Substitution: Videos & Practice Problems

To evaluate integrals involving composite functions, the substitution method is essential. By choosing an appropriate substitution, such as $u=x+3$, and finding $\mathrm{du}=\mathrm{dx}$, we can simplify the integral. For definite integrals, bounds can be transformed using the substitution, allowing for straightforward evaluation. This method enhances understanding of integrals and their applications in calculus, reinforcing concepts like the fundamental theorem of calculus and the power rule.

Indefinite Integrals

Since we've been working through evaluating various integrals, let's take a look at another integral here. The integral of x squared plus one cubed times two x d x. Now looking at this integral, we can see that we have this function inside of another function then being multiplied by yet another function. So we don't currently have the tools that we need in order to evaluate this integral as is. Now we could go ahead and multiply this x squared plus one out three times and distribute that two x throughout, but that would be a rather tedious process.

And luckily, there's a much easier way to evaluate this integral. Now earlier on in calculus, we were able to evaluate the derivatives of composite functions, that's one function inside of another function, using the chain rule. And now that we're faced with evaluating integrals that also include composite functions, we're going to use a process called substitution, which you may also hear referred to as u substitution or change of variables. Now, integrals are the reverse process of finding the derivative. And just like that, substitution is basically the reverse process of using the chain rule.

I'm going to walk you through this substitution process here step by step. So let's go ahead and jump back to our example here. Our integral of x squared plus one cubed times two x d x. Now we already identified here that we have this function x squared plus one contained inside of another function because it's being cubed. And looking at this integral as a whole, we can see that it takes a very particular form where we have this composite function, f(g(x)), then being multiplied by g'(x).

That's the derivative of that inside function. Now whenever we have an integral of this form, we can use substitution to make it much easier to evaluate, allowing us to rewrite it as just f(u) du, where u is just a placeholder variable that we're going to use to make our substitution. Now we can see that this clearly looks much easier to evaluate than this mess on the left over here, but the question then becomes, how do we get there? We clearly need to know what is u and what is du. Now when making a substitution, we get to choose what we want u to be.

But we have to be careful to make the right choice. U will most often be our inside function, either inside a parenthesis, under a radical, or in an exponent. So let's come back down here to our example and choose u. Now looking at this, I can see that this x squared plus one is contained in parentheses and is then being cubed. Because it's contained in those parentheses, that's a good clue that this is what we should choose to be u.

So if u is this x squared plus one, what then is du? Well, if I go ahead and take the derivative here with respect to x, so du dx, that will give me here two x. Then if I go ahead and move this dx to the other side, that means that du is then equal to two x dx, which is exactly what I have in my integral here. So I have this u being cubed, and then it's being multiplied by du. So making this substitution here, subbing that u in for that x squared plus one, allows me to rewrite this integral as simply u cubed du.

I = 1 4 u 4 + C

Now this is much easier to evaluate here because we can simply use the power rule. I can go ahead and just add one to that exponent and then divide by my new exponent here, leaving me with one fourth u to the power of four and then plus my constant of integration c here because this is an indefinite integral. So here, we've successfully performed our substitution and gotten our integral here. But because our original function was in terms of x, we want to put this back in terms of x, which we can easily do because we know that u is simply x squared plus one. So I can rewrite this as a one fourth times x squared plus one to the power of four plus that constant of integration c, and this is my final answer here, having evaluated my original integral using the process of substitution.

Now in this particular example, we saw that when we chose u as x squared plus one, du just nicely appeared in our integral. But this will not always be the case. So let's actually work through another example here, breaking this down step by step so that you are well equipped to handle any integral that gets thrown your way. We actually already use these same exact steps when solving our example up above, and here we're just putting them into words. Now the integral that we're tasked with evaluating here is the integral of four of the square root of four x minus one d x.

So let's go ahead and get started with our first step here the same way that we did up above by choosing u and finding du. Now remember, u is going to be that inside function, either contained in parentheses under a radical or in an exponent. Now in this case, since we have this four x minus one underneath this radical, that's a good indicator that this is what we should choose to be u. So if u is equal to that four x minus one, what then is du? Or remember that du is going to be equal to the derivative of that function that we just chose as u then multiplied by dx.

So here, the derivative of four x minus one is four. Then we multiply that by dx to get our full du as being four times dx. Now step one done, we can go ahead and move on to step two where we're going to rewrite our integral only in terms of that u and du that we just found in step one. Now looking at my integral here, I know that I have this u contained underneath this radical, but then I just have dx by itself. And I know that du is not just dx, it's four dx.

So we're off by a constant here, and we need to find a way to fix this. Now whenever our du is off by a constant, we can go ahead and just multiply by that constant in order to make our substitution work. So if I go ahead and multiply this by four, I now have that four dx, which is the du that I'm looking for. But because I don't actually want to change the value of my integral, whenever I multiply by a constant, I also need to go ahead and multiply by that constant's reciprocal to effectively cancel it out. So since here I'm multiplying by four, I also need to flip that to the bottom of a fraction and multiply by one fourth.

Now one fourth times four is just one. So I'm not actually changing the value of my integral. I'm just rewriting it in order to get my substitution to work. So now that I have this four x minus one as u and I have this four times dx as du, I can rewrite this integral fully in terms of u and du. Now I can go ahead and pull this one fourth out front since it's just a constant, rewriting my integral as one fourth times the integral of the square root of u du.

So now step two is all done and I can move on to step three where we're just going to integrate here with respect to u. Now looking at this, the square root of u, remember that if I'm taking the square root that's the same thing as raising to the power of one half, so we can just apply the power rule here when finding our antiderivative, adding one to that exponent and then dividing by that new exponent, that one half plus one. Now keeping that constant out front, I have that one fourth. And then having applied the power rule, I'm dividing here by one half plus one, which is three halves. Dividing by three halves is the same thing as multiplying by two thirds.

So I have two thirds u to the power of three halves and then, of course, plus that constant of integration c since this is an indefinite integral. Now we can go ahead and c

Indefinite Integrals Example 1

In this problem, we are asked to use substitution to evaluate the given integral. Then, after we do that, we want to then check our answer by differentiating. So, let's go ahead and get started here by using substitution to evaluate this integral, the indefinite integral of 2xx3x7 5 6xx21dx. Now using substitution here, we're just going to jump right into our steps. Our very first step is going to be to choose u and find du.

Now, remember, u is going to be an inside function, typically in parentheses, under a radical, in an exponent, or something like that. In this case, we can see that there are two functions that are enclosed within parentheses, so you may be confused as to which one should be u. We can see that one of these functions is raised to a power whereas the other one is not. So that's usually a good clue that this is the inside function that we should choose as u. So, let's go ahead and see how that works out.

Choosing u to be that 2xx3x7. Then, once we have u, we want to find du, which is going to be the derivative of u times dx. So the derivative of this 2xx3x7 using the power rule is going to be 6xx21, and then we multiply that by dx. So we can already see based on our integral that this is going to work out rather nicely. So with step one done, we can go ahead and rewrite our integral only in terms of this u and du.

So, I can see that if I chose this as a u, that then made this du, that 6xx21 dx. So because of that, we can rewrite this integral as a u5 du. Now with step two done, we can move on to step three where we just want to go ahead and integrate with respect to u. Here we can do that rather easily using the power rule, adding one to that exponent and then dividing by that new exponent. That gives me a 16u6 and then, of course, my constant of integration c because this is an indefinite integral.

So with that step three done, we move on to our final step in finding this integral, which is replacing u with that original function of x. Now remember, we chose u to be that 2xx3x7, so that's what we want to replace u as here to get our final answer. This is going to give us a 16(2xx3x7)6 plus my constant of integration c. So this is my final answer here, having evaluated that integral using a substitution. Now remember, in this specific example, we were told that we want to go ahead and check our answer by differentiating our result here.

If we did this correctly, when we take the derivative of this function here, we should get the original function that was in our integrand. So let's go ahead and find this derivative,

Indefinite Integrals Example 2

In this problem, we're asked to evaluate the indefinite integral of the cosine of t times the sine raised to the ninety-ninth power of t dt. Now, it may be a bit unclear at first, just looking at this, how exactly we can go about evaluating this integral, but we can actually use substitution to evaluate this integral rather easily. So let's see how exactly that works. We're going to start with our first step for substitution here, which is choosing u and finding du. Now we know that u should be an inside function, but when working with trig functions, it can be a bit unclear what exactly constitutes an inside function.

So we have to think about what's going on in our problem. Now, we have this cosine of t, and then we have this sine that's raised to the ninety-ninth power. So I know that the sine of t is being raised to a power, which makes it a good candidate for u. So let's see what happens if we choose u to be that sine of t. If u is equal to the sine of t, that then makes du equal to the derivative of this times dt.

Now, in this case, we are working with t's as opposed to x's, so the function of u is g of t and here we have the derivative of that with respect to t multiplied by t. It's just whatever particular variable you're working with in your integral, in this case, t. So back over here, what is du? Well, the derivative of the sine of t is the cosine of t, and then multiply here by dt to get that full du. With that step one done, moving into step two, rewriting our integral only in terms of u and du, taking a look at my integral here, I'm going to go ahead and rearrange what order this is written in.

If I instead write this as a sine to the power of 99 t times the cosine of t dt, I can see that I have my u right here, the sine of t, and then I have du, cosine of t dt, right here for that du. So that means if I want to rewrite this in terms of u and du, I can simply rewrite it as u to the power of 99 times du. Now this just becomes a straightforward power rule when evaluating this integral. So moving into step three here, integrating with respect to u, just using the power rule, if I add one to that exponent and then divide by that new exponent, that gives me one over one hundred times u to the power of 100 plus my constant of integration c because this is an indefinite integral. So with that evaluated, we can move on to our final step here, which is replacing u with our original function, in this case, t.

So we know that we chose u to be that sine of t, so that's what we want to replace u with here. So this is one over 100 times the sine of t to the power of 100 plus c. Now the way that we currently have this written might seem a little bit weird and that is because we typically put the power of trig functions in between the trig function and what we're taking the trig function of. So we can rewrite this in a bit more familiar form as one over 100 times the sine⁻ᴼᴼʰ of t plus that constant of integration c, and this is my final answer here. Feel free to let us know if you have any questions, and I'll see you in the next one.

Evaluate the indefinite integral.
${\int }_{}^{}3t\sqrt{t^2+7}dt$

• A. $t^{\frac{3}{2}}+C$
• B. ${(t^2+7)}^{\frac{2}{3}}+C$
• C. ${(t^2+7)}^{\frac{3}{2}}+C$
• D. $t^{\frac{2}{3}}+C$

Answer: (C)

Evaluate the indefinite integral.
${\int }_{}^{}\frac{1}{{(3x+2)}^5}dx$

• A. $-\frac{1}{12{(3x+2)}^4}+C$
• B. $-\frac{1}{5{(3x+2)}^4}+C$
• C. $\frac{1}{12{(3x+2)}^4}+C$
• D. $\frac{1}{5{(3x+2)}^4}+C$

Answer: (A)

Evaluate the indefinite integral.
${\int }_{}^{}\theta \bullet se{c}^2(5{\theta }^2+1)d\theta$

• A. $10\tan (5{\theta }^2+1)+C$
• B. $\frac{1}{10}\tan (5{\theta }^2+1)+C$
• C.
  $10se{c}^2(5\theta +1)+C$
• D. $\frac{1}{20}{\theta }^2\tan (5{\theta }^2+1)+C$

Answer: (B)

Substitution With an Extra Variable

Earlier, we learned how to evaluate an integral using the process of substitution. And when making a substitution, we saw that du didn't always just appear nicely in our integral, and it could be off by some factor of a constant. And as it turns out, our integral may also be off by some extra variable, like some extra factor of x or x squared or x plus or minus a constant, and we need to account for that when making our substitution. And that's exactly what I'm going to show you how to do here by just adding one extra step within those steps that we've already learned. So let's go ahead and get started here by jumping right into our example where we're asked to evaluate the integral by making a substitution.

And here we have the integral of x times the square root of x plus three dx. So getting started with our steps the same way we have before, we want to go ahead and choose u and find du. Now looking at my integral here, since I have this x plus three underneath a radical, that is a good indicator that that's what I should choose to be u. And if u is equal to x plus three, that then means that du is just equal to dx because the derivative of x plus three is just one. So with that U and DU that I found in my integral here, we can go ahead and move on to step two, which is rewriting our integral only in terms of that U and DU.

Now based on u and du here, if I wanted to go ahead and rewrite my integral, I would have the integral of x times the square root of u du. Now looking at this, this is clearly not only in terms of u and du because we have this extra factor of x here. So how can we get rid of this x and put this whole integral just in terms of u and du? Well, since I know what u is as a function of x, it's just equal to x plus three, I can go ahead and just rearrange this equation. So instead of having u in terms of x, I can get x in terms of u.

So if I go ahead and just subtract three from both sides here, I am then left with x being equal to u minus three. And if x is equal to u minus three, I can go ahead and just replace that in my integral. So rewriting this integral, this is then the integral of (u minus three) times the square root of u du. And now this is only in terms of u and du, just like we wanted. So we can now proceed on to step three, solving this as usual.

Now remember that step three is to integrate here with respect to u, and I'm going to go ahead and clean up this integral a little bit to make it easier to apply our power rule. Remember that the square root of u is the same thing as u to the power of one half. I'm going to go ahead and distribute that into my parentheses here. So this is going to be the integral of u11/2 minus three times u12 integrated of course with respect to u. Now applying the power rule here to get my antiderivative u32, I'm going to add one to that exponent, which gives me 52.

If I divide by 52, that's the same thing as multiplying by two fifths. So this is two fifths u52 minus three, keeping that constant on the outside there, just paying attention to this u to the power of 12. Again, using the power rule, adding one to 12 gives me 32. If I divide by 32, that's the same thing as multiplying by two thirds. So that's two thirds u32 And then of course, plus my constant of integration c because this is an indefinite integral.

Now looking at this, I can see that those threes are going to cancel and I have fully integrated this with respect to u. So we can move on to that final step here, replacing u with g of x. Remember that u is equal to x plus three. So I'm going to plug that back in for you here to get my final answer back in terms of x. So replacing that here, this is going to be two fifths times (x plus three) to the power of five halves minus two thirds times (x plus three) to the power of three halves plus that constant of integration c.

And this is my final answer here having completed that final step. Now we're going to continue getting practice with substitution coming up next because now that we have this extra step, we can solve a ton of different integrals. I'll see you in the next video.

Evaluate the indefinite integral.
${\int }_{}^{}x{(5+x)}^{79}dx$

• A. $\frac{1}{81}{(5+x)}^{81}-\frac{1}{16}{(5+x)}^{80}+C$
• B. $\frac{1}{81}{x}_{}^{81}-\frac{1}{16}{x}^{80}+C$
• C. $\frac{1}{81}{(5+x)}^{81}-\frac{1}{80}{(5+x)}^{80}+C$
• D. $\frac{1}{81}{x}^{81}-\frac{1}{80}{x}^{80}+C$

Answer: (A)

Evaluate the indefinite integral.
${\int }_{}^{}\frac{t}{\sqrt{t-2}}dt$

• A. $\frac{2}{3}{t}^{\frac{3}{2}}+4t^{\frac{1}{2}}+C$
• B. $\frac{2}{3}{(t-2)}^{\frac{2}{3}}+2+C$
• C. $\frac{2}{3}{(t+2)}^{\frac{3}{2}}+4{(t+2)}^{\frac{1}{2}}+C$
• D. $\frac{2}{3}{(t-2)}^{\frac{3}{2}}+4{(t-2)}^{\frac{1}{2}}+C$

Answer: (D)

Evaluate the indefinite integral.
${\int }_{}^{}\frac{x}{{(x-6)}^5}dx$

• A. $-\frac{x}{6{(x-6)}^6}-\frac{3x}{6{(x-6)}^6}+C$
• B. $-\frac{1}{3{(x-6)}^6}-\frac{3}{2{(x-6)}^6}+C$
• C. $-\frac{1}{3{(x-6)}^3}-\frac{3}{2{(x-6)}^4}+C$
• D. $\frac{1}{3{(x-6)}^3}+\frac{3}{2{(x-6)}^4}+C$

Answer: (C)

Evaluate the indefinite integral.
${\int }^{}sin\theta {(1+\sin \theta )}^{84}\cos \theta d\theta$

• A. $-\frac{1}{85}{(\sin \theta )}^{85}-\frac{1}{86}{(\sin \theta )}^{86}$
• B. $\frac{1}{86}{(\cos \theta )}^{86}-\frac{1}{85}{(\cos \theta )}^{85}+C$
• C. $\frac{1}{86}{(1+\sin \theta )}^{86}-\frac{1}{85}{(1+\sin \theta )}^{85}+C$
• D. $\frac{1}{86}{(1+\sin \theta )}^{86}-\frac{1}{86}{(1+\sin \theta )}^{86}+C$

Answer: (C)

Definite Integrals

Earlier in the course, we learned how to evaluate indefinite integrals using the process of substitution. And now, we'll need to use substitution to evaluate definite integrals as well. Luckily, we'll be able to use a ton of the same steps that we know from working with our indefinite integrals. The only difference is our integral now has bounds. It turns out that there are two different ways that we could choose to address the bounds of our definite integral.

And I'm going to walk you through both of those methods here. So let's go ahead and get started by jumping right into our example. We're asked to evaluate the definite integral using substitution. We're given the integral from zero to two of xx2+1 cubed times two x dx. We've worked with this exact integral before.

The only difference is it's now bounded from zero to two. So, how exactly do we address those bounds, and where does that fit into our steps? Well, for this first method, we're actually going to start by completely ignoring our bounds and treating this as though it's just an indefinite integral. Getting right into our first couple of steps here, we know that we need to choose u and find du so that we can rewrite our integral just in terms of them.

Having worked with this exact integral before, we know that if we choose u to be x2+1, that then makes du 2x dx, allowing us to rewrite our integral as the integral of u3 du. Having completed those first two steps, we can move on to step number three where we want to go ahead and integrate here with respect to u. This is going to give me 14u4 using the power rule. Then I need to add that constant of integration C because this is currently an indefinite integral. With that third step done, we move on to step four, which is replacing u with that original function of x.

In this case, x2+1. Plugging that x2+1 in for u here, we get 14(x2+1)4 and then plus that constant of integration C. If this were actually an indefinite integral, we would be completely done at this point. But remember, this is not an indefinite integral and it is in fact bounded from zero to two. So here's where our bounds come in.

I'm going to go ahead and get out of the way as we add an additional step to our steps box here in order to address those bounds. At this point, we want to go ahead and evaluate the antiderivative that we found at those original bounds. In this case, zero and two. So looking at our antiderivative here, that 14x2+1 to the power of four, I don't need to worry about that constant of integration C here because this is a definite integral, and I want to go ahead and bound it from zero to two. We can use the fundamental theorem of calculus here to plug in those bounds and get our final answer.

This is going to be 14((2)2+1)4-(0)2+1)4, plugging in my upper and lower bounds. When I work this out algebra I will end up getting a final answer here of 156 having evaluated this definite integral using a substitution. For this first method, we wanted to treat this as an indefinite integral, doing our substitution the way that we know how, putting our function back in terms of x, and then applying those original bounds. But remember that there is another way we could choose to address our bounds. So let's take a look at our second method here.

Now we're faced with the same exact definite integral here from zero to two of xx2+1 cubed times two x dx. Having worked with this integral so many times before, we already know that if we choose u as x2+1 that makes du two x dx, allowing us to rewrite our integral here as the integral of u3 du. At this point in our previous method, we were completely ignoring the bounds of our integral, but that's not what we're going to do here. Instead, we're going to transform our bounds along with the rest of our integral. So having completed step one already here, as we are rewriting our integral only in terms of u and du, we're going to add one additional part to this step here where we're going to address our bounds.

We want to go ahead and transform our bounds using the substitution that we made, plugging our original bounds into that function of x. So whereas our original bounds were from a to b, as we rewrite our new integral in terms of u and we have that f of u du, this will now be bounded from g of a to g of b, having plugged that a and b into that function of x. In this case, since we chose u as x2+1, I want to plug zero and two into that. So g of zero here, plugging that in, is going to be 02+1, which is just one. And then g of two, plugging that two in, that's 22+1, which gives me five.

So this one and five represent my brand-new transformed bounds. So this integral of u3 du is bounded from one to five. Now we have fully rewritten our entire integral in terms of u and du, including those bounds. And we can go ahead and now move on to step three. We want to integrate here with respect to u as we've done before.

Having worked with this so many times, we already know that this will come out to 14u4, and we don't have to worry about our constant of integration because we have bounds. This is bounded from one to five. And since we have those brand-new bounds, moving on to that fourth step here, we no longer need to put this back in terms of x. So we're going to completely get rid of that step here and instead add one final step to finish evaluating this integral by going ahead and evaluating this antiderivative at these brand-new bounds that we found. So going ahead and plugging these in here using the fundamental theorem of calculus, this is going to be 14(54-14).

Now, this will work out algebraically to give us an answer of 156 for this definite integral, which is the same answer that we got above in method one. Now between these two methods, there is not one best method, and they'll both always work. But no matter what method you choose to use, make sure you know exactly how to apply your bounds. In that first method, we treat it as an indefinite integral, doing our substitution, putting it back in terms of x, and applying those original bounds. In our second method here, we put everything in terms of u when doing our substitution so that we could keep it in terms of u and evaluate our antiderivative at those brand-new transformed bounds.

Now we're going to continue getting practice with substitution coming up next. I'll see you there.

Evaluate the definite integral.
${\int }_0^1\frac{t}{\sqrt{t^2+1}}dt$

• A. 1.414
• B. 0.414
• C. 1
• D. 0.707

Answer: (B)

Evaluate the definite integral.
${\int }_1^2(x-3){(x^2-6x)}^{2}dx$

• A. -64.5
• B. 533
• C. -2.33
• D. -1.17

Answer: (A)

Evaluate the definite integral.
${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\tan y$ $se{c}^{2}ydy$

• A. 0.308
• B. $\pi$
• C. 0
• D. $\frac{\pi }{4}$

Answer: (C)

Here’s what students ask on this topic:

What is the substitution method in calculus, and how is it used to evaluate integrals?

The substitution method in calculus is a technique used to simplify the evaluation of integrals, especially those involving composite functions. It involves choosing a substitution, typically denoted as $u$, to replace a part of the integrand, and then finding $du$ to replace $dx$. This transforms the integral into a simpler form, often allowing the use of basic integration rules like the power rule. For definite integrals, the bounds are also transformed using the substitution, facilitating straightforward evaluation. This method is crucial for understanding integrals and their applications, reinforcing concepts such as the fundamental theorem of calculus.

How do you choose the correct substitution for an integral?

Choosing the correct substitution for an integral involves identifying the 'inside function' within the integrand. This function is often found inside parentheses, under a radical, or in an exponent. Once identified, it is set as $u$. The derivative of this function, multiplied by $dx$, gives $du$. The goal is to rewrite the integral entirely in terms of $u$ and $du$. If the substitution doesn't perfectly match the integrand, adjustments like multiplying by constants or rearranging equations may be necessary. This process simplifies the integral, making it easier to evaluate.

What are the steps to evaluate a definite integral using substitution?

To evaluate a definite integral using substitution, follow these steps: First, choose a substitution $u$ and find $du$. Next, rewrite the integral in terms of $u$ and $du$, transforming the bounds by substituting the original limits into the function of $u$. Then, integrate with respect to $u$. Finally, evaluate the antiderivative at the transformed bounds using the fundamental theorem of calculus. This method allows for a straightforward evaluation of definite integrals, ensuring accuracy and efficiency.

How does the substitution method relate to the chain rule in calculus?

The substitution method is essentially the reverse process of the chain rule in calculus. While the chain rule is used to differentiate composite functions, substitution is used to integrate them. In differentiation, the chain rule involves taking the derivative of the outer function and multiplying it by the derivative of the inner function. In integration, substitution involves identifying the inner function and its derivative, then transforming the integral into a simpler form. This relationship highlights the interconnectedness of differentiation and integration, reinforcing the concept that integration is the reverse operation of differentiation.

What are common mistakes to avoid when using substitution in integrals?

Common mistakes when using substitution in integrals include choosing an incorrect substitution, failing to transform the bounds in definite integrals, and neglecting to replace the original variable after integration. It's crucial to identify the correct 'inside function' for substitution and ensure that $du$ matches the integrand. For definite integrals, bounds must be transformed using the substitution. After integrating, always replace $u$ with the original variable to express the final answer in terms of the original variable. Attention to these details ensures accurate and efficient evaluation of integrals.