### Video Transcript

In this lesson, weβll 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, weβll 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, itβs important to be able to find the antiderivative. But our formulae donβt tell us how to evaluate integrals such as the integral of π₯ to the fifth power times π₯ to the sixth power plus nine to the seventh power with respect to π₯. To find this integral, we use a special strategy of introducing something extra, a new variable.

This is called integration by substitution. And itβs 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 π of π₯ and its derivative π prime of π₯. And as itβs often the case, itβs sensible just to have a look at an example of how this works.

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

This is not a polynomial thatβs 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 π of π₯ and its derivative π prime of π₯. If we look carefully, we see that π₯ to the fifth power is a scalar multiple of the derivative of π₯ 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 π’ is equal to π of π₯ is a differentiable function whose range is some interval π and π is continuous on this interval, then the integral of π of π of π₯ multiplied by π prime of π₯ with respect to π₯ is equal to the integral of π of π’ with respect to π’. Weβre going to let π’ be equal to the function that we originally defined π of π₯. So, π’ is equal to π₯ to the sixth power plus nine.

Now, this is great, as when we differentiate this function π’ with respect to π₯, we see that dπ’ by dπ₯ is equal to six π₯ to the fifth power. In integration by substitution, we think of the dπ’ and dπ₯ as differentials. And we can alternatively write this as dπ’ equals six π₯ to the fifth power dπ₯. Notice that whilst dπ’ by dπ₯ 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 dπ’ is equal to π₯ to the fifth power dπ₯.

And now, letβs look back to our original integral. We see that we can replace π₯ to the fifth power dπ₯ with a sixth dπ’. And we replace π₯ to the sixth power plus nine with π’. And we now see that the integral weβre evaluating is the integral of a sixth of π’ to the seventh power dπ’. If we so choose, we can take this factor of a sixth outside of the integral sign, and then weβre evaluating a sixth of the integral of π’ to the seventh power with respect to π’.

The integral of π’ to the seventh power is π’ to the eighth power divided by eight plus, since it is an indefinite integral, π the constant of integration. We distribute our parentheses. And we see that the integral is equal to one over 48 times π’ to the eighth power plus πΆ. And notice Iβve chosen this to be a capital πΆ 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 π₯. So, we look to our original definition of π’. And we said that π’ was equal to π₯ to the sixth power plus nine. And we then see that our integral is equal to one over 48 times π₯ to the sixth power plus nine to the eighth power plus πΆ.

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

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

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

Letβs try π’ equals eight π₯ plus nine. This means that dπ’ by dπ₯ is equal to eight. And we can treat dπ’ and dπ₯ as differentials. Remember, dπ’ by dπ₯ is not a fraction but we certainly treat it like one when performing integration by substitution. We can say that dπ’ is equal to eight dπ₯. Or, equivalently, an eighth dπ’ is equal to dπ₯. Now, this isnβt instantly helpful. As if we replace dπ₯ with an eighth dπ’ and eight π₯ plus nine with π’, weβll still have part of our function, thatβs the eight π₯, which is in terms of π₯.

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 π₯ is equal to π’ minus nine. Then, our integral becomes π’ minus nine times π’ squared multiplied by an eighth dπ’. Letβs 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 π’ cubed is π’ to the fourth power divided by four. The integral of negative nine π’ squared is negative nine π’ cubed divided by three. And we mustnβt forget πΆ, our constant of integration. We can simplify nine π’ cubed divided by three to three π’ cubed. But we mustnβt forget to replace π’ with eight π₯ plus nine in our final step.

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

Determine the integral of 48 minus six π₯ over the fifth root of 16 minus two π₯ dπ₯.

Itβs 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 π₯ is three times 16 minus two π₯. This is a hint to us that weβre going to need to use integration by substitution to evaluate this indefinite integral.

Weβre going to let π’ be equal to 16 minus two π₯. Weβve chosen this part for our substitution as 16 minus two π₯ is the inner function in a composite function. We differentiate π’ with respect to π₯, and we see that dπ’ by dπ₯ is equal to negative two. Now, remember, dπ’ by dπ₯ 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 dπ’ is equal to dπ₯.

Letβs substitute what we now have into our original integral. We saw that 48 minus six π₯ is equal to three times 16 minus two π₯. So, the numerator becomes three π’. The denominator becomes the fifth root of π’. And we replace dπ₯ with negative a half dπ’. Letβs take out a factor of negative three over two. And weβll write our denominator as π’ to the fifth power. Weβre dividing π’ to the power of one by π’ to the power of one-fifth.

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

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

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

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

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

And we see that dπ’ by dπ₯, the derivative of π’ with respect to π₯, is negative 24π₯ cubed plus 30 sin six π₯. Remember, dπ’ by dπ₯ 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 dπ’ is equal to negative 24π₯ cubed plus 30 sin six π₯ dπ₯. We, therefore, replace negative 24π₯ cubed plus 30 sin six π₯dπ₯ with dπ’. And we replace negative six π₯ to the fourth power minus five cos six π₯ with π’.

And we see that our integral becomes really nice. Itβs the integral of π’ to the fifth power dπ’. Well, the antiderivative of π’ to the fifth power is π’ to the sixth power over six. So, the integral of π’ to the fifth power dπ’ is π’ to the sixth power over six plus the constant of integration π. Remember though, our integralβs in terms of π₯, so we replace π’ with negative six π₯ to the fourth power minus five cos of six π₯.

And weβve evaluated our integral. Itβs a sixth of negative six π₯ to the fourth power minus five cos of six π₯ to the sixth power plus π.

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 π₯ times the cube root of the natural log of π₯ dπ₯.

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

Remember, dπ’ by dπ₯ 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 dπ’ is equal to one over π₯dπ₯. Letβs 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 π₯ dπ₯ with dπ’. And we can replace the natural log of π₯ with π’.

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

And Iβve changed it from a lowercase π to a capital πΆ, as weβve 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 π₯. So, we replace π’ with the natural log of π₯. And we see that our answer is negative eleven-eighths times the natural log of π₯ to the power of four-thirds plus πΆ.

In this video, weβve seen that we can introduce a substitution to find the integral of more complicated functions. Weβve learned that we usually try to choose π’ to be some factor of the integrand whose differential also occurs, albeit some scalar multiple of it. If thatβs not possible though, we try choosing π’ 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.