WEBVTT
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In this video, weβll learn how to use integration by parts to find the integral of a product of functions.
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Because of the fundamental theorem of calculus, we can integrate a function if we know its antiderivative.
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And by this stage, you should feel confidence in evaluating the integral of polynomial functions as well as trigonometric and exponential functions.
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Itβs not entirely necessary to be confident in applying integration by substitution to access this video.
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Though it would be beneficial.
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Every differentiation rule has a corresponding integration rule.
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For example, integration by substitution is the rule that corresponds to the chain rule for differentiation.
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Weβre now going to be looking at integration by parts.
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This rule for integration corresponds to the product rule for differentiation.
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The product rule states that, for two differentiable functions π and π, the derivative of their product is π times π prime plus π times π prime.
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That is, π times the derivative of π plus π times the derivative of π.
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By integrating both sides of this equation, we obtain π times π to be equal to the integral of π times π prime plus π prime times π, evaluated with respect to π₯.
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Now we know that the integral of the sum of two functions is equal to the sum of the integral of each function.
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And we can rewrite π times π as shown.
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We rearrange.
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And we obtain the formula for integration by parts.
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We see that the indefinite integral of π times π prime is equal to π times π minus the indefinite integral of π prime times π, evaluated with respect to π₯.
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This is sometimes alternatively written using functions π’ and π£ and Leibniz notation.
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Such that the integral of π’ times dπ£ by dπ₯ evaluated with respect to π₯ is equal to π’π£ minus the integral of π£ times dπ’ by dπ₯ evaluated with respect to π₯.
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Weβll begin by considering a fairly straightforward example of the application of this formula.
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Use integration by parts to evaluate the integral of π₯ times sin π₯ with respect to π₯.
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The function weβre looking to integrate is the product of two functions.
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Thatβs π₯ and sin π₯.
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This indicates that we might need to use integration by parts to evaluate our integral.
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Remember the formula here says the integral of π’ times dπ£ by dπ₯ is equal to π’π£ minus the integral of π£ times dπ’ by dπ₯.
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If we compare this formula to our integrand, we see that weβre going to need to decide which function is π’.
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And which function is dπ£ by dπ₯.
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So how do we decide this?
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Well, our aim is going to be to ensure that the second integral we get over here is a little simpler.
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We therefore want π’ to be a function that either becomes simpler when differentiated or helps to simplify the integrand when multiplied by the function π£.
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It should be quite clear that, out of π₯ and sin π₯, the function that becomes simpler when differentiated is π₯.
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So we let π’ be equal to π₯ and dπ£ by dπ₯ be equal to sin π₯.
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The derivative of π’ with respect to π₯ is one.
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And what do we do with dπ£ by dπ₯?
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Well, weβre going to need to find π£.
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So we find the antiderivative of sin of π₯, which is of course negative cos of π₯.
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Letβs substitute what we have into our formula.
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We see that our integral is equal to π₯ times negative cos of π₯ minus the integral of negative cos of π₯ times one dπ₯.
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This simplifies to negative π₯ cos of π₯.
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And then we take the factor of negative one outside of our integral.
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And we see that we add the integral of cos of π₯ evaluated with respect to π₯.
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The antiderivative of cos of π₯ is sin of π₯.
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And of course, because this is an indefinite integral, we must add that constant of integration π.
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And we see that the solution is negative π₯ cos of π₯ plus sin of π₯ plus π.
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Now itβs useful to remember that we can check our answer by differentiating.
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If we do, we indeed obtain π₯ times sin π₯ as required.
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Weβre now going to look at a common example.
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Thatβs the integral of the natural log of π₯.
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Integrate the natural log of π₯ dπ₯ by parts using π’ equals the natural log of π₯ and dπ£ equals dπ₯.
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Well, this is great.
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Weβve been told to use integration by parts.
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And weβve also been told to let π’ be equal to the natural log of π₯ and dπ£ be equal to dπ₯.
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So we recall the formula for integration by parts.
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The integral of π’ times dπ£ by dπ₯ is equal to π’π£ minus the integral of π£ times dπ’ by dπ₯.
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Now we can rewrite dπ£ equals dπ₯ slightly.
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And we can say that if dπ£ is equal to dπ₯, then dπ£ by dπ₯ must be equal to one.
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Our job is going to be to find the missing parts of the formula.
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dπ’ by dπ₯ is fairly straightforward.
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The derivative of the natural log of π₯ is one over π₯.
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And if we integrate both sides of the equation dπ£ equals dπ₯, we obtain π£ as being equal to π₯.
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Substituting everything into the formula, and we get π₯ times the natural log of π₯ minus the integral of π₯ times one over π₯ dπ₯.
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Well, this integral simplifies really nicely.
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Weβre actually going to be integrating one with respect to π₯.
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The integral of one is, of course, simply π₯.
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And of course, since this is an indefinite integral, we must remember to add that constant of integration π.
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So we obtain the integral of the natural log of π₯ dπ₯ as being π₯ times the natural log of π₯ minus π₯ plus π.
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And integration by parts was really effective for integrating the natural log of π₯.
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Because the derivative of the natural log of π₯ is much simpler than the original function the natural log of π₯.
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In our next example, weβre going to consider how integration by parts can be used to find the integral of a quotient.
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Determine the indefinite integral of two π to the power of π₯ times π₯ over three times π₯ plus one all squared, evaluated with respect to π₯.
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It might not be instantly obvious how we are going to evaluate this.
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However, thereβs no clear substitution we can make.
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And itβs certainly not something we can evaluate in our heads.
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So that tells us we might need to use integration by parts.
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Remember, this says the integral of π’ times dπ£ by dπ₯ is equal to π’π£ minus the integral of π£ times dπ’ by dπ₯.
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Now, if we compare this formula to our integrand, we see that we need to decide which function is π’.
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And which function is dπ£ by dπ₯.
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So how do we decide this?
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Well, our aim is to make sure that the second integral we get over here is a little simpler.
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We therefore want π’ to be a function that either becomes simpler when differentiated or helps us to simplify the integrand when multiplied by this π£.
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Now it isnβt hugely obvious what we should choose π’ to be.
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So a little trial and error might be in order.
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Letβs rewrite our integrand as two-thirds π₯ times π to the power of π₯ times one over π₯ plus one squared.
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And in fact, letβs take the constant factor of two-thirds outside of the integral.
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Letβs try π’ equals π₯ times π to the power of π₯ and dπ£ by dπ₯ as being one over π₯ plus one all squared, which Iβve written as π₯ plus one to the power of negative two.
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To find dπ’ by dπ₯, weβre going to need to use the product rule.
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If we do, we see that dπ’ by dπ₯ is equal to π₯ times the derivative of π to the power of π₯ plus π to the power of π₯ times the derivative of π₯.
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Well, the derivative of π to the power of π₯ is π to the power of π₯.
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And the derivative of π₯ is one.
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So weβve obtained dπ’ by dπ₯ to be equal to π₯π to the power of π₯ plus π to the power of π₯.
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We can use the reverse of the chain rule to work out the antiderivative of π₯ plus one to the power of negative two.
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Itβs negative π₯ plus one to the negative one.
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Rewriting π₯ plus one to the power of negative one as one over π₯ plus one.
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And we can substitute everything we have into our formula for integration by parts.
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Now over here, this next step is important.
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We factor by π to the power of π₯.
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And then we can see we can divide through by π₯ plus one.
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And our second integrand becomes negative π to the π₯.
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We take out that factor of negative one.
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And we know that the integral of π of the power of π₯ is simply π to the power of π₯.
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And so we have two-thirds of π₯ times π to the power of π₯ over negative π₯ plus one plus π to the power of π₯ plus π.
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We rewrite this first fraction slightly and then multiply the numerator and the denominator of π to the power of π₯ by π₯ plus one.
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So that we can add these fractions.
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And we see that the sum of these fractions is negative π₯ times π to the power of π₯ plus π₯ times π to the power of π₯, which is zero, plus π to the power of π₯ over π₯ plus one.
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Our final step is to distribute the parentheses.
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And we get two π to the power of π₯ over three times π₯ plus one.
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And Iβve changed our constant of integration to capital πΆ.
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Since our original constant of integration was multiplied by two-thirds.
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Now, in this example, it was a little bit tricky to figure out what we needed π’ to be.
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But there is a bit of a trick that can help us identify how to choose the function for π’.
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Itβs the letters L-I-A-T-E or LIATE.
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Essentially, you set π’ to the first term you see in the list.
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L is logarithm.
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I is inverse trig.
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A stands for algebraic.
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T stands for trigonometric.
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And finally, you look for an exponential function.
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Now this rule doesnβt cover everything.
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No rule can.
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But it does work remarkably well and can be a good starting point.
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In our next example, weβll see how sometimes integration by parts might be required twice.
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Evaluate the definite integral between the limits of zero and one of π₯ squared times π to the power of π₯ dπ₯.
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In this example, we have the product of two functions.
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Thatβs a hint that we might need to use integration by parts.
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This says that the integral of π’ times dπ£ by dπ₯ is equal to π’π£ minus the integral of π£ times dπ’ by dπ₯.
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So how do we decide what weβre going to let π’ be equal to?
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Well, remember, we want to ensure that this second integral over here is a little simpler.
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Therefore, weβre going to want π’ to be a function that either becomes simpler when differentiated or helps to simplify the integrand when multiplied by this π£.
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It makes no sense for us to let π’ be equal to π to the power of π₯.
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Since the derivative of π to the power of π₯ is just π to the power of π₯.
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So, instead, weβre going to let π’ be equal to π₯ squared.
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And dπ£ by dπ₯ is therefore π to the power of π₯.
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dπ’ by dπ₯ is then two π₯.
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And the antiderivative of π to the power of π₯ is π to the power of π₯.
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So π£ is π to the power of π₯.
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And we obtain the integral to be equal to π₯ squared π to the power of π₯ minus the integral of two π₯π to the power of π₯.
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Now notice, Iβve used this kind of funny half bracket here.
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This is just a way of reminding ourselves that weβre dealing with a definite integral.
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And weβre going to need to evaluate both parts of it between the limits of zero and one.
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But how do we evaluate the integral of two π₯ times π to the power of π₯?
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Well, weβre going to need to use integration by parts again.
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For much the same reason, we once again choose dπ£ by dπ₯ to be equal to π to the power of π₯.
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And then π’ is equal to two π₯.
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So we see that dπ’ by dπ₯ is equal to two.
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And π£ is equal to π to the power of π₯.
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So letβs just quickly evaluate the integral of two π₯π to the power of π₯.
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Itβs two π₯π to the power of π₯ minus the integral of two times π to the power of π₯.
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Well, the integral of two π to the power of π₯ is just two π to the power of π₯.
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And Iβve put plus π in brackets because the integral weβve just done is an indefinite integral.
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But actually the one weβre really going to be doing is between the limits of zero and one.
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Replacing the integral of two π₯π to the power of π₯ with two π₯π to the power of π₯ minus two π to the π₯.
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And we obtain the integral of π₯ squared π to the power of π₯ to be equal to π₯ squared π to the power of π₯ minus two π₯π to the power of π₯ plus two π to the power of π₯.
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Now, weβre going to need to evaluate this between the limits of zero and one.
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Substituting zero and one and finding their difference.
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And we have π to the power of one minus two π to the power of one plus two π to the power of one.
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All these disappear and minus two.
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And weβre done.
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The integral evaluated between zero and one of π₯ squared times π to the power of π₯ is π minus two.
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In our final example, weβre going to look at how integration by parts can help us to evaluate the integral of an inverse trigonometric function.
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Calculate the definite integral evaluated between zero and one of the inverse tan of π₯ with respect to π₯.
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When integrating inverse trigonometric functions, we plump for using integration by parts.
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And the formula for that is as shown.
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And weβre going to do something a little bit strange.
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Weβre going to rewrite our integrand as one times the inverse trigonometric function.
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Thatβs one times the inverse tan of π₯.
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We then let π’ be equal to the inverse tan of π₯.
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Remember, we know how to find the derivative of this.
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And we let dπ£ by dπ₯ be equal to one.
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The derivative of the inverse tangent function is one over one plus π₯ squared.
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So thatβs dπ’ by dπ₯.
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And the antiderivative of one is simply π₯.
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So π£ is equal to π₯.
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Great!
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Substituting everything we have into the formula for integration by substitution.
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And we see that the integral of the inverse tan of π₯ is π₯ times the inverse tan of π₯ minus the integral of π₯ over one plus π₯ squared.
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Now weβre actually going to need to use integration by substitution here to evaluate the integral of π₯ over one plus π₯ squared.
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Now, usually, when performing integration by substitution, we introduce a new variable π’.
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π’ already has a meaning in this example, though.
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So weβre going to let π‘ be equal to one plus π₯ squared, which means that dπ‘ by dπ₯ is equal to two π₯.
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Now, dπ‘ by dπ₯ is, of course, not a fraction.
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But we treat it a little like one when working with integration by substitution.
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And we see that we can equivalently say that a half dπ‘ equals π₯ dπ₯.
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So actually, weβre going to be working out the integral of one over two π‘ or taking the constant factor of a half outside the integral.
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So we can just actually integrate one over π‘.
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But weβre going to need to do something with these limits.
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Weβre going to use the substitution π‘ equals one plus π₯ squared.
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And the first limit, that weβre interested in, is one.
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So we substitute π₯ equals one.
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And we get one plus one squared, which is two.
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Down here, weβre substituting π₯ equals zero.
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Thatβs one plus zero squared, which is equal to one.
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So weβre going to evaluate the integral of one over π‘ between the limits of one and two.
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The integral of one over π‘ is simply the natural log of π‘.
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So we can evaluate the natural log of π‘ between the limits of one and two by substituting them in and finding their difference.
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Thatβs the natural log of two minus the natural log of one.
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And since the natural log of one is zero, we found this integral to be equal to a half times the natural log of two.
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We can pop this back into our original integral.
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And we see that weβre going to need to evaluate π₯ times the inverse tan of π₯ between the limits of zero and one.
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Well, thatβs one times the inverse tan of one minus zero, which is equal to π by four.
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And weβre done.
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The integral of the inverse tan of π₯ between the limits of zero and one is π by four minus the natural log of two over two.
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In this video, weβve seen that integration by parts is the corresponding rule for integration to the product rule for differentiation.
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We saw that using Leibniz notation, the formula is the integral of π’ times dπ£ by dπ₯ equals π’π£ minus the integral of π£ times dπ’ by dπ₯.
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We saw that, generally, we choose our function π’, with an aim of ensuring that the second integral we get is a little simpler.
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But we also saw that the acronym LIATE can help us decide which function is going to be π’.
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We finally saw that we can use this to evaluate the integrals of the product of functions, quotients, and reciprocal functions.
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And sometimes it might be applied more than once.