2.1: Integration by Parts

Learning Objectives

• Recognize when to use Integration by Parts.
• Use the Integration by Parts formula to solve integration problems.
• Use the Integration by Parts formula for definite integrals.

By now we have a fairly thorough procedure for how to evaluate many basic integrals. However, although we can integrate \( \int x \sin (x^2)\,dx\) by using the substitution, \(u=x^2\), something as simple looking as \( \int x\sin x\,\,dx\) defies us. Many students want to know whether there is a Product Rule for integration. There is not, but there is a technique based on the Product Rule for differentiation that allows us to exchange one integral for another. We call this technique Integration by Parts.

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The Integration by Parts Formula

If, \(h(x)=f(x)g(x)\), then by using the Product Rule, we obtain

\[h^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x). \label{eq1} \]

Although at first it may seem counterproductive, let’s now integrate both sides of Equation \ref{eq1}:

\[ \int h^{\prime}(x)\,\,dx= \int (g(x)f^{\prime}(x)+f(x)g^{\prime}(x))\,\,dx. \nonumber \]

This gives us

\[ h(x)=f(x)g(x)= \int g(x)f^{\prime}(x)\,dx+ \int f(x)g^{\prime}(x)\,\,dx. \nonumber \]

Now we solve for \( \int f(x)g^{\prime}(x)\,\,dx\):

\[ \int f(x)g^{\prime}(x)\,dx=f(x)g(x)− \int g(x)f^{\prime}(x)\,\,dx. \nonumber \]

By making the substitutions \(u=f(x)\) and \(v=g(x)\), which in turn make \(du=f^{\prime}(x)\,dx\) and \(dv=g^{\prime}(x)\,dx\), we have the more compact form

\[ \int u\,dv=uv− \int v\,du. \nonumber \]

Theorem: Integration by Parts

Let \(u=f(x)\) and \(v=g(x)\) be functions with continuous derivatives. Then, the Integration by Parts formula (also known as IbP) for the integral involving these two functions is:

\[ \int u\,dv=uv− \int v\,du. \label{IBP} \]

The advantage of using the Integration by Parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The following example illustrates its use.

Example \(\PageIndex{1}\): Using Integration by Parts

Use Integration by Parts with \(u=x\) and \(dv=\sin x\,\,dx\) to evaluate

\[ \int x\sin x\,\,dx. \nonumber \]

Solution

By choosing \(u=x\), we have \(du=1\,\,dx\). Since \(dv=\sin x\,\,dx\), we get

\[v= \int \sin x\,\,dx=−\cos x. \nonumber \]

It is handy to keep track of these values in a table, as follows:

\( u = x \)	\( dv = \sin x \, dx \)
\( du = 1\, dx \)	\( v = \int \sin x \, dx = - \cos x \)

Applying the Integration by Parts formula (Equation \ref{IBP}) results in

\[ \begin{array}{rclr}
\displaystyle \int x\sin x\,\,dx & = & (x)(−\cos x) − \displaystyle \int (−\cos x)(1\,\,dx) & \left( \text{Substitute} \right) \\
& = & −x\cos x+ \displaystyle \int \cos x\,\,dx & \left(\text{Simplify} \right) \\
\end{array} \nonumber \]

Then use

\[ \int \cos x\,\,dx =\sin x+C \nonumber \]

to obtain

\[ \int x\sin x\,\,dx =−x\cos x+\sin x+C. \nonumber \]

Analysis

At this point, there are probably a few items that need clarification. First of all, you may be curious about what would have happened if we had chosen \(u=\sin x\) and \(dv=x,\ dx\). If we had done so, then we would have \(du=\cos x \, dx\) and \(v=\frac{1}{2}x^2\). Thus, after applying Integration by Parts (Equation \ref{IBP}), we have

\[ \int ​x\sin x\,\,dx=\dfrac{1}{2}x^2\sin x− \int \dfrac{1}{2}x^2\cos x\,\,dx. \nonumber \]

Unfortunately, with the new integral, we are in no better position than before. It is important to keep in mind that when we apply Integration by Parts, we may need to try several choices for \(u\) and \(dv\) before finding a choice that works.

Second, you may wonder why, when we find \(v= \int ​\sin x\,\,dx=−\cos x\), we do not use \(v=−\cos x+K\). To see that it makes no difference, we can rework the problem using \(v=−\cos x+K\):

\[ \begin{array}{rcl}
\displaystyle \int x\sin x\,\,dx & = & (x)(−\cos x+K)− \int (−\cos x+K)(1\,\,dx) \\
& = & −x\cos x+Kx+ \displaystyle \int \cos x\,\,dx− \int K\,\,dx \\
& = & −x\cos x+Kx+\sin x−Kx+C \\
& = & −x\cos x+\sin x+C. \\
\end{array} \nonumber \]

As you can see, it makes no difference in the final solution.

Last, we can check to make sure that our antiderivative is correct by differentiating \(−x\cos x+\sin x+C:\)

\[ \begin{array}{rcl}
\dfrac{d}{\,dx}(−x\cos x+\sin x+C) & = & \cancel{(−1)\cos x} + (−x)(−\sin x) + \cancel{\cos x} \\
& = & x\sin x \\
\end{array} \nonumber \]

Therefore, the antiderivative checks out.

Exercise \(\PageIndex{1}\)

Evaluate \( \int ​xe^{2x}\,dx\) using the Integration by Parts formula (Equation \ref{IBP}) with \(u=x\) and \(dv=e^{2x}\,\,dx\).

Hint

Find \(du\) and \(v\), and use the previous example as a guide.

Answer

\[ \int ​xe^{2x}\,\,dx=\dfrac{1}{2}xe^{2x}−\dfrac{1}{4}e^{2x}+C \nonumber \]

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Choosing \( u \) and \( dv \)

The natural question to ask at this point is: How do we know what to choose for \(u\) and \(dv\)?

Most authors will immediately give you a trick to picking the "right" choice for \( u \); however, the trick they give is not flawless. In fact, it often fails. Therefore, I think it best to develop an approach to Integration by Parts from the other direction - by asking, "what is a good choice for \( dv \)?"

Our choice for \( dv \) will immediately inform us as to what \( u \) needs to be. Since we are going to have to integrate \( dv \), you definitely want to choose it to be something you actually can integrate. This is so important, it bears repeating:

Integration by Parts boils down to selecting a factor, preferably the largest factor, of the integrand that you can integrate either by direct integration or by the Substitution Method.

That's it! The rest of this subsection discusses why we make certain choices for \( dv \) and \( u \), but in the end it's all about asking yourself, "What's the largest factor of this integrand I can actually integrate?"

Now, think of all of the "traditional" functions you have encountered up to this point in mathematics, and ask yourself which ones you can easily integrate.

Basic Function Type	Easily Integrable?
Algebraic (e.g., \( x^3 \), \( \sqrt{x} \), and \( \frac{1}{x} \))	Yes
Exponential (e.g., \( e^x \) and \( 2^x \) )	Yes
Logarithmic (e.g., \( \ln(x) \) and \( \log_3(x) \))	No
Trigonometric (e.g., \( \sin(x) \) and \( \sec^2(x) \), but not \( \sec(x) \))	Yes (in some cases)
Inverse Trigonometric (e.g., \( \tan^{-1}(x) \) and \( \csc^{-1}(x) \))	No

Since we do not have integration formulas that allow us to integrate simple Logarithmic functions and Inverse trigonometric functions,¹ it makes sense that they should not be chosen for \(dv\). On the other hand, Exponential and Trigonometric functions are, in general, easy to integrate and make good choices for \(dv\). Finally, we can always integrate basic Algebraic functions, but their antiderivatives require slightly more work than the straightforward antiderivatives of the exponential and trigonometric functions. Therefore, if given the option we would rather integrate exponential or trigonometric functions over algbraic functions.

Given that discussion, we have the following tactic.

Tactic for Choosing \( dv \)

Given an integral where you wish to use Integration by Parts, the order of preference for choosing \( dv \) is as follows:

1. Exponential functions
2. Trigonometric functions
3. Algebraic functions
4. Inverse trigonometric functions
5. Logarithmic functions

As was mentioned previously, most authors approach this conversation by stating what your choice of \( u \) should be, rather than \( dv \). To accommodate students with instructors who take this same approach, we could reverse the direction of our tactic for choosing \( dv \) and, instead, create the commonly-taught tactic for choosing \( u \). If logarithms are the worst choice for \( dv \), then they will be our first choice for \( u \), and so on. This gives us the following tactic for choosing \( u \).

Tactic for Choosing \( u \)

Given an integral where you wish to use Integration by Parts, the order of preference for choosing \( u \) is as follows:

1. Logarithmic functions
2. Inverse trigonometric functions
3. Algebraic functions
4. Trigonometric functions
5. Exponential functions

This last tactic gives a common mnemonic, LIATE, to take some of the guesswork out of our choices for \( u \). The type of function in the integral that appears first in the list should be our first choice of \(u\).² When we have chosen \(u\), \(dv\) is selected to be the remaining part of the integrand.

Caution

I tend to stick with finding \( dv \) first instead of finding \( u \) so that you get a more natural, and understandable, approach to Integration by Parts. All of the examples will reflect this.

Thus, if an integral contains the product of a logarithmic function and an algebraic function, we would choose \( dv \) to be the algebraic function (because we don't know how to integrate a logarithm as of yet). Therefore, we would choose \( u \) to be the logarithmic function. Following the typical textbook tactic, we see our choice for \( u \) is correct because L comes before A in LIATE.

The integral in Example \(\PageIndex{1}\) has a trigonometric function, \(\sin x\), and an algebraic function, \(x\). Both of these are easily integrable, but we would rather integrate \( \sin x \) over \( x \) because the antiderivative of trigonometric functions are simple (as opposed to algebraic functions which require raising powers and dividing by new powers). Thus, we would choose \( dv = \sin x \, dx \) and \( u = x \). Again, this coincides with the decision we would have arrived at using the common textbook tactic because A comes before T in LIATE.

There is a second, more important reason as to why we are selecting to differentiate the algebraic function and integrate the trigonometric function in Example \( \PageIndex{1} \). When we tried to go the opposite direction (choosing to integrate the algebraic function and differentiate the trigonometric function), we arrived at an integral that got worse as we ended with an algebriac function of higher power and we still had a trigonometric function. A hidden goal when dealing with products of polynomials and other functions is to remove the polynomial. This can only be done if we take derivatives of the polynomial. While this might seem confusing right now, things will get clearer as you use Integration by Parts.

Finally, here is the rule of thumb I tell my students:

Rule of Thumb: Choosing \( dv \) and \( u \)

When using Integration by Parts:

• Choose \( dv \) to be the largest factor of the integrand that you can integrate, either directly or by using the Substitution Method - \( u \) will be the rest of the integrand. You will often need to rewrite the integrand to see this largest factor and, remember, \( \int f(x) \, dx = \int f(x) \cdot 1 \, dx \). This is especially useful if you cannot integrate any obvious factor within the integrand.
• If you can integrate all factors within the integrand, then switch gears and instead choose \( u \) to be the factor whose derivative (or eventual, higher-order derivative) changes forms or becomes a constant.
• If you have to use Integration by Parts more than once while evaluating an integral, be sure to stay with the same "function type" choice for all Integrations by Parts.

The best way to understand this Rule of Thumb is through examples.

Example \(\PageIndex{2}\): Using Integration by Parts

Evaluate \[ \int \dfrac{\ln x}{x^3}\,\,dx. \nonumber \]

Solution

It is helpful, at times, to begin by rewriting the integral:

\[ \int \dfrac{\ln x}{x^3}\,\,dx= \int ​x^{−3}\ln x\,\,dx. \nonumber \]

Consider the factors of the integrand - \( x^{-3} \) and \( \ln(x) \). Which one can we easily integrate? The obvious answer here is \( x^{-3} \). Hence, we choose \( dv = x^{-3} \, dx \) (don't forget the \( dx \)) and we let \( u \) be the rest of the integrand. That is, \( u = \ln(x) \).

For the LIATE Folks

We choose \(u=\ln x\) and \( dv = x^{-3} \, dx \), since L comes before A in LIATE.

Next, since \(u=\ln x,\) we have \(du=\frac{1}{x}\,dx\). Also, \(v= \int ​x^{−3}\,dx=−\frac{1}{2}x^{−2}\). Summarizing,

\( u = \ln(x) \)	\( dv = x^{-3} \, dx \)
\( du = \dfrac{1}{x} \, dx \)	\( v = -\dfrac{1}{2} x^{-2} \)

Substituting into the Integration by Parts formula (Equation \ref{IBP}) gives

\[ \begin{array}{rcl}
\displaystyle \int \dfrac{\ln x}{x^3}\,dx & = & \int ​x^{−3}\ln x\,dx \\
& = & \left(\ln x\right)\left(−\dfrac{1}{2}x^{−2}\right)− \displaystyle \int ​\left(−\dfrac{1}{2}x^{−2}\right)\left(\dfrac{1}{x}\,dx\right) \\
& = & −\dfrac{1}{2}x^{−2}\ln x+ \displaystyle \int \dfrac{1}{2}x^{−3}\,dx \\
& = & −\dfrac{1}{2}x^{−2}\ln x−\dfrac{1}{4}x^{−2}+C \\
& = & −\dfrac{1}{2x^2}\ln x−\dfrac{1}{4x^2}+C \\
\end{array} \nonumber \]

Exercise \(\PageIndex{2}\)

Evaluate \[ \int ​x \ln x \, dx. \nonumber \]

Hint

Use \(u=\ln x\) and \(dv=x\,\,dx\).

Answer

\[ \int ​x\ln x \,\,dx=\dfrac{1}{2}x^2\ln x−\dfrac{1}{4}x^2+C \nonumber \]

Subsection Footnotes

¹ The capital, bold letters are purposeful here, we will see why momentarily.

² Note that we put LI at the beginning of the mnemonic; however, we could just as easily have started with IL, since these two types of functions won’t appear together in an Integration by Parts problem and they are both terrible choices for \( dv \).

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More Complex Uses of Integration by Parts

In some cases, as in the next two examples, it may be necessary to apply Integration by Parts more than once.

Example \(\PageIndex{3}\): Applying Integration by Parts More Than Once

Evaluate \[ \int ​x^2e^{3x}\,dx. \nonumber \]

Solution

Since we can easily integrate both \( x^2 \) and \( e^{3x} \), we switch gears and ask ourselves which one has a derivative (or eventual, higher-order derivative) that will change forms. Thus, we choose \( u = x^2 \) because a couple of derivatives will result in the function completely devolving into a constant function. Therefore, \( dv = e^{3x} \, dx \).

\( u = x^2 \)	\( dv = e^{3x} \, dx \)
\( du = 2x \, dx \)	\( v = \dfrac{1}{3} e^{3x} \)

For the LIATE Folks

Choose \(u=x^2\) and \(dv=e^{3x}\,dx\) because A comes before E in LIATE.

Substituting into Equation \ref{IBP} produces

\[ \int x^2e^{3x}\,dx = \dfrac{1}{3}x^2e^{3x} − \int \dfrac{2}{3}xe^{3x}\,dx = \dfrac{1}{3}x^2e^{3x} − \dfrac{2}{3} \int xe^{3x}\,dx. \label{3A.2} \]

We still cannot integrate \( \displaystyle \int xe^{3x}\,dx\) directly, but the integral now has a lower power on \(x\). We can evaluate this new integral by using Integration by Parts again. Since we have already started the Integration by Parts process on this integral, we stick with the same "function type" choices for \( u \) and \( dv \). That is, we chose \( u \) to be the algebraic function on our first Integration by Parts, so our choice of \( u \) on this new Integration by Parts must also be an algebraic function.

For the LIATE Folks

Choose \(u=x\) and \(dv=e^{3x}\,dx\) because A comes before E in LIATE.

\( u = x \)	\( dv = e^{3x} \, dx \)
\( du = dx \)	\( v = \dfrac{1}{3} e^{3x} \)

Substituting back into Equation \ref{3A.2} yields

\[ \int ​x^2e^{3x}\,dx = \dfrac{1}{3}x^2e^{3x}− \dfrac{2}{3} \left(\dfrac{1}{3}xe^{3x}− \int \dfrac{1}{3}e^{3x}\,dx\right). \nonumber \]

After evaluating the last integral and simplifying, we obtain

\[ \int x^2e^{3x}\,dx=\dfrac{1}{3}x^2e^{3x}−\dfrac{2}{9}xe^{3x}+\dfrac{2}{27}e^{3x}+C. \nonumber \]

Example \( \PageIndex{3} \) showcases an important difference between LIATE and having a more natural understanding of Integration by Parts using the Rule of Thumb. When using LIATE, we chose \( u = x^2 \) by rote memorization without understanding the consequences of the choice. The Rule of Thumb, on the other hand, delivered an extra bit of "hidden" knowledge - since we chose \( u = x^2 \) because it changes form after two derivatives, we could have immediately recognized that this integral would require two integrations (in this case, two applications of Integration by Parts).

Example \(\PageIndex{4}\): Applying Integration by Parts When LIATE Does not Quite Work

Evaluate

\[ \int ​t^3e^{t^2}\, dt. \nonumber \]

Solution

This is a great example of having many approaches to a problem that are all correct; however, it is best to choose the approach that is not only the most elegant, but the one that will repay us with useable knowledge for the future.

Approach #1 (Ineffective): Using LIATE

For the LIATE-lovers out there, this one's for you. If we use a strict interpretation of the mnemonic LIATE to make our choice of \(u\), we end up with \(u=t^3\) and \(dv=e^{t^2} \, dt\).

\( u = t^3 \)	\( dv = e^{t^2} \, dt \)
\( du = 3t^2 \, dt \)	\( v = \ldots \left( \text{we're stuck} \right) \)

Unfortunately, we cannot continue with this approach because we cannot integrate \( e^{t^2} \).

Approach #2: Starting with a Substitution

Using the Substitution Method, we could let \( w = t^2 \) so that \( dw = 2 t \, dt \). This means \( \frac{1}{2} dw = t \, dt \). Hence,

\[ \begin{array}{rclr}
\displaystyle \int t^3 e^{t^2} \, dt & = & \displaystyle \int t^2 e^{t^2} t \, dt & \\
& = & \dfrac{1}{2} \displaystyle \int w e^w \, dw & \left( \text{Substituting }w = t^2 \implies \dfrac{1}{2} dw = t \, dt \right) \\
\end{array} \nonumber \]

We can integrate both \( w \) and \( e^w \), so we focus on choosing \( u \) to be the factor whose derivative (or eventual derivative) changes form. This is \( w \). Thus, we choose \( u = w \) and \( dv = e^w \, dw \).

\( u = w \)	\( dv = e^w \, dw \)
\( du = dw \)	\( v = e^w \)

Using Integration by Parts, we get

\[ \begin{array}{rclr}
\displaystyle \int t^3 e^{t^2} \, dt & = & \displaystyle \int t^2 e^{t^2} t \, dt & \\
& = & \dfrac{1}{2} \displaystyle \int w e^w \, dw & \left( \text{Substituting }w = t^2 \implies \dfrac{1}{2} dw = t \, dt \right) \\
& = & \dfrac{1}{2} \left( w e^w - \displaystyle \int e^w \, dw \right) & \left( \text{IBP} \right) \\
& = & \dfrac{1}{2} \left( w e^w - e^w + C_1\right) & \\
& = & \dfrac{1}{2} \left( t^2 e^{t^2} - e^{t^2} + C_1 \right) & \left( \text{Resubstituting }w = t^2 \right) \\
& = & \dfrac{1}{2} t^2 e^{t^2} - \dfrac{1}{2} e^{t^2} + C & \left( \dfrac{1}{2} C_1 \text{ is still a constant, so we just call it }C \right) \\
\end{array} \nonumber \]

Approach #3 (Most Efficient): Starting with IBP by Choosing \( dv \) to be the Largest Integrable Factor

If we rewrite the given integral as

\[ \int t^2 \left( t e^{t^2} \right) \, dt, \nonumber \]

we can see that both \( t^2 \) and \( t e^{t^2} \) are integrable. Therefore, we choose \( dv \) to be the "larger" (read as, "more complex") of these two factors. Hence, we choose \( dv = t e^{t^2} \, dt \) and \( u = t^2 \).³

\( u = t^2 \)	\( dv = t e^{t^2} \, dt \)
\( du = 2 t \, dt \)	\( v = \dfrac{1}{2} e^{t^2} \) (Substitution Method)

Therefore, we get

\[ \begin{array}{rclr}
\displaystyle \int t^3e^{t^2} \, dt & = & \displaystyle \int t^2 \left(t e^{t^2}\right) \, dt & \\
& = & \dfrac{1}{2} t^2 e^{t^2} - \displaystyle \int t e^{t^2} \, dt. & \left( \text{IBP} \right)\\
& = & \dfrac{1}{2} t^2 e^{t^2} - \dfrac{1}{2} e^{t^2} + C & \left( \text{this is the result of integrating }dv \right) \\
\end{array} \nonumber \]

There are a couple of things I want to mention about Example \( \PageIndex{4} \). First, notice that LIATE failed us here, but a natural understanding of our goal - to find the largest factor of the integrand that we can integrate - still led us to a solution. This is why I personally do not teach LIATE in my classes (again, I only include it here because you are going to hear of it from students in other classes and those students might have the illusion that it always works).

The second thing to mention is that Approach #3 in this example demonstrates the level of comfort with integration for which you want to strive. There is nothing wrong with Approach #2, but you definitely want to build the skill set to be able to perform complex integrations like this. The only way you are going to build that skill, however, is through lots of practice!

Caution

Integration by Parts (and most other integration techniques) is almost a completely heuristic method. That is, you will only learn how to properly identify when to use a method, and what pitfalls to avoid, by practicing... a lot.

Example \(\PageIndex{5}\): Applying Integration by Parts More Than Once

Evaluate

\[ \int ​\sin (\ln x)\,dx. \nonumber \]

Solution

Again, LIATE completely fails us here, but the Rule of Thumb is still helpful. We begin by rewriting the integral as

\[ \int ​\sin (\ln x) \cdot 1 \,dx. \nonumber \]

We can integrate \( 1 \), but not \( \sin(\ln(x)) \), so we let \( dv = 1 \, dx \).

\( u = \sin(\ln(x)) \)	\( dv = 1 \, dx \)
\( du = \dfrac{\cos(\ln(x))}{x} \, dx \)	\( v = x \)

Therefore, we have

\[ \int ​\sin \left(\ln (x)\right) \,dx = x \sin (\ln (x)) − \int \cos (\ln (x))\,dx. \nonumber \]

Unfortunately, this process leaves us with a new integral that is very similar to the original. However, let’s see what happens when we apply Integration by Parts again.

Since we chose \( u \) to be the composition of the trigonometric function and the logarithmic function on our first pass of the Integration by Parts method, we choose the same style of function for \( u \) on our second pass.

\( u = \cos(\ln(x)) \)	\( dv = 1 \, dx \)
\( du = -\dfrac{\sin(\ln(x))}{x} \, dx \)	\( v = x \)

Substituting, we have

\[ \int ​\sin (\ln (x))\,dx = x \sin (\ln x)−\left(x \cos (\ln (x)) - \int -​\sin (\ln (x))\,dx\right). \nonumber \]

After simplifying, we obtain

\[ \int ​\sin (\ln (x))\,dx=x\sin (\ln (x))−x \cos (\ln (x))− \int ​\sin (\ln (x))\,dx. \nonumber \]

The last integral is now the same as the original. It may seem that we have simply gone in a circle, but now we can actually evaluate the integral. To see how to do this more clearly, substitute \(I= \int ​\sin (\ln (x))\,dx.\) Thus, the equation becomes

\[I=x \sin (\ln (x))−x \cos (\ln (x))−I. \nonumber \]

First, add \(I\) to both sides of the equation to obtain

\[2I=x \sin (\ln (x))−x \cos (\ln (x)). \nonumber \]

Next, divide by 2:

\[I=\dfrac{1}{2}x \sin (\ln (x))−\dfrac{1}{2}x \cos (\ln (x)). \nonumber \]

Substituting \(I= \int ​\sin (\ln (x))\,dx\) again, we have

\[ \int \sin (\ln (x)) \,dx=\dfrac{1}{2}x \sin (\ln (x))−\dfrac{1}{2}x \cos (\ln (x)). \nonumber \]

From this we see that \(\frac{1}{2}x \sin (\ln (x))−\frac{1}{2}x \cos (\ln (x))\) is an antiderivative of \(\sin (\ln (x))\). For the most general antiderivative, add \(C\):

\[ \int \sin (\ln (x)) \,dx=\dfrac{1}{2}x \sin (\ln (x))−\dfrac{1}{2}x \cos (\ln (x))+C. \nonumber \]

Analysis

If this method feels a little strange at first, we can check the answer by differentiation:

\[\begin{array}{rcl}
\dfrac{d}{\,dx}\left(\dfrac{1}{2}x \sin (\ln x)−\dfrac{1}{2}x\cos (\ln x)\right) & = & \dfrac{1}{2}(\sin (\ln x))+\cos (\ln x) \cdot \dfrac{1}{x} \cdot \dfrac{1}{2}x−\left(\dfrac{1}{2}\cos (\ln x)−\sin (\ln x) \cdot \dfrac{1}{x} \cdot \dfrac{1}{2}x\right) \\
& = & \sin (\ln x). \\
\end{array} \nonumber \]

Exercise \(\PageIndex{5}\)

Evaluate

\[ \int x^2\sin x\,dx. \nonumber \]

Hint

This is similar to Examples \(\PageIndex{3A}\) - \(\PageIndex{3C}\).

Answer

\[ \int x^2\sin x\,dx=−x^2\cos x+2x\sin x+2\cos x+C \nonumber \]

Subsection Footnotes

³ Another way we could have made our choice for \( u \) and \( dv \) here is to note that, since both \( t^2 \) and \( t e^{t^2} \) are integrable, we switch gears and choose \( u \) to be the factor that will eventually change forms given enough derivatives. This will be \( u = t^2 \) because after two derivatives, it will become a constant function.

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Integration by Parts for Definite Integrals

Now that we have used Integration by Parts successfully to evaluate indefinite integrals, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.

Theorem: Integration by Parts for Definite Integrals

Let \(u=f(x)\) and \(v=g(x)\) be functions with continuous derivatives on [\(a,b\)]. Then

\[ \int ^b_a u\,dv=uv\Big|^b_a− \int ^b_a v\, du \nonumber \]

Example \(\PageIndex{6}\): Finding the Area of a Region

Find the area of the region bounded above by the graph of \(y=\tan^{−1}x\) and below by the \(x\)-axis over the interval [\(0,1\)].

Solution

This region is shown in Figure \(\PageIndex{1}\). To find the area, we must evaluate

\[ \int ^1_0 \tan^{−1}x\, \,dx. \nonumber \]

Figure \(\PageIndex{1}\): To find the area of the shaded region, we have to use Integration by Parts.

Since we can integrate \( 1 \), this will be our choice for \( dv \).

\( u = \tan^{-1}(x) \)	\( dv = 1 \, dx \)
\( du = \dfrac{1}{1 + x^2} \, dx \)	\( v = x \)

After applying the Integration by Parts formula (Equation \ref{IBP}) we obtain

\[ \text{Area} = x \tan^{−1} x \bigg|^1_0 − \int ^1_0 \dfrac{x}{x^2+1} \,dx. \nonumber \]

Use \(u\)-substitution to obtain

\[ \int ^1_0\dfrac{x}{x^2+1}\,dx = \dfrac{1}{2}\ln \left(x^2+1\right) \bigg|^1_0. \nonumber \]

Thus,

\[\text{Area} = x \tan^{−1}x \bigg|^1_0− \dfrac{1}{2}\ln \left( x^2+1 \right) \bigg|^1_0 = \left(\dfrac{ \pi }{4}−\dfrac{1}{2}\ln 2\right) \,\text{units}^2. \nonumber \]

At this point it might not be a bad idea to do a "reality check" on the reasonableness of our solution. Since \(\frac{ \pi }{4}−\frac{1}{2}\ln 2 \approx 0.4388\,\text{units}^2,\) and from Figure \(\PageIndex{1}\) we expect our area to be slightly less than \(0.5\,\text{units}^2,\) this solution appears to be reasonable.

Example \(\PageIndex{7}\): Finding a Volume of Revolution

Find the volume of the solid obtained by revolving the region bounded by the graph of \(f(x)=e^{−x},\) the \(x\)-axis, the \(y\)-axis, and the line \(x=1\) about the \(y\)-axis.

Solution

The best option to solving this problem is to use the Method of Cylindrical Shells. Begin by sketching the region to be revolved, along with a typical rectangle (Figure \(\PageIndex{2}\)).

Figure \(\PageIndex{2}\): We can use the shell method to find a volume of revolution.

To find the volume using shells, we must evaluate

\[2 \pi \int ^1_0xe^{−x}\,dx. \label{4B.1} \]

Both \( x \) and \( e^{-x} \) are easily integrable, so we instead resort to choosing for \( u \) the function whose form eventually changes through differentiation - this is \( x \). Thus, we let \(u=x\) and \(dv=e^{−x} \, dx\).

\( u = x \)	\( dv = e^{-x} \, dx \)
\( du = dx \)	\( v = -e^{-x} \)

Using the Shell Method formula and substituting these values in, we obtain

\[ \begin{array}{rcl}
\text{Volume} & = & 2 \pi \displaystyle \int ^1_0 x e^{−x} \, dx \\
& = & 2 \pi \left(−xe^{−x}\bigg|^1_0 + \displaystyle \int ^1_0 e^{−x} \, dx \right) \\
& = & -2 \pi \left(xe^{−x}\bigg|^1_0 - \displaystyle \int ^1_0e^{−x}\,dx \right) \\
& = & -2 \pi \left( e^{-1} - 0 + e^{-x}\bigg|^1_0\right) \\
& = & -2 \pi \left( e^{-1} + e^{-1} - 1 \right) \\
& = & -2 \pi \left( 2e^{-1} - 1 \right) \\
& = & 2 \pi - \dfrac{4 \pi}{e}\,\text{units}^3. \\
\end{array} \nonumber \]

Analysis

Again, it is a good idea to check the reasonableness of our solution. We observe that the solid has a volume slightly less than that of a cylinder of radius \(1\) and height of \(1/e\) added to the volume of a cone of base radius \(1\) and height of \(1−\dfrac{1}{e}.\) Consequently, the solid should have a volume a bit less than

\[ \pi (1)^2\dfrac{1}{e}+\left(\dfrac{ \pi }{3}\right)(1)^2\left(1−\dfrac{1}{e}\right)=\dfrac{2 \pi }{3e}+\dfrac{ \pi }{3} \approx 1.8177\,\text{units}^3. \nonumber \]

Since \(2 \pi −\dfrac{4 \pi }{e} \approx 1.6603,\) we see that our calculated volume is reasonable.

Exercise \(\PageIndex{7}\)

Evaluate

\[ \int ^{ \pi /2}_0x\cos x\,dx. \nonumber \]

Hint

Use Equation \ref{IBP} with \(u=x\) and \(dv=\cos x\,dx.\)

Answer

\[ \int ^{ \pi /2}_0x\cos x\,dx = \dfrac{ \pi }{2}−1 \nonumber \]

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Key Concepts

• The Integration by Parts formula (Equation \ref{IBP}) allows the exchange of one integral for another, possibly easier, integral.
• Integration by Parts applies to both definite and indefinite integrals.

Key Equations

• Integration by Parts formula

\( \int u\,dv=uv− \int v\,du\)

• Integration by Parts for definite integrals

\( \int ^b_au\,dv=uv\Big|^b_a− \int ^b_av\,du\)

Glossary

Integration by Parts

a technique of integration that allows the exchange of one integral for another using the formula \( \int ​u\,dv=uv− \int ​v\,du\)