# Video: Integration by Substitution: Indefinite Integrals

In this video, we will learn how to use integration by substitution for indefinite integrals.

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### Video Transcript

In this lesson, we’ll learn how to use integration by substitution to evaluate indefinite integrals. At this stage, you should feel comfortable finding the antiderivative of a variety of functions, including polynomials, trigonometric, and logarithmic functions. In this lesson, we’ll look at how to apply these rules to find the antiderivative, or the integral, for more complicated functions.

Because of the fundamental theorem of calculus, it’s important to be able to find the antiderivative. But our formulae don’t tell us how to evaluate integrals such as the integral of 𝑥 to the fifth power times 𝑥 to the sixth power plus nine to the seventh power with respect to 𝑥. To find this integral, we use a special strategy of introducing something extra, a new variable.

This is called integration by substitution. And it’s sometimes referred to as the reverse chain rule. The first step is often to get the integral into this form. Note we have a function 𝑔 of 𝑥 and its derivative 𝑔 prime of 𝑥. And as it’s often the case, it’s sensible just to have a look at an example of how this works.

Determine the integral of 𝑥 to the fifth power multiplied by 𝑥 to the sixth power plus nine all to the seventh power with respect to 𝑥

This is not a polynomial that’s nice to integrate using our standard rules for finding the antiderivative. And we certainly do not want to distribute our parentheses and find the antiderivative for each term. Instead, we spot that the integral is set up in this form. We have some function 𝑔 of 𝑥 and its derivative 𝑔 prime of 𝑥. If we look carefully, we see that 𝑥 to the fifth power is a scalar multiple of the derivative of 𝑥 to the power of six plus nine. And this means we can use integration by substitution to evaluate our indefinite integral.

The substitution rule says that if 𝑢 is equal to 𝑔 of 𝑥 is a differentiable function whose range is some interval 𝑖 and 𝑓 is continuous on this interval, then the integral of 𝑓 of 𝑔 of 𝑥 multiplied by 𝑔 prime of 𝑥 with respect to 𝑥 is equal to the integral of 𝑓 of 𝑢 with respect to 𝑢. We’re going to let 𝑢 be equal to the function that we originally defined 𝑔 of 𝑥. So, 𝑢 is equal to 𝑥 to the sixth power plus nine.

Now, this is great, as when we differentiate this function 𝑢 with respect to 𝑥, we see that d𝑢 by d𝑥 is equal to six 𝑥 to the fifth power. In integration by substitution, we think of the d𝑢 and d𝑥 as differentials. And we can alternatively write this as d𝑢 equals six 𝑥 to the fifth power d𝑥. Notice that whilst d𝑢 by d𝑥 is definitely not a fraction, we do treat it a little like one in this process. We divide through by six, and we see that a sixth d𝑢 is equal to 𝑥 to the fifth power d𝑥.

And now, let’s look back to our original integral. We see that we can replace 𝑥 to the fifth power d𝑥 with a sixth d𝑢. And we replace 𝑥 to the sixth power plus nine with 𝑢. And we now see that the integral we’re evaluating is the integral of a sixth of 𝑢 to the seventh power d𝑢. If we so choose, we can take this factor of a sixth outside of the integral sign, and then we’re evaluating a sixth of the integral of 𝑢 to the seventh power with respect to 𝑢.

The integral of 𝑢 to the seventh power is 𝑢 to the eighth power divided by eight plus, since it is an indefinite integral, 𝑐 the constant of integration. We distribute our parentheses. And we see that the integral is equal to one over 48 times 𝑢 to the eighth power plus 𝐶. And notice I’ve chosen this to be a capital 𝐶 because our original constant of integration has been multiplied by one-sixth.

But, remember, we were originally looking to evaluate our integral in terms of 𝑥. So, we look to our original definition of 𝑢. And we said that 𝑢 was equal to 𝑥 to the sixth power plus nine. And we then see that our integral is equal to one over 48 times 𝑥 to the sixth power plus nine to the eighth power plus 𝐶.

In this example, we saw that we should try and choose 𝑢 to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it. If that’s not possible though, we try choosing 𝑢 to be some more complicated part of the integrand. This might be the inner function in a composite function or similar. Let’s have a look at an example of this form.

Determine the integral of eight 𝑥 times eight 𝑥 plus nine squared with respect to 𝑥 by using the substitution method.

In this example, we’ve been very explicitly told to use the substitution method to evaluate this integral. Usually, we will look to choose our substitution 𝑢 to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it. Here though, it’s not instantly obvious what that might be. Instead, then, we try choosing 𝑢 to be some more complicated part of the function, perhaps the inner function in a composite function.

Let’s try 𝑢 equals eight 𝑥 plus nine. This means that d𝑢 by d𝑥 is equal to eight. And we can treat d𝑢 and d𝑥 as differentials. Remember, d𝑢 by d𝑥 is not a fraction but we certainly treat it like one when performing integration by substitution. We can say that d𝑢 is equal to eight d𝑥. Or, equivalently, an eighth d𝑢 is equal to d𝑥. Now, this isn’t instantly helpful. As if we replace d𝑥 with an eighth d𝑢 and eight 𝑥 plus nine with 𝑢, we’ll still have part of our function, that’s the eight 𝑥, which is in terms of 𝑥.

But if we look back to our substitution, we see that we can rearrange this. We subtract nine from both sides, and we see that eight 𝑥 is equal to 𝑢 minus nine. Then, our integral becomes 𝑢 minus nine times 𝑢 squared multiplied by an eighth d𝑢. Let’s take this eighth outside of the integral and then distribute the parentheses, and we see we have a simple polynomial that we can integrate.

The integral of 𝑢 cubed is 𝑢 to the fourth power divided by four. The integral of negative nine 𝑢 squared is negative nine 𝑢 cubed divided by three. And we mustn’t forget 𝐶, our constant of integration. We can simplify nine 𝑢 cubed divided by three to three 𝑢 cubed. But we mustn’t forget to replace 𝑢 with eight 𝑥 plus nine in our final step.

When we do, we see that our integral is equal to an eighth times eight 𝑥 plus nine to the fourth power divided by four minus three times eight 𝑥 plus nine cubed plus 𝐶. When we distribute our parentheses, we see we have our solution. It’s one over 32 times eight 𝑥 plus nine to the fourth power minus three-eighths times eight 𝑥 plus nine cubed plus 𝐶.

Determine the integral of 48 minus six 𝑥 over the fifth root of 16 minus two 𝑥 d𝑥.

It’s not instantly obvious how we need to evaluate this integral. However, if we look carefully, we see that the numerator is a scalar multiple of the inner function on the denominator such that 48 minus six 𝑥 is three times 16 minus two 𝑥. This is a hint to us that we’re going to need to use integration by substitution to evaluate this indefinite integral.

We’re going to let 𝑢 be equal to 16 minus two 𝑥. We’ve chosen this part for our substitution as 16 minus two 𝑥 is the inner function in a composite function. We differentiate 𝑢 with respect to 𝑥, and we see that d𝑢 by d𝑥 is equal to negative two. Now, remember, d𝑢 by d𝑥 is not a fraction, but we treat it a little like one when performing integration by substitution. And we can see that this is equivalent to saying negative one-half d𝑢 is equal to d𝑥.

Let’s substitute what we now have into our original integral. We saw that 48 minus six 𝑥 is equal to three times 16 minus two 𝑥. So, the numerator becomes three 𝑢. The denominator becomes the fifth root of 𝑢. And we replace d𝑥 with negative a half d𝑢. Let’s take out a factor of negative three over two. And we’ll write our denominator as 𝑢 to the fifth power. We’re dividing 𝑢 to the power of one by 𝑢 to the power of one-fifth.

So, we subtract one-fifth from one, and we’re left with 𝑢 to the four-fifths. The antiderivative of 𝑢 to the four-fifths is 𝑢 to the nine-fifths divided by nine-fifths. That’s the same as five-ninths times 𝑢 to the nine-fifths. And, remember, this is an indefinite integral, so we add that constant of integration 𝑐.

When we distribute the parentheses, we have negative five-sixths times 𝑢 to the nine-fifths plus capital 𝐶. Since our original constant has been multiplied by negative three over two, and since we were evaluating an integral in terms of 𝑥, we must replace 𝑢 with 16 minus two 𝑥. And we see that our integral is negative five-sixths times 16 minus two 𝑥 to the nine-fifths plus 𝐶.

In our previous two examples, we saw that we can perform integration by substitution, even if it’s not instantly obvious what that process might look like. We’ll now see how we can use the process to integrate a more complicated trigonometric function.

Determine the integral of negative 24𝑥 cubed plus 30 sin six 𝑥 times negative six 𝑥 to the fourth power minus five cos of six 𝑥 to the fifth power with respect to 𝑥.

To evaluate this integral, we need to spot that negative 24𝑥 cubed plus 30 sin six 𝑥 is the derivative of the inner part of this composite function negative six 𝑥 to the fourth power minus five cos six 𝑥. This tells us we can use integration by substitution to evaluate this integral. We’ll let 𝑢 be the inner function in our composite function, and then we use the general result for the derivative of cos 𝑎𝑥.

And we see that d𝑢 by d𝑥, the derivative of 𝑢 with respect to 𝑥, is negative 24𝑥 cubed plus 30 sin six 𝑥. Remember, d𝑢 by d𝑥 is not a fraction, but we do treat it a little like one when performing integration by substitution. And we see that this is equivalent to saying that d𝑢 is equal to negative 24𝑥 cubed plus 30 sin six 𝑥 d𝑥. We, therefore, replace negative 24𝑥 cubed plus 30 sin six 𝑥d𝑥 with d𝑢. And we replace negative six 𝑥 to the fourth power minus five cos six 𝑥 with 𝑢.

And we see that our integral becomes really nice. It’s the integral of 𝑢 to the fifth power d𝑢. Well, the antiderivative of 𝑢 to the fifth power is 𝑢 to the sixth power over six. So, the integral of 𝑢 to the fifth power d𝑢 is 𝑢 to the sixth power over six plus the constant of integration 𝑐. Remember though, our integral’s in terms of 𝑥, so we replace 𝑢 with negative six 𝑥 to the fourth power minus five cos of six 𝑥.

And we’ve evaluated our integral. It’s a sixth of negative six 𝑥 to the fourth power minus five cos of six 𝑥 to the sixth power plus 𝑐.

In our final example, we will look at how we can use integration by substitution to integrate a logarithmic function.

Determine the integral of negative 11 over six 𝑥 times the cube root of the natural log of 𝑥 d𝑥.

In order to evaluate this integral, we need to spot that the derivative of the natural log of 𝑥 is one over 𝑥 and that a part of our function is a scalar multiple of one over 𝑥. This tells us we can use integration by substitution to evaluate our integral. We let 𝑢 be equal to the natural log of 𝑥. And we’ve seen that d𝑢 by d𝑥 is, therefore, one over 𝑥.

Remember, d𝑢 by d𝑥 is not a fraction but we treat it a little like one when performing integration by substitution. And we see that this is equivalent to saying d𝑢 is equal to one over 𝑥d𝑥. Let’s substitute what we now have into our integral. If we take out a factor of negative eleven-sixths, we see that we can replace one over 𝑥 d𝑥 with d𝑢. And we can replace the natural log of 𝑥 with 𝑢.

To make this easy to integrate, we recall that the cube root of 𝑢 is the same as 𝑢 to the power of one-third. And we know that the antiderivative of 𝑢 to the power of one-third is 𝑢 to the power of four-thirds divided by four-thirds, or three-quarters 𝑢 to the power of four-thirds. We distribute our parentheses, and we see that our integral is equal to negative eleven-eighths times 𝑢 to the power of four-thirds plus capital 𝐶.

And I’ve changed it from a lowercase 𝑐 to a capital 𝐶, as we’ve multiplied our original constant of integration by negative eleven-sixths, thereby changing the number. It is, of course, important to remember that our original integral was in terms of 𝑥. So, we replace 𝑢 with the natural log of 𝑥. And we see that our answer is negative eleven-eighths times the natural log of 𝑥 to the power of four-thirds plus 𝐶.

In this video, we’ve seen that we can introduce a substitution to find the integral of more complicated functions. We’ve learned that we usually try to choose 𝑢 to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it. If that’s not possible though, we try choosing 𝑢 to be some complicated part of the integrand. This might be the inner function in a composite function or similar. We also saw that this method can be used to integrate functions involving roots, trigonometric functions, and logarithmic functions.