WEBVTT
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In this lesson, weβll learn how to use integration by substitution to evaluate indefinite integrals.
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At this stage, you should feel comfortable finding the antiderivative of a variety of functions, including polynomials, trigonometric, and logarithmic functions.
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In this lesson, weβll look at how to apply these rules to find the antiderivative, or the integral, for more complicated functions.
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Because of the fundamental theorem of calculus, itβs important to be able to find the antiderivative.
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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₯.
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To find this integral, we use a special strategy of introducing something extra, a new variable.
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This is called integration by substitution.
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And itβs sometimes referred to as the reverse chain rule.
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The first step is often to get the integral into this form.
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Note we have a function π of π₯ and its derivative π prime of π₯.
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And as itβs often the case, itβs sensible just to have a look at an example of how this works.
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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₯
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This is not a polynomial thatβs nice to integrate using our standard rules for finding the antiderivative.
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And we certainly do not want to distribute our parentheses and find the antiderivative for each term.
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Instead, we spot that the integral is set up in this form.
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We have some function π of π₯ and its derivative π prime of π₯.
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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.
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And this means we can use integration by substitution to evaluate our indefinite integral.
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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’.
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Weβre going to let π’ be equal to the function that we originally defined π of π₯.
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So, π’ is equal to π₯ to the sixth power plus nine.
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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.
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In integration by substitution, we think of the dπ’ and dπ₯ as differentials.
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And we can alternatively write this as dπ’ equals six π₯ to the fifth power dπ₯.
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Notice that whilst dπ’ by dπ₯ is definitely not a fraction, we do treat it a little like one in this process.
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We divide through by six, and we see that a sixth dπ’ is equal to π₯ to the fifth power dπ₯.
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And now, letβs look back to our original integral.
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We see that we can replace π₯ to the fifth power dπ₯ with a sixth dπ’.
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And we replace π₯ to the sixth power plus nine with π’.
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And we now see that the integral weβre evaluating is the integral of a sixth of π’ to the seventh power dπ’.
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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’.
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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.
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We distribute our parentheses.
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And we see that the integral is equal to one over 48 times π’ to the eighth power plus πΆ.
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And notice Iβve chosen this to be a capital πΆ because our original constant of integration has been multiplied by one-sixth.
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But, remember, we were originally looking to evaluate our integral in terms of π₯.
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So, we look to our original definition of π’.
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And we said that π’ was equal to π₯ to the sixth power plus nine.
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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.
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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.
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If thatβs not possible though, we try choosing π’ to be some more complicated part of the integrand.
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This might be the inner function in a composite function or similar.
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Letβs have a look at an example of this form.
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Determine the integral of eight π₯ times eight π₯ plus nine squared with respect to π₯ by using the substitution method.
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In this example, weβve been very explicitly told to use the substitution method to evaluate this integral.
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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.
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Here though, itβs not instantly obvious what that might be.
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Instead, then, we try choosing π’ to be some more complicated part of the function, perhaps the inner function in a composite function.
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Letβs try π’ equals eight π₯ plus nine.
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This means that dπ’ by dπ₯ is equal to eight.
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And we can treat dπ’ and dπ₯ as differentials.
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Remember, dπ’ by dπ₯ is not a fraction but we certainly treat it like one when performing integration by substitution.
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We can say that dπ’ is equal to eight dπ₯.
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Or, equivalently, an eighth dπ’ is equal to dπ₯.
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Now, this isnβt instantly helpful.
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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₯.
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But if we look back to our substitution, we see that we can rearrange this.
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We subtract nine from both sides, and we see that eight π₯ is equal to π’ minus nine.
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Then, our integral becomes π’ minus nine times π’ squared multiplied by an eighth dπ’.
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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.
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The integral of π’ cubed is π’ to the fourth power divided by four.
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The integral of negative nine π’ squared is negative nine π’ cubed divided by three.
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And we mustnβt forget πΆ, our constant of integration.
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We can simplify nine π’ cubed divided by three to three π’ cubed.
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But we mustnβt forget to replace π’ with eight π₯ plus nine in our final step.
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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.
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When we distribute our parentheses, we see we have our solution.
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Itβs one over 32 times eight π₯ plus nine to the fourth power minus three-eighths times eight π₯ plus nine cubed plus πΆ.
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Determine the integral of 48 minus six π₯ over the fifth root of 16 minus two π₯ dπ₯.
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Itβs not instantly obvious how we need to evaluate this integral.
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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₯.
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This is a hint to us that weβre going to need to use integration by substitution to evaluate this indefinite integral.
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Weβre going to let π’ be equal to 16 minus two π₯.
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Weβve chosen this part for our substitution as 16 minus two π₯ is the inner function in a composite function.
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We differentiate π’ with respect to π₯, and we see that dπ’ by dπ₯ is equal to negative two.
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Now, remember, dπ’ by dπ₯ is not a fraction, but we treat it a little like one when performing integration by substitution.
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And we can see that this is equivalent to saying negative one-half dπ’ is equal to dπ₯.
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Letβs substitute what we now have into our original integral.
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We saw that 48 minus six π₯ is equal to three times 16 minus two π₯.
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So, the numerator becomes three π’.
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The denominator becomes the fifth root of π’.
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And we replace dπ₯ with negative a half dπ’.
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Letβs take out a factor of negative three over two.
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And weβll write our denominator as π’ to the fifth power.
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Weβre dividing π’ to the power of one by π’ to the power of one-fifth.
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So, we subtract one-fifth from one, and weβre left with π’ to the four-fifths.
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The antiderivative of π’ to the four-fifths is π’ to the nine-fifths divided by nine-fifths.
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Thatβs the same as five-ninths times π’ to the nine-fifths.
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And, remember, this is an indefinite integral, so we add that constant of integration π.
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When we distribute the parentheses, we have negative five-sixths times π’ to the nine-fifths plus capital πΆ.
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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₯.
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And we see that our integral is negative five-sixths times 16 minus two π₯ to the nine-fifths plus πΆ.
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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.
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Weβll now see how we can use the process to integrate a more complicated trigonometric function.
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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₯.
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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₯.
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This tells us we can use integration by substitution to evaluate this integral.
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Weβll let π’ be the inner function in our composite function, and then we use the general result for the derivative of cos ππ₯.
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And we see that dπ’ by dπ₯, the derivative of π’ with respect to π₯, is negative 24π₯ cubed plus 30 sin six π₯.
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Remember, dπ’ by dπ₯ is not a fraction, but we do treat it a little like one when performing integration by substitution.
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And we see that this is equivalent to saying that dπ’ is equal to negative 24π₯ cubed plus 30 sin six π₯ dπ₯.
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We, therefore, replace negative 24π₯ cubed plus 30 sin six π₯dπ₯ with dπ’.
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And we replace negative six π₯ to the fourth power minus five cos six π₯ with π’.
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And we see that our integral becomes really nice.
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Itβs the integral of π’ to the fifth power dπ’.
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Well, the antiderivative of π’ to the fifth power is π’ to the sixth power over six.
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So, the integral of π’ to the fifth power dπ’ is π’ to the sixth power over six plus the constant of integration π.
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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₯.
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And weβve evaluated our integral.
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Itβs a sixth of negative six π₯ to the fourth power minus five cos of six π₯ to the sixth power plus π.
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In our final example, we will look at how we can use integration by substitution to integrate a logarithmic function.
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Determine the integral of negative 11 over six π₯ times the cube root of the natural log of π₯ dπ₯.
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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₯.
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This tells us we can use integration by substitution to evaluate our integral.
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We let π’ be equal to the natural log of π₯.
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And weβve seen that dπ’ by dπ₯ is, therefore, one over π₯.
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Remember, dπ’ by dπ₯ is not a fraction but we treat it a little like one when performing integration by substitution.
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And we see that this is equivalent to saying dπ’ is equal to one over π₯dπ₯.
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Letβs substitute what we now have into our integral.
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If we take out a factor of negative eleven-sixths, we see that we can replace one over π₯ dπ₯ with dπ’.
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And we can replace the natural log of π₯ with π’.
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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.
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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.
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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.
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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.
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It is, of course, important to remember that our original integral was in terms of π₯.
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So, we replace π’ with the natural log of π₯.
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And we see that our answer is negative eleven-eighths times the natural log of π₯ to the power of four-thirds plus πΆ.
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In this video, weβve seen that we can introduce a substitution to find the integral of more complicated functions.
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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.
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If thatβs not possible though, we try choosing π’ to be some complicated part of the integrand.
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This might be the inner function in a composite function or similar.
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We also saw that this method can be used to integrate functions involving roots, trigonometric functions, and logarithmic functions.