Integration by Parts

Calculus 2

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Overview

To reverse the chain rule, we have the method of u-substitution.

To reverse the product rule, we have the method of integration by parts.

We begin by deriving the integration by parts (IBP) formula from the product rule:

\begin{aligned} \frac{d}{dx} \left[f(x) \cdot g(x) \right] &= f(x)g'(x) + f'(x)g(x)\\ f(x) \cdot g(x) &= \int ( f(x)g'(x) + f'(x)g(x) ) dx\\ &= \int f(x)g'(x)dx + \int f'(x)g(x)dx\\ \int f(x)g'(x)dx &= f(x)g(x) - \int f'(x)g(x)dx\\ \end{aligned}

Letting $u=f(x) \Rightarrow du = f'(x)dx$ and $v=g(x) \Rightarrow dv = g'(x)dx$ , our equation becomes:

\begin{aligned} \int f(x)g'(x)dx &= f(x)g(x) - \int f'(x)g(x)dx\\ \int u dv &= uv - \int v du \end{aligned}

Therefore, we get the following formula: $\int u \ dv = uv - \int v \ du$

Keep in mind that this technique is supposed to make integrating easier, not harder. This means that the integral $\int v \ du$ should be easier than $\int u \ dv$ .

To make sure this occurs, we must make the right choice for $u$ and $dv$ . While this requires a bit of trial and error, we have general guidelines as to which functions work best for $u$ and $dv$ :

This means that $u$ and $dv$ should be functions of different type. That is: a log and a polynomial or a trig function and an exponential or . . . the list goes on.

Focus on choosing a $u$ that is the easier to differentiate and a $dv$ that is easier to integrate. Use LIPET and LIATE (or ILATE) as a guide.

Example 1

Compute $\int x e^x \ dx$ .

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$xe^x-e^x+C$

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To compute $\int x e^x \ dx$ , we begin by making our choices for $u$ and $dv$ (and then calculating $du$ and $v$ ):

\begin{aligned} u &= x & \ \ \ dv &= e^xdx \\ du &= dx & \ \ \ v&= e^x \end{aligned}

We then compute the integral by plugging everything into the integration by parts formula:

\begin{aligned} \int x e^x \ dx &= xe^x - \int e^x dx \\ &= xe^x - e^x +C \end{aligned}

Example 2

Compute $\int x \cos x \ dx$ .

View Answer

$x\sin x + \cos x +C$

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To compute $\int x \cos x \ dx$ , we begin by making our choices for $u$ and $dv$ (and then calculating $du$ and $v$ ):

\begin{aligned} u &= x & \ \ \ dv &= \cos x dx \\ du &= dx & \ \ \ v&= \sin x \end{aligned}

We then compute the integral by plugging everything into the integration by parts formula:

\begin{aligned} \int x \cos x \ dx &= x \sin x - \int \sin x dx \\ &= x \sin x + \cos x + C \end{aligned}

Multiple Rounds of IBP

Sometimes, we have to apply the integration by parts formula more than once to fully solve the problem. We will learn to determine whether multiple rounds are needed simply by inspecting the original integral. For example, polynomial terms that have a degree larger than 1 will require multiple derivatives (and therefore multiple $u \to du$ rounds) to reduce that polynomial and make the resulting integral ( $v \ du$ ) possible.

This repetition of IBP is analogous to doing multiple rounds of L'Hopital's Rule to solve an indeterminate limit. It is important to maintain the same type of function for $u$ and for $dv$ when doing multiple rounds.

Example 3

Compute $\int x^2 e^x \ dx$ using the integration by parts formula.

View Answer

$e^x(x^2-2x+2)+C$

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To compute $\int x^2 e^x \ dx$ , we begin by making our choices for $u$ and $dv$ (and then calculating $du$ and $v$ ):

\begin{aligned} u &= x^2 & \ \ \ dv &= e^x dx \\ du &= 2xdx & \ \ \ v&= e^x \end{aligned}

Plugging everything into the integration by parts formula, we get:

\begin{aligned} \int x^2 e^x \ dx &= x^2e^x - \int e^x \cdot 2xdx\\ &= x^2e^x - \int 2xe^xdx \end{aligned}

Notice how the new integral $\int 2xe^xdx$ requires us to perform another round of integration by parts (two functions of different type -- polynomial and exponential). We should have guessed this from the beginning, as $x^2$ requires two rounds of differentiation to become a constant and allow for the integration of itself multiplied by $e^x$ .

When performing the next round of IBP, we must choose new values for $u$ and $dv$ . At this step, it is very import to choose the same type of function for $u$ and for $dv$ . That is, $u$ should be another polynomial and $dv$ should be another exponential, as those were the functions we chose for $u$ and $dv$ during our first round of IBP. Therefore:

\begin{aligned} u &= 2x & \ \ \ dv &= e^x dx \\ du &= 2dx & \ \ \ v&= e^x \end{aligned}

Building on from our first round of integration, we get:

\begin{aligned} \int x^2 e^x \ dx &= x^2e^x - \int 2xe^xdx\\ &= x^2e^x - (2xe^x - \int e^x \cdot 2dx)\\ &= x^2e^x - 2xe^x + 2 \int e^x dx\\ &= x^2e^x -2xe^x + 2e^x +C\\ &= e^x(x^2-2x+2)+C \end{aligned}

To help speed things up and bookkeep a bit better, we can use the tabular method (sometimes referred to as a derivative-integral (DI) table).

With the tabular method, you want to follow these general steps:

Create a table with two columns: $u$ (or D) column and $dv$ (or I) column
Write down your choices of $u$ and $dv$ on the first row
Take the derivative of your $u$ until you get 0
Take the integral of your $dv$ until you get to the row you stopped at in Step 3
Put alternating $+/-$ signs, starting with +, on each row you have
Multiply all diagonal terms and add them up

The alternating sign in Step 5 takes into account the negative signs that get introduced in each round of IBP ( $uv \boxed{-} \int v du$ ).

Example 4

Compute $\int x^2 e^x \ dx$ again, this time using the tabular method.

View Answer

$(x^2-2x+2)e^x +C$

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Following the general steps above, we should generate the following table:

Solution coming soon.

Functions That Never Terminate

We might run into some integrals for which the functions in the integrand never terminate, no matter how many times we integrate or differentiate them (such as exponential and trigonometric functions). In these cases, we look for a constant multiple of the original integral.

Example 5

Compute $\int e^x \sin x \ dx$ .

View Answer

$\frac{1}{2} e^x (\sin x - \cos x) +C$

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Definite Integrals

So far, we've only seen IBP applied to indefinite integrals (ones with no bounds/limits of integration). Let's take a look at how to perform these integrals if we had bounds.

Keep in mind we have two options:

Compute the antiderivative without the bounds, then use the Fundamental Theorem of Calculus (FTC) to evaluate
Evaluate on the bounds as we go

Generally, it is easiest to do Option 1.

Example 6

Compute $\int_1^e x \ln x \ dx$ .

View Answer

$\frac{1}{4}(e^2+1)$

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To solve $\int_1^e x \ln x \ dx$ , we begin by making our choices for $u$ and $dv$ (and then calculating $du$ and $v$ ):

\begin{aligned} u &= \ln x & \ \ \ dv &= xdx \\ du &= \frac{1}{x}dx & \ \ \ v&= \frac{x^2}{2} \end{aligned}

We then solve the integral via integration by parts (using either Option 1 or Option 2):

Option 1: Integrate then evaluate

\begin{aligned} \int_1^e x \ln x \ dx &= \left[ \ln x \frac{x^2}{2} - \int \frac{x^2}{2} \frac{1}{x} dx \right]_1^e \\ &= \left[ \frac{x^2}{2} \ln x - \int \frac{x}{2} dx \right]_1^e \\ &= \left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right]_1^e \\ &= \left( \frac{e^2}{2} \ln(e) - \frac{e^2}{4} \right) - \left( \cancel{\frac{1}{2} \ln(1)} - \frac{1}{4} \right)\\ &= \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4}\\ &= \frac{2e^2 - e^2 + 1}{4}\\ &= \frac{e^2 + 1}{4}\\ &= \frac{1}{4}(e^2+1)\\ \end{aligned}

Option 2: Evaluate as you integrate

\begin{aligned} \int_1^e x \ln x \ dx &= \left[ \ln x \frac{x^2}{2} \right]_1^e - \int_1^e \frac{x^2}{2} \frac{1}{x} dx \\ &= \left( \frac{e^2}{2} \ln(e) - \cancel{\frac{1}{2} \ln(1)} \right) - \int_1^e \frac{x}{2} dx \\ &= \left( \frac{e^2}{2} \right) - \left[ \frac{x^2}{4} \right]_1^e\\ &= \frac{e^2}{2} - \left( \frac{e^2}{4} - \frac{1}{4} \right)\\ &= \frac{2e^2 - e^2 -1}{4}\\ &= \frac{e^2 + 1}{4}\\ &= \frac{1}{4}(e^2+1)\\ \end{aligned}

As you can see, both answers come out to be the same no matter which option you choose.

You can also use the tabular method for definite integrals that require multiple rounds of IBP. To do so, simply solve the integral as if it was an indefinite integral and then evaluate the final expression obtained from the table on the given bounds using the FTC.

Trickier Integrals

Sometimes, we might run into integrals that don't seem to fit the typical "two functions of different type" form for IBP we've seen so far. That doesn't necessarily mean that IBP can't help us to solve the integral.

Challenge 1

Compute $\int \ln x \ dx$ .

View Answer

$x \ln x - x + C$

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To solve $\int \ln x \ dx$ , we begin by making our choices for $u$ and $dv$ (and then calculating $du$ and $v$ ). We have to be creative here as there doesn't seem to be two functions in the problem.

\begin{aligned} u &= \ln x & \ \ \ dv &= dx \\ du &= \frac{1}{x}dx & \ \ \ v&= x \end{aligned}

We then compute the integral by plugging everything into the integration by parts formula:

\begin{aligned} \int \ln x dx &= \ln x \cdot x - \int x \cdot \frac{1}{x}dx\\ &= x \ln x - \int dx \\ &= x \ln x - x + C \end{aligned}

Other times, the integral fits the form but it's just a pain to do!

Challenge 2

Compute $\int x \arctan x \ dx$ .

View Answer

$\frac{x^2}{2} \arctan x - \frac{1}{2}(x - \arctan x)+C$

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To solve $\int x \arctan x \ dx$ , we begin by making our choices for $u$ and $dv$ (and then calculating $du$ and $v$ ):

\begin{aligned} u &= \arctan x & \ \ \ dv &= xdx \\ du &= \frac{1}{x^2+1}dx & \ \ \ v&= \frac{x^2}{2} \end{aligned}

We then compute the integral by plugging everything into the integration by parts formula:

\begin{aligned} \int x \arctan x \ dx &= \arctan x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x^2+1}dx\\ &= \frac{x^2}{2} \arctan x - \frac{1}{2} \int \frac{x^2}{x^2+1}dx \end{aligned}

The resulting integrand is an improper fraction, and so we will apply what we learned from Partial Fraction Decomposition: perform long division to make it proper.

$\frac{x^2}{x^2+1} = \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2}$

Therefore, we get:

\begin{aligned} \int x \arctan x \ dx &= \frac{x^2}{2} \arctan x - \frac{1}{2} \int \frac{x^2}{x^2+1}dx\\ &= \frac{x^2}{2} \arctan x - \frac{1}{2} \int \left( 1 - \frac{1}{1+x^2} \right) dx\\ &= \frac{x^2}{2} \arctan x - \frac{1}{2}(x - \arctan x)+C\\ \end{aligned}

Recap

Integration by parts is a great technique to use when the integrand consists of two different types of functions (i.e. logarithmic and exponential, trigonometric and polynomial, etc.).

Practice Problems

Problem 1

Compute $\int_{\pi/2}^{\pi} x \cos x \ dx$ .

View Answer

$-\frac{2+\pi}{2}$

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Solution coming soon.

More coming soon . . .

Partial Fraction Decomposition

Trigonometric Integrals