Integrating Using Integration by Parts

#6.11 Integrating Using Integration by Parts

Welcome back to AP Calculus with Fiveable! In this study guide, we'll delve into the technique of Integration by Parts. Integration by Parts is a powerful method used to integrate the product of two functions, and it often comes in handy when dealing with more complex integrals. We have a few techniques such as u-substitution and Riemann sums in our calculus toolbox, so let's keep building those integration skills! 🧱

#🔄 Integration by Parts Basics

Take a look at the following integral:

$$\int x^2 \sin(x),dx$$

We can’t use any of our current integration tools to evaluate it: substitution fails, and we only know how to integrate $x^2$ and $\sin(x)$ separately. Why don’t we try to use that to our advantage?

For the integral above, we will be using a method called Integration by Parts, which is based on the product rule for differentiation. It’s essentially the reverse process!

Here is the product rule, as $u$ and $v$ representing two different functions.

$$\frac{d}{dx} uv = \textcolor{blue}{u}\cdot \textcolor{pink}{v'} +\textcolor{red}{ v}\cdot \textcolor{teal}{u'}$$

If we try to reverse the process, we take the integral of all of those terms. It will look like the following:

$$uv = \int\textcolor{blue}{u}\cdot \textcolor{pink}{v'} +\int\textcolor{red}{ v}\cdot \textcolor{teal}{u'}$$

When we rearrange the terms, we get the following rule for Integration by Parts:

$$\int \textcolor{blue}{u} , \textcolor{pink}{dv} = uv - \int \textcolor{red}{v} , \textcolor{teal}{du}$$

Where:

• $\textcolor{blue}{u}$ and $\textcolor{pink}{dv}$ are selected parts of the integrand.
• $\textcolor{teal}{du}$ is the derivative of $\textcolor{blue}{u}$ with respect to the variable of integration.
• $\textcolor{red}{v}$ is the antiderivative (integral) of $\textcolor{pink}{dv}$.

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#📚 Application and Strategy

To successfully apply Integration by Parts, follow these steps:

1. Select $u$ and $dv$ carefully. Choose the variables in a way that simplifies the integral or makes it easier to integrate! We typically use the LIATE mnemonic to quickly choose $u$ because they tend to efficiently simplify the integral.
2. Differentiate and Integrate. Compute $du$ and $v$ by finding the derivative of $u$ and the antiderivative of $dv$.
3. Apply the Formula. Plug the values of $u$, $dv$, $v$, and $du$ into the Integration by Parts formula.
4. Evaluate the New Integral. The resulting integral on the right side, with $v$ and $du$ may still need simplification. Repeat the Integration by Parts process if necessary!
5. Solve for the Original Integral. If you end up with a similar integral on both sides of the equation, solve for the original integral with algebraic manipulation.

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#🧮 Integration by Parts Practice Problems

Let's reinforce our understanding with some practice problems:

#Integration by Parts Example 1

Evaluate the following integral using integration by parts.

$$\int xe^x , dx$$

First, let's go through the LIATE acronym to make an educated guess on the best possible expression to use for $u$. Since Algebra comes before Exponential, we should start by choosing $u = x$, and then set $dv = e^x$.

Now let's differentiate and integrate to collect all the necessary information.

• $u = x$
• $du = dx$
• $v = e^x$
• $dv = e^x ,dx$ Now we can apply Integration by Parts:

$$\int x \cdot e^x , dx = x \cdot e^x - \int e^x , dx$$

Now let's evaluate the new integral and simplify.

$$\int x \cdot e^x , dx = x \cdot e^x - e^x + C$$

Therefore,

$$\int xe^x , dx = x e^x - e^x + C$$

That was a great start. Let’s try another one!

#Integration by Parts Example 2

Evaluate the following integral:

$$\int \ln(x)$$

When we look at this integral, it doesn't match the format that we’re used to seeing when working with integration by parts. Let’s do a little algebraic manipulation to make it fit!

If we multiply the integrand by $1$, we get $\int 1\cdot \ln(x)$. Now, we have two distinct functions and can use LIATE to determine which to designate as $u$.

Since Logarithmic is more of a priority than Algebra, we can set $u = \ln(x)$ and $dv = 1$. Then, let’s work on getting the information for $v$ and $du$.

• $u = \ln(x)$
• $du = \frac{1}{x} ,dx$
• $v = x$
• $dv = 1 ,dx$ Now that we have all of the necessary information, we can work through integration by parts!

$$\int \ln(x) = x\ln(x)- \int x\cdot \frac{1}{x} dx$$

We can recognize that the integral on the right simplifies to give us: $x\ln(x)- \int dx$.

Let's evaluate the integral.

$$\int dx = x$$

Therefore,

$$\int \ln(x) = x\ln(x)- x + C$$

Amazing! Keep up the good work 🤩

#Integration by Parts Example 3

Evaluate the Integral.

$$\int x^2 \cos(x) , dx$$

We’ve worked through a few of these problems by now, so we got the rhythm down. Let’s start by picking $u$ and $dv$ using LIATE, and then solving for $du$ and $v$.

• $u = x^2$
• $du = 2x dx$
• $v = \sin x$
• $dv = \cos x$ Now we can use the Integration by Parts equation!

$$\int x^2 \cos(x) , dx = x^2 \cdot \sin(x) - \int 2x \cdot \sin(x) , dx$$

We have to apply integration by parts again to the remaining integral: $\int 2x \cdot \sin(x) , dx$

Choose:

• $u = 2x$
• $du=2dx$
• $v =-\cos x$
• $dv = \sin x$ And then we can do integration by parts for the second integral:

$$\int 2x \cdot \sin(x) , dx = -2x \cdot \cos(x) - \int (-2) \cdot \cos(x) , dx$$

Now, substitute this back into the original equation. We’re almost there! 🏎️

$$\int x^2 \cdot \cos(x) , dx = x^2 \cdot \sin(x) + 2x \cdot \cos(x) + 2 \cdot \int \cos(x) , dx$$

Finally, integrate the remaining term:

$$\int \cos(x) , dx = \sin(x)$$

Once we substitute this back into the equation, we get

$$\int x^2 \cdot \cos(x) , dx = x^2 \cdot \sin(x) + 2x \cdot \cos(x) + 2 \cdot \sin(x) + C$$

Therefore,

$$\int x^2 \cos(x) , dx = x^2 \sin(x) + 2x \cos(x) + 2 \sin(x) + C$$

You did amazing! 🙌

#Integration by Parts Challenge Question

Evaluate the following integral:

$$\int e^x \cos x$$

Try working through this question and see if you can get the answer below.

Hint: Try using integration of parts twice!

$$\int e^x \cos x = \frac{e^x\sin x + e^x\cos x}{2} + C$$

If you do integration by parts twice, keeping $e^x$ as the same variable, you will get an integral that matches the original. Use some algebraic manipulation to get both integrals on the same side, and then divide by two! If you can solve this one, you're an integration by parts expert!

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#🌟 Closing

Fantastic job! 🥳 Integration by Parts is a valuable tool in your calculus toolkit, allowing you to tackle a wide range of integrals. Remember to carefully choose $u$ and $dv$ to simplify the integration process. Keep practicing, and you'll become a master at integrating using Integration by Parts!

Encouraging GIF of Ice Cream

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