Lesson Video: Integration by Parts | Nagwa Lesson Video: Integration by Parts | Nagwa

# Lesson Video: Integration by Parts Mathematics

In this video, we will learn how to use integration by parts to find the integral of a product of functions.

18:38

### Video Transcript

Integration by Parts

In this video, we will learn how to use integration by parts to find the integral of a product of two functions. Weβll do this by reversing the product rule for derivatives and by using the fact that integration gives us the most general antiderivative of a function.

So, letβs start by recalling what the product rule for differentiation tells us. It tells us if we have two differentiable functions π of π₯ and π of π₯, then the derivative of π of π₯ multiplied by π of π₯ with respect to π₯ is equal to π prime of π₯ times π of π₯ plus π of π₯ multiplied by π prime of π₯. And this rule helps us differentiate the product of two functions. We need to reverse this rule to get an equivalent rule in terms of integration. This will help us integrate the product of two functions.

We can do this by noting π of π₯ multiplied by π of π₯ is an antiderivative of the right-hand side of the equation: π prime of π₯ times π of π₯ plus π of π₯ times π prime of π₯. And now, since integration gives us the most general antiderivative of a function, we can rewrite this as an integral result. We get the indefinite integral of π prime of π₯ times π of π₯ plus π of π₯ times π prime of π₯ with respect to π₯ is π of π₯ multiplied by π of π₯.

However, we canβt yet use this result to evaluate the integral of a product of two functions because our integrand contains two terms. But letβs instead take the integral of each term in the integrand separately. This gives us the indefinite integral of π prime of π₯ times π of π₯ with respect to π₯ plus the indefinite integral of π of π₯ times π prime of π₯ with respect to π₯ is equal to π of π₯ multiplied by π of π₯.

We can now find an equation for the product of two functions by rearranging to make this the subject. This then gives us that the indefinite integral of π of π₯ times π prime of π₯ with respect to π₯ is equal to π of π₯ multiplied by π of π₯ minus the indefinite integral of π prime of π₯ times π of π₯ with respect to π₯. And this is called the integration by parts method.

And at first, we might notice something interesting. We want to use this method to evaluate the integral of the product of two functions. However, in our formula, weβre using the integration of the product of two functions. And so it might not seem that this formula will be useful. However, we can see weβre differentiating a different function inside of our integrand. And as weβll see, being able to differentiate one of the factors of our integrand often makes the integral easier to evaluate.

And before we move on to some examples, itβs also worth noting the integration by parts formula is often written in Leibnizβs notation. However, this time, weβll call our original functions π’ of π₯ and π£ of π₯. This gives us the indefinite integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯. Letβs now see an example of applying integration by parts to evaluate the integral of a product of two functions.

Use integration by parts to evaluate the indefinite integral of π₯ times the sin of π₯ with respect to π₯.

In this question, we are asked to evaluate the indefinite integral of the product of two functions: an algebraic function π₯ and a trigonometric function sin of π₯. And this should hint to us that we could try using integration by parts to evaluate this integral, even if we were not told to use this method in the question. So, letβs start by recalling the formula for integration by parts.

It tells us the indefinite integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯. To apply this to evaluate our integral, weβre going to need to choose which of the factors in our integrand needs to be π’ and which one should be dπ£ by dπ₯. So, we need to determine how weβre going to choose which factor will be π’ and which will be dπ£ by dπ₯.

To do this, we need to know when weβre using this formula, the most difficult part of this expression to evaluate is the integral. So, we want to choose our functions to make this integral as easy to evaluate as possible. And usually, we want to do this by choosing π’ which makes dπ’ by dπ₯ simpler. However, sometimes, we need to choose our function π’ such that it cancels when multiplied by part of π£. In either case, letβs look at our functions π₯ and sin of π₯.

We know when we differentiate π₯, its degree will decrease. However, if we were to differentiate the sin of π₯, we would just get another trigonometric function. This would not be any simpler. So, weβll set our function π’ to be equal to π₯ and dπ£ by dπ₯ to be equal to the sin of π₯. Now, to apply integration by parts, weβre going to need to find an expression for dπ’ by dπ₯ and π£.

So, letβs start by finding dπ’ by dπ₯. Thatβs the derivative of π₯ with respect to π₯, which is just equal to one. We also want to find an expression for π£ from dπ£ by dπ₯. And we can recall π£ will be an antiderivative of this function. So, we know that π£ is going to be equal to the indefinite integral of the sin of π₯ with respect to π₯, which we can recall is negative the cos of π₯ plus the constant of integration. However, we donβt need to add a constant of integration in this case.

To see this, consider what would happen if we replace π£ by π£ plus πΆ in our integration by parts formula. In the first term, we would get π’ multiplied by the constant of integration πΆ. However, in our second term, we get πΆ times dπ’ by dπ₯. Negative the integral of this is negative π’ time πΆ, so these two terms cancel. So, itβs another case of constants of integration canceling, so we donβt need to include this constant.

Letβs now substitute our expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯ into our integration by parts formula. We get the indefinite integral of π₯ times the sin of π₯ with respect to π₯ is equal to π₯ times negative the cos of π₯ minus the indefinite integral of negative cos of π₯ multiplied by one with respect to π₯. The first term simplifies to give us negative π₯ times the cos of π₯. And in our second term, we take out the factor of negative one. This means we add the indefinite integral of the cos of π₯ with respect to π₯.

Now, all thatβs left to do is evaluate this indefinite integral. We know that the sin of π₯ is an antiderivative of cos of π₯. So, the indefinite integral of cos of π₯ is sin of π₯ plus the constant of integration πΆ. Therefore, we were able to show the indefinite integral of π₯ times the sin of π₯ with respect to π₯ is negative π₯ cos of π₯ plus sin of π₯ plus πΆ. And it is worth noting we can check our answer by differentiating it with respect to π₯ and checking we get π₯ multiplied by the sin of π₯.

In our next example, weβre going to use integration by parts to evaluate the integral of the natural logarithm function.

Integrate the natural logarithm of π₯ with respect to π₯ by parts using π’ is equal to the natural logarithm of π₯ and dπ£ is equal to dπ₯.

In this question, we want to evaluate an indefinite integral by using integration by parts. We can do this by first recalling integration by parts tells us the indefinite integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯.

Usually when weβre using integration by parts, we first need to determine which function we set as π’ and which one we set as dπ£ by dπ₯. However, weβre already told this information in the question. Weβre told to set π’ equal to natural logarithm of π₯. And weβre told in terms of differentials, dπ£ is equal to dπ₯. This means that dπ£ by dπ₯ is equal to one. We can then see if we set π’ equal to the natural logarithm of π₯ and dπ£ by dπ₯ equal to one, then in our integration by parts formula we get the indefinite integral of the natural logarithm of π₯ with respect to π₯.

So to apply this, weβre going to need to find expressions for π£ and dπ’ by dπ₯. Letβs start by differentiating π’ with respect to π₯. Thatβs the derivative of the natural logarithm with respect to π₯, which we know is the reciprocal function dπ’ by dπ₯ is one over π₯. Since dπ£ by dπ₯ is equal to one, the derivative of π£ with respect to π₯ is equal to one. In other words, π£ is an antiderivative of one. And we know the derivative of π₯ with respect to π₯ is one, so weβll set π£ equal to π₯.

Now, we just substitute all of these expressions into our integration by parts formula. We get the indefinite integral of the natural logarithm of π₯ with respect to π₯ is equal to π₯ times the natural logarithm of π₯ minus the indefinite integral of π₯ multiplied by one over π₯ with respect to π₯. And we can simplify this; π₯ multiplied by one over π₯ is just equal to one. So, weβre just left with π₯ times the natural logarithm of π₯ minus the indefinite integral of one with respect to π₯. And we can evaluate this integral. Itβs equal to π₯ plus the constant of integration πΆ.

Therefore, we were able to show the indefinite integral of the natural logarithm of π₯ with respect to π₯ is equal to π₯ times the natural logarithm of π₯ minus π₯ plus πΆ. And integration by parts was really useful for helping us evaluate this integral because the derivative of the natural logarithm of π₯ is a much simpler expression than its integral.

Thus far, when weβve used integration by parts, weβve either been given our function dπ’ and dπ£ by dπ₯ or weβve needed to choose them ourselves. And itβs quite difficult to choose these functions ourselves since itβs not always obvious which function to choose as π’. But there is one method we can use to help us choose our function π’; itβs called the LIATE method. And this tells us to choose our function π’ based on which of five possible types of functions appears first in our integrand.

The L stands for logarithmic, the I stands for inverse trigonometric functions, the A stands for algebraic, the T stands for trigonometric, and finally the E stands for exponential. We just choose our function π’ based on which of these five functions appears first in our integrand. And it is worth noting the LIATE method does not always give us the best function π’. However, it does usually work. Letβs now see an example of applying the LIATE method to evaluate an integral by parts.

Determine the indefinite integral of three π₯ plus four all squared times π to the power of π₯ with respect to π₯.

In this question, we are asked to evaluate the indefinite integral of the product of two functions. So, weβre going to attempt to do this by using integration by parts. And we can recall integration by parts tells us the indefinite integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯.

So now, we need to choose our functions π’ and dπ£ by dπ₯. And weβre going to choose our function π’ by using the LIATE method. We check our integrand for each of the five types of functions in turn to determine our function π’. First, we start with logarithmic functions. We can see thereβs no logarithmic function in our integrand. So, we move on to the next type of function, inverse trig. Once again, we can see thereβs no inverse trigonometric functions in our integrand. So, weβll move on to the third type of function, algebraic, and we can see that three π₯ plus four all squared is a polynomial. So, itβs an algebraic function. So, weβll set π’ to be three π₯ plus four all squared. And this means dπ£ by dπ₯ is going to be equal to the remaining factor π to the π₯.

Now, to apply integration by parts, weβre going to need to find expressions for dπ’ by dπ₯ and π£. Letβs start with dπ’ by dπ₯. Weβre going to want to differentiate π’ with respect to π₯. And thereβs a few different ways of doing this. For example, π’ is the composition of functions, so we could do this by using the chain rule. However, our outer exponent is only two. So, itβs probably easier to just distribute two over our parentheses. We get nine π₯ squared plus 24π₯ plus 16. This then allows us to differentiate this term by term by using the power rule for differentiation. We get dπ’ by dπ₯ is equal to 18π₯ plus 24.

Letβs now find π£. We can see that dπ£ by dπ₯ is equal to π the π₯. So, π£ is going to be an antiderivative of π to the π₯. π£ is equal to the indefinite integral of π to the π₯ with respect to π₯. And we know this is equal to π to the power of π₯ plus the constant of integration. However, we donβt need the constant of integration in this case, we can now substitute these expressions into our integration by parts formula. We get the indefinite integral of three π₯ plus four all squared times π to the power of π₯ with respect to π₯ is equal to three π₯ plus four squared times π to the power of π₯ minus the indefinite integral of π to the power of π₯ multiplied by 18π₯ plus 24 with respect to π₯.

And at this stage, we might be worried weβve chosen the wrong function π’ since weβve ended up with the integral of a product of two functions. However, we can notice something interesting. If we differentiate 18π₯ plus 24 with respect to π₯, weβll get a constant. So, we can evaluate this integral by applying integration by parts a second time. To do this, letβs start by clearing some space and then choose our function π’. Weβll do this once again by using the LIATE method. We can see thereβs no logarithmic functions and no inverse trigonometric functions. However, there is an algebraic function.

So, weβll set π’ to be 18π₯ plus 24 and dπ£ by dπ₯ to be π to the power of π₯. We now want to find dπ’ by dπ₯ and π£. Since π’ is a linear function, its derivative is the coefficient of π₯. dπ’ by dπ₯ is 18. And π£ is an antiderivative of π to the power of π₯. π£ is just π to the power of π₯. We can now substitute these values into our integration by parts formula to evaluate the second indefinite integral. This then gives us three π₯ plus four squared times π to the power of π₯ minus 18π₯ plus 24 times π to the power of π₯ minus the indefinite integral of π to the power of π₯ times 18 with respect to π₯. And we can simplify this.

First, we can distribute the negative over our parentheses, which gives us the following. Next, we can take 18 out of our integral. Now, we can evaluate our final indefinite integral. The indefinite integral of π to the power of π₯ with respect to π₯ is π to the power of π₯ plus the constant of integration, which then leaves us with the following expression.

Now we could leave our answer like this. However, we can also notice all three of these terms share a factor of π to the power of π₯. So, if we take out the shared factor of π to the power of π₯, we get the following expression, and we can simplify the algebraic expression inside of our parentheses. And if we were to do this, we would get our final answer. The indefinite integral of three π₯ plus four all squared times π to the power of π₯ with respect to π₯ is equal to π to the power of π₯ times nine π₯ squared plus six π₯ plus 10 plus the constant of integration πΆ.

Letβs now see an example of using indefinite integration to evaluate the integral of the product of a trigonometric function and an exponential function.

By setting π’ equal to π to the power of π₯ and dπ£ equal to cos of π₯ dπ₯, evaluate the indefinite integral of π to the power of π₯ multiplied by the cos of π₯ with respect to π₯ by integrating by parts.

In this question, we are asked to evaluate an indefinite integral by using integration by parts. And we can recall integration by parts tells us the indefinite integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯. And at this point, we usually need to choose our functions π’ and dπ£ by dπ₯ to apply integration by parts. However, weβre told what to choose in the question.

Weβre told to choose π’ equal to π to the power of π₯. And weβre told in terms of differentials, dπ£ is equal to cos of π₯ dπ₯. This is equivalent to saying that dπ£ by dπ₯ is equal to cos of π₯. So, π’ is π to the power of π₯, and dπ£ by dπ₯ is cos of π₯. We need to use these to determine dπ’ by dπ₯ and π£. First, dπ’ by dπ₯ is the derivative of π to the power of π₯ with respect to π₯, which is just equal to itself.

Next, π£ is the indefinite integral of the cos of π₯ with respect to π₯. Thatβs the sin of π₯ plus the constant of integration. However, in this case, we donβt need the constant. We can now substitute these expressions into our integration by parts formula. This then gives us π to the power of π₯ multiplied by the sin of π₯ minus the indefinite integral of the sin of π₯ times π to the power of π₯ with respect to π₯.

And at this point, we might be worried. Weβve not made our indefinite integral any easier to evaluate by using this integration by parts. Integrating sin of π₯ times π to the power of π₯ is just as difficult as integrating π to the power of π₯ times the cos of π₯. However, we can notice something interesting if we try and evaluate this integral by using integration by parts. If we once again set our function π’ to be the exponential function π to the power of π₯ and our function dπ£ by dπ₯ to be the trigonometric function sin π₯, then π’ will be π to the power of π₯ and π£ is going to be negative cos of π₯.

So, this indefinite integral ends up being the indefinite integral of π to the power of π₯ times cos of π₯ with respect to π₯. This is the integral weβre trying to evaluate, and weβll be able to use this to evaluate our integral. So, letβs start by setting π’ equal to π to the power of π₯ and dπ£ by dπ₯ equal to the sin of π₯. We differentiate π’ with respect to π₯. We get dπ’ by dπ₯ is equal to π to the power of π₯. And the indefinite integral of the sin of π₯ with respect to π₯ is negative the cos of π₯ plus πΆ. So, π£ is negative cos of π₯.

We can now substitute these expressions into our integration by parts formula to evaluate the indefinite integral. This then gives us π to the π₯ times sin π₯ minus π to the π₯ multiplied by negative cos of π₯ minus the indefinite integral of negative cos of π₯ times π to the power of π₯ with respect to π₯. And we can then simplify this. Letβs start by distributing the negative over our parentheses and rearranging. This gives us π to the π₯ times sin π₯ plus π to the π₯ times cos of π₯ minus the indefinite integral of π to the π₯ multiplied by cos of π₯ with respect to π₯.

But this integral is exactly the same as the integral we have on the left-hand side of the equation. So, letβs add the indefinite integral of π to the power of π₯ times the cos of π₯ with respect to π₯ to both sides of the equation. This gives us two times the indefinite integral of π to the power of π₯ times cos of π₯ with respect to π₯ is equal to π to the power of π₯ times sin π₯ plus π to the power of π₯ times cos of π₯ plus the constant of integration πΆ.

And it is worth reiterating we do need to add a constant of integration here. We can now solve for the integral by dividing both sides of the equation through by one-half. And of course, dividing our constant of integration by two will still leave us with a constant. So, weβll just call this πΆ. This then gives us our final answer. The indefinite integral of π to the power of π₯ multiplied by the cos of π₯ with respect to π₯ is equal to one-half times π to the π₯ sin π₯ plus π to the π₯ cos of π₯ plus the constant of integration πΆ.

Letβs now go over the key points of this video. First, we showed that integration by parts is the corresponding rule for the product rule. In other words, if we consider the product rule in terms of indefinite integration, we get integration by parts. In particular, in terms of Leibnizβs notation, integration by parts tells us the indefinite integral of π’ times dπ£ by dπ₯ with respect to π₯ is equal to π’ times π£ minus the indefinite integral of π£ multiplied by dπ’ by dπ₯ with respect to π₯.

And we saw that in general itβs a good idea to choose our function π’ so that when we differentiate it, we make our second integral as easy as possible to evaluate. And we also saw one method of choosing our function π’ by using the LIATE acronym. We choose π’ to be the first function in this list of five which appears in our integrand. However, it is worth noting the LIATE acronym does not always give us the best function π’, so we should always be careful of this.

Finally, itβs also worth noting we saw we could integrate certain functions, such as the natural logarithm of π₯, by using integration by parts. We just write it as one times the natural logarithm of π₯ and then apply integration by parts. And finally, we saw that sometimes we need to apply integration by parts multiple times to evaluate the indefinite integral.

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