynomial expressions, this often means that we choose the higher degree of two polynomials, since the derivative of polynomials always results in a polynomial of lower degree. In this example, therefore, the natural choice would be u = x4+2x2+5, and therefore du = (4x3+8x)dx = 4(x3+2x)dx. This expression of du contains the term (x3+2x)dx, which is the portion of the integral that is not replaced by our choice of 'u'. However, it also contains a constant term 4, but this is ok, because the constant can be moved in front of the integral, and our replacements for u and du now exactly cover the expressions of x and dx. We cannot ignore the constant term, but it does not cause a problem. Here is the final solution:


Occasionally, you might find after choosing a u and solving for du that you have a power of x remaining. For example, consider the following example:

In this case, you can choose u=2x+1, and then du=2dx. But what about the x that remains? We can solve this by solving the u equation for x!

This allows us to distribute the u½ thru the term (u+1), and then we simply continue the integration using the power rule:

Definite Integrals - transforming the limits

In all of our examples above, the integrals have been indefinite integrals - in other words, integrals without limits of integration (the "a" and "b" in the statement "the integral from a to b"). When there are limits, and we need to use U-Substitution, there are a few things we need to keep in mind:

  1. The limits are not the same when you write the integral with u's instead of x's
  2. The limits, therefore, should either be dropped (as if we are performing an indefinite integral), or written in terms of u
  3. If you choose to write the limits in terms of u, then simply apply the equation you assigned to u to determine the new limits
  4. If you choose to write the limits in terms of u, then after performing the integration, you can directly substitute the u limits into the u-version of the integral - there is no need to re-substitute x's back in for the u's! Otherwise, you must evaluate the integral with the original x limits.

That last point can help to save a lot of time when you are performing U-subtitution with definite integrals. The next example demonstrates that both methods provide the same answer:

Notice how, when plugging in the original x limits, the first step is exactly the same operation ("x+1") that you perform when transforming the limits. This is why the two methods are exactly equivalent, but the 'limits transformation' method simply allows you to avoid having to resubstitute the original x expression back into the integral, which helps to eliminate mistakes. Also notice that when writing the integral in Method 2, I did not write the limits of integration even for the x integral, because it wouldn't be proper to say that a definite integral of x is equal to an indefinite integral of u.

Pattern-matching your du

When choosing your u, remember that your du must also be in the equation, and must exactly account for the remaining variable terms of x. You cannot add new variables, nor can you (in most cases) add additive constants to the integral to make it match. (You can more easily add multiplicative constants as needed by simultaneously applying the reciprocal so that the resulting equation is the same). That does not necessarily mean that the intregral can't be done, but the U-substitution technique is strict in that you must exactly match up the u and du to the x and dx terms. Here's an example of a problem that can't be done directly with U-Substitution, but can be performed with a little bit of work ahead of time.

Your initial instinct for the first step to solve this problem may be to set u=x²+1. However, if you do that, then du=2x·dx, and therefore the resulting substitution would look like this:

Returning to the idea of the 'most important rule', clearly, this integral does not match any of the known integration rules because of that little '+1' that is stuck in the numerator. You might argue that we could split this integral into two integrals, one for du/u, and one for 1/u, but then the second integral doesn't even have a du with which to integrate, so that doesn't work.

But as I mentioned initially, the fact that U-Substitution doesn't work directly doesn't mean that the integration can't be performed. It just takes a bit of algebraic manipulation first:

After performing this algebraic step, the first integral can now be handled easily by U-substitution (the detailed steps are left to the reader as an exercise, hint: u=x²+1), and the second integral is a known integration rule, so no U-Substitution is necessary:


Use U-substitution to evaluate each of the following integrals and confirm that the equation is true. You may need to use additional techniques discussed above or other math identities to solve some of these.

Calculus > U Substitution


  1. Overview and Basic Example
  2. U Substitution for Trigonometric Functions
  3. U Substitution for Definite Integrals
  4. U Substitution for Exponential Functions

1. Overview and Basic Example

Integration by substitution is one of the first techniques you use in integral calculus. All the technique is doing is taking a rather complicated integral and turning it — using algebra — into integrals you can recognize and easily integrate. U substitution requires strong algebra skills and knowledge of rules of differentiation. Why? Because you’ll need to be able to look at the integral and see where a little algebra might get the form into one you can easily integrate — and as integration is really reverse-differentiation, knowing your rules of differentiation will make the task much easier. For example, the following sample problem uses the integral 2x(x2+3)70. Recognizing that if you differentiate x2 +3, you get 2x, is the key to successful u substitution.

Sample problem: Integrate 2x(x2+3)70 using integration by substitution.

Step 1: Choose a term to substitute for u. Pick a term that when you substitute u in, it makes it easily to integrate. In this example, replacing (x2) with u makes the function look more familiar for integrating:

Step 2: Take the derivative of the u function. This particular function uses the power rule, so:
du = 2x dx

Step 3: Rewrite the problem using the “u” and “du” you derived in Steps 1 and 2:
= ∫2x(u)70dx
which can be rewritten as:
∫(u)702x dx
substituting du from Step 2:

Step 4: Integrate the function, using the power function rule for integration.
∫(u)70du=u7171 + C

Step 5: Resubstitute your original term (from Step 1) in place of u:
u7171 + C = (X2 + 3)7171 + C

Tip: When using integration by substitution, always look for terms that are derivatives of each other. In the above example, the derivative of x2 is 2x.

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2. U Substitution for Trigonometric Functions

U substitution is one way you can find integrals for trigonometric functions. With u substitution, you substitute “u” to simplify the process of integration, re-substituting the original term at the end of the process. The “trick” to successfully using u substitution is to be very familiar with the rules of integration, because you are going to be picking terms to substitute that leave you with something that can easily be integrated using the general rules.

U Substitution Trigonometric Functions: Example

Sample problem #1: Integrate ∫sin 3x dx.

Step 1: Select a term for “u.” Look for substitution that will result in a more familiar equation to integrate. Substituting u for 3x will leave an easier term to integrate (sin u), so:
u = 3x

Step 2: Differentiate u:
du = 3 dx
Or (rewriting using algebra — necessary because you need to replace “dx”, not 3 dx:
⅓ du = dx

Step 3: Replace all forms of x in the original equation:
Substituting for u: ∫ sin 3x dx = ∫ sin u dx

Substituting for dx: ∫ sin u dx = ∫ sin u ⅓ du

Step 4: Rewrite, bringing the constant in front of the integral symbol (so that you can easily integrate):
∫ sin u ⅓ du = ⅓ ∫ sin u du

Step 5: Integrate using the usual rules of integration:
⅓ ∫ sin u du = ⅓ (-cos u) + C = -⅓ cos u + C

Step 6: Re-substitute for u:
“u” is left in the equation, so: ⅓ cos u + C = ⅓ cos 3x + C

That’s it!

Sample problem #2: Integrate ∫ 5 sec 4x dx

Step 1: Pick a term to substitute for u:
u = 4x

Step 2: Differentiate, using the usual rules of differentiation.
du = 4 dx
¼ du = dx (using algebra to rewrite, as you need to substitute dx on its own, not 4x)

Step 3: Substitute u and du into the equation:
∫ 5 sec 4x tan 4x dx = 5 ∫ sec u tan u ¼ du = 54 ∫ sec u tan u du

Step 4: Integrate, using the usual rules of integration. For this problem, integrate using the rule D(sec x) = sec x tan x:
54∫ sec u tan u du = 54 sec u + C

Step 5: Resubstitute for u:
54 sec u + C = 54 sec 4x + C

Tip: If you don’t know the rules by heart, compare your function to the general rules of integration and look for familiar looking integrands before you attempt to substitute anything for u.

That’s all there is to U Substitution for Trigonometric Functions!

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U Substitution for Definite Integrals

In general, a definite integral is a good candidate for u substitution if the equation contains both a function and that function’s derivative. When evaluating definite integrals, figure out the indefinite integral first and then evaluate for the given limits of integration.

Sample problem: Evaluate:

Step 1: Pick a term for u. Choose sin x for this sample problem, because the derivative is cos x.
u = sin x.

Step 2: Find the derivative of u:
du = cos x dx

Step 3: Substitute u and du into the function:

Step 4: Integrate the function from Step 3:

Step 5: Evaluate at the given limits:

That’s it!

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4. U Substitution for Exponential Functions

With u substitution, you algebraically simplify a function so that its antiderivative can be easily recognized. U substitution is just like it sounds — you substitute in the variable u to perform the integration, which simplifies the process. At the end of your calculations, you re-substitute in your original terms for the u.

Sample question: Find the integral for the exponential function ex+1ex using u substitution.

Step 1: Rewrite your function using algebra to get it in a form where you can easily find an integral:
ex+1ex = ∫(exex+1ex) = ∫(1+e-x)dx

Step 2: Split up the function into separate parts:
∫(1+e-x)dx = ∫1dx + ∫e-xdx

Step 3: Pick u and find the derivative of u. For this example, pick “-x” in e-x:
u = -x
du =-1*dx

Step 4: Find a way to get rid of the symbol dx using your second substitution in Step 3. Using algebra:
du =-1*dx

Step 5: Substitute the “u”, and “du” from Steps 3 and 4 into the equation.
∫1dx + ∫eu(-1)du

Step 6: Solve the integrals:
∫1dx + ∫eu(-1)du = x – eu + C

Step 7: Resubstitute your terms back into the function. u=-x, so:
x – eu + C = x – e-x + C

That’s it!


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