Integration-by-Parts Formula

Learning Outcomes

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.

If, $h\left(x\right)=f\left(x\right)g\left(x\right)$ , then by using the product rule, we obtain ${h}^{\prime }\left(x\right)={f}^{\prime }\left(x\right)g\left(x\right)+{g}^{\prime }\left(x\right)f\left(x\right)$ . Although at first it may seem counterproductive, let’s now integrate both sides of this equation: $\displaystyle\int {h}^{\prime }\left(x\right)dx=\displaystyle\int \left(g\left(x\right){f}^{\prime }\left(x\right)+f\left(x\right){g}^{\prime }\left(x\right)\right)dx$ .

This gives us

h (x) = f (x) g (x) = \int g (x) f^{'} (x) d x + \int f (x) g^{'} (x) d x

$h\left(x\right)=f\left(x\right)g\left(x\right)=\displaystyle\int g\left(x\right){f}^{\prime }\left(x\right)dx+\displaystyle\int f\left(x\right){g}^{\prime }\left(x\right)dx$ .

Now we solve for $\displaystyle\int f\left(x\right){g}^{\prime }\left(x\right)dx:$

\int f (x) g^{'} (x) d x = f (x) g (x) - \int g (x) f^{'} (x) d x

$\displaystyle\int f\left(x\right){g}^{\prime }\left(x\right)dx=f\left(x\right)g\left(x\right)-\displaystyle\int g\left(x\right){f}^{\prime }\left(x\right)dx$ .

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

\int u d v = u v - \int v d u

$\displaystyle\int udv=uv-\displaystyle\int vdu$ .

Integration by Parts

Let $u=f\left(x\right)$ and $v=g\left(x\right)$ be functions with continuous derivatives. Then, the integration-by-parts formula for the integral involving these two functions is:

\int u d v = u v - \int v d u

$\displaystyle\int udv=uv-\displaystyle\int vdu$ .

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: Using Integration by Parts

Use integration by parts with $u=x$ and $dv=\sin{x}dx$ to evaluate $\displaystyle\int x\sin{x}dx$ .

Show Solution

Solution

By choosing $u=x$ , we have $du=1dx$ . Since $dv=\sin{x}dx$ , we get $v=\displaystyle\int \sin{x}dx=\text{-}\cos{x}$ . It is handy to keep track of these values as follows:

\begin{matrix} u & = & x & d v & = & sin x d x d u & = & 1 d x & v & = & \int sin x d x = - cos x . \end{matrix}

$\begin{array}{ccccccc}\hfill u& =\hfill & x\hfill & & \hfill dv& =\hfill & \sin{x}dx\hfill \\ \hfill du& =\hfill & 1dx\hfill & & \hfill v& =\hfill & \displaystyle\int \sin{x}dx=\text{-}\cos{x}. \end{array}$

Applying the integration-by-parts formula results in

\begin{matrix} \int x sin x d x & = (x) (- cos x) - \int (- cos x) (1 d x) & Substitute. = - x cos x + \int cos x d x & Simplify. = - x cos x + sin x + C . & Use \int cos x d x = sin x + C . \end{matrix}

$\begin{array}{cccc}\hfill {\displaystyle\int x\sin{x}dx}& =\left(x\right)\left(\text{-}\cos{x}\right)-{\displaystyle\int \left(\text{-}\cos{x}\right)\left(1dx\right)}\hfill & & \text{Substitute.}\hfill \\ & =\text{-}x\cos{x}+{\displaystyle\int \cos{x}dx}\hfill & & \text{Simplify.}\hfill \\ & =\text{-}x\cos{x}+\sin{x}+C.\hfill & & \text{Use}{\displaystyle\int \cos{x}dx=\sin{x}+C.}\end{array}$

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 $dv=x$ . If we had done so, then we would have $du=\cos{x}$ and $v=\frac{1}{2}{x}^{2}$ . Thus, after applying integration by parts, we have ${\displaystyle\int }^{\text{ }}x\sin{x}dx=\frac{1}{2}{x}^{2}\sin{x}-{\displaystyle\int }^{\text{ }}\frac{1}{2}{x}^{2}\cos{x}dx$ . 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 $v={\displaystyle\int }^{\text{ }}\sin{x}dx=\text{-}\cos{x}$ , we do not use $v=\text{-}\cos{x}+K$ . To see that it makes no difference, we can rework the problem using $v=\text{-}\cos{x}+K\text{:}$

\begin{matrix} \int x sin x d x & = (x) (- cos x + K) - \int (- cos x + K) (1 d x) = - x cos x + K x + \int cos x d x - \int K d x = - x cos x + K x + sin x - K x + C = - x cos x + sin x + C . \end{matrix}

$\begin{array}{cc}{\displaystyle\int}x\sin{x}dx\hfill & =\left(x\right)\left(\text{-}\cos{x}+K\right)-{\displaystyle\int}\left(\text{-}\cos{x}+K\right)\left(1dx\right)\hfill \\ \hfill & =\text{-}x\cos{x}+Kx+{\displaystyle\int}\cos{x}dx-{\displaystyle\int}Kdx\hfill \\ \hfill & =\text{-}x\cos{x}+Kx+\sin{x}-Kx+C\hfill \\ \hfill & =\text{-}x\cos{x}+\sin{x}+C.\hfill \end{array}$

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 $\text{-}x\cos{x}+\sin{x}+C\text{:}$

\begin{matrix} \frac{d}{d x} (- x cos x + sin x + C) & = (- 1) cos x + (- x) (- sin x) + cos x = x sin x . \end{matrix}

$\begin{array}{cc}\frac{d}{dx}\left(\text{-}x\cos{x}+\sin{x}+C\right)\hfill & =\left(-1\right)\cos{x}+\left(\text{-}x\right)\left(\text{-}\sin{x}\right)+\cos{x}\hfill \\ \hfill & =x\sin{x}.\hfill \end{array}$

Therefore, the antiderivative checks out.

Interactive

Watch this video for an introduction to integration by parts and visit this website for examples of integration by parts.

Try It

Evaluate ${\displaystyle\int }^{\text{ }}x{e}^{2x}dx$ using the integration-by-parts formula with $u=x$ and $dv={e}^{2x}dx$ .

Hint

Find $du$ and $v$ , and use the previous example as a guide.

Show Solution

${\displaystyle\int }^{\text{ }}x{e}^{2x}dx=\frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}+C$

Watch the following video to see the worked solution to the above Try It

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “3.1 Integration by Parts” here (opens in new window).

The natural question to ask at this point is: How do we know how to choose $u$ and $dv?$ Sometimes it is a matter of trial and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym stands for Logarithmic Functions, Inverse Trigonometric Functions, Algebraic Functions, Trigonometric Functions, and Exponential Functions. This mnemonic serves as an aid in determining an appropriate choice for $u$ .

The type of function in the integral that appears first in the list should be our first choice of $u$ . For example, if an integral contains a and an , we should choose $u$ to be the logarithmic function, because L comes before A in LIATE. The integral in the previous example has a trigonometric function $\text{(}\sin{x}\text{)}$ and an algebraic function $\left(x\right)$ . Because A comes before T in LIATE, we chose $u$ to be the algebraic function. When we have chosen $u$ , $dv$ is selected to be the remaining part of the function to be integrated, together with $dx$ .

Why does this mnemonic work? Remember that whatever we pick to be $dv$ must be something we can integrate. Since we do not have integration formulas that allow us to integrate simple logarithmic functions and inverse trigonometric functions, it makes sense that they should not be chosen as values for $dv$ . Consequently, they should be at the head of the list as choices for $u$ . Thus, we put LI at the beginning of the mnemonic. (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.) The exponential and trigonometric functions are at the end of our list because they are fairly easy to integrate and make good choices for $dv$ . Thus, we have TE at the end of our mnemonic. (We could just as easily have used ET at the end, since when these types of functions appear together it usually doesn’t really matter which one is $u$ and which one is $dv$ .) Algebraic functions are generally easy both to integrate and to differentiate, and they come in the middle of the mnemonic.

To use the by-parts technique successfully, it is helpful to first review the derivative rules of several familiar transcendental functions.

Recall: Derivative rules for transcendental functions

$\frac{d}{dx} (\sin x) = \cos x$
$\frac{d}{dx} (\cos x) = -\sin x$
$\frac{d}{dx} (\ln x) = \frac{1}{x}$
$\frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx} (\arctan x) = \frac{1}{1+x^2}$

example: Using integration by parts

Evaluate $\displaystyle\int \frac{\text{ln}x}{{x}^{3}}dx$ .

Show Solution

Begin by rewriting the integral:

\int \frac{ln x}{x^{3}} d x = \int x^{- 3} ln x d x

$\displaystyle\int \frac{\text{ln}x}{{x}^{3}}dx={\displaystyle\int }^{\text{ }}{x}^{-3}\text{ln}xdx$ .

Since this integral contains the algebraic function ${x}^{-3}$ and the logarithmic function $\text{ln}x$ , choose $u=\text{ln}x$ , since L comes before A in LIATE. After we have chosen $u=\text{ln}x$ , we must choose $dv={x}^{-3}dx$ .

Next, since $u=\text{ln}x$ , we have $du=\frac{1}{x}dx$ . Also, $v={\displaystyle\int }^{\text{ }}{x}^{-3}dx=-\frac{1}{2}{x}^{-2}$ . Summarizing,

\begin{matrix} u & = & ln x & d v & = & x^{- 3} d x d u & = & \frac{1}{x} d x & v & = & \int x^{- 3} d x = - \frac{1}{2} x^{- 2} . \end{matrix}

$\begin{array}{ccccccc}\hfill u& =\hfill & \text{ln}x\hfill & & \hfill dv& =\hfill & {x}^{-3}dx\hfill \\ \hfill du& =\hfill & \frac{1}{x}dx\hfill & & \hfill v& =\hfill & {\displaystyle\int }^{\text{ }}{x}^{-3}dx=-\frac{1}{2}{x}^{-2}.\end{array}$

Substituting into the integration-by-parts formula gives

\begin{matrix} \int \frac{ln x}{x^{3}} d x & = \int x^{- 3} ln x d x = (ln x) (- \frac{1}{2} x^{- 2}) - \int (- \frac{1}{2} x^{- 2}) (\frac{1}{x} d x) = - \frac{1}{2} x^{- 2} ln x + \int \frac{1}{2} x^{- 3} d x & Simplify . = - \frac{1}{2} x^{- 2} ln x - \frac{1}{4} x^{- 2} + C & Integrate . = - \frac{1}{2 x^{2}} ln x - \frac{1}{4 x^{2}} + C . & Rewrite with positive integers. \end{matrix}

$\begin{array}{ccccc}\hfill \displaystyle\int \frac{\text{ln}x}{{x}^{3}}dx& ={\displaystyle\int }^{\text{ }}{x}^{-3}\text{ln}xdx=\left(\text{ln}x\right)\left(\text{-}\frac{1}{2}{x}^{-2}\right)-{\displaystyle\int }^{\text{ }}\left(\text{-}\frac{1}{2}{x}^{-2}\right)\left(\frac{1}{x}dx\right)\hfill & & & \\ & =-\frac{1}{2}{x}^{-2}\text{ln}x+{\displaystyle\int }^{\text{ }}\frac{1}{2}{x}^{-3}dx\hfill & & & \text{Simplify}.\hfill \\ & =-\frac{1}{2}{x}^{-2}\text{ln}x-\frac{1}{4}{x}^{-2}+C\hfill & & & \text{Integrate}.\hfill \\ & =-\frac{1}{2{x}^{2}}\text{ln}x-\frac{1}{4{x}^{2}}+C.\hfill & & & \text{Rewrite with positive integers.}\end{array}$

try it

Evaluate ${\displaystyle\int }^{\text{ }}x\text{ln}xdx$ .

Hint

Use $u=\text{ln}x$ and $dv=xdx$ .

Show Solution

$\frac{1}{2}{x}^{2}\text{ln}x-\frac{1}{4}{x}^{2}+C$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for this segmented clip of “3.1 Integration by Parts” here (opens in new window).

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

example: Applying integration by parts more than once

Evaluate ${\displaystyle\int }^{\text{ }}{x}^{2}{e}^{3x}dx$ .

Show Solution

Using LIATE, choose $u={x}^{2}$ and $dv={e}^{3x}dx$ . Thus, $du=2xdx$ and $v=\displaystyle\int {e}^{3x}dx=\left(\frac{1}{3}\right){e}^{3x}$ . Therefore,

\begin{matrix} u & = & x^{2} & d v & = & e^{3 x} d x d u & = & 2 x d x & v & = & \int e^{3 x} d x = \frac{1}{3} e^{3 x} . \end{matrix}

$\begin{array}{ccccccc}\hfill u& =\hfill & {x}^{2}\hfill & & \hfill dv& =\hfill & {e}^{3x}dx\hfill \\ \hfill du& =\hfill & 2xdx\hfill & & \hfill v& =\hfill & \displaystyle\int {e}^{3x}dx=\frac{1}{3}{e}^{3x}.\end{array}$

Substituting into the integration-by-parts formula produces

\int x^{2} e^{3 x} d x = \frac{1}{3} x^{2} e^{3 x} - \int \frac{2}{3} x e^{3 x} d x

$\displaystyle\int {x}^{2}{e}^{3x}dx=\frac{1}{3}{x}^{2}{e}^{3x}-\displaystyle\int \frac{2}{3}x{e}^{3x}dx$ .

We still cannot integrate $\displaystyle\int \frac{2}{3}x{e}^{3x}dx$ directly, but the integral now has a lower power on $x$ . We can evaluate this new integral by using integration by parts again. To do this, choose $u=x$ and $dv=\frac{2}{3}{e}^{3x}dx$ . Thus, $du=dx$ and $v=\displaystyle\int \left(\frac{2}{3}\right){e}^{3x}dx=\left(\frac{2}{9}\right){e}^{3x}$ . Now we have

\begin{matrix} u & = & x & d v & = & \frac{2}{3} e^{3 x} d x d u & = & d x & v & = & \int \frac{2}{3} e^{3 x} d x = \frac{2}{9} e^{3 x} . \end{matrix}

$\begin{array}{ccccccc}\hfill u& =\hfill & x\hfill & & \hfill dv& =\hfill & \frac{2}{3}{e}^{3x}dx\hfill \\ \hfill du& =\hfill & dx\hfill & & \hfill v& =\hfill & \displaystyle\int \frac{2}{3}{e}^{3x}dx=\frac{2}{9}{e}^{3x}.\end{array}$

Substituting back into the previous equation yields

\int x^{2} e^{3 x} d x = \frac{1}{3} x^{2} e^{3 x} - (\frac{2}{9} x e^{3 x} - \int \frac{2}{9} e^{3 x} d x)

${\displaystyle\int }^{\text{ }}{x}^{2}{e}^{3x}dx=\frac{1}{3}{x}^{2}{e}^{3x}-\left(\frac{2}{9}x{e}^{3x}-{\displaystyle\int }^{\text{ }}\frac{2}{9}{e}^{3x}dx\right)$ .

After evaluating the last integral and simplifying, we obtain

\int x^{2} e^{3 x} d x = \frac{1}{3} x^{2} e^{3 x} - \frac{2}{9} x e^{3 x} + \frac{2}{27} e^{3 x} + C

$\displaystyle\int {x}^{2}{e}^{3x}dx=\frac{1}{3}{x}^{2}{e}^{3x}-\frac{2}{9}x{e}^{3x}+\frac{2}{27}{e}^{3x}+C$ .

Example: Applying Integration by Parts When LIATE Doesn’t Quite Work

Evaluate ${\displaystyle\int }^{\text{ }}{t}^{3}{e}^{{t}^{2}}dt$ .

Show Solution

If we use a strict interpretation of the mnemonic LIATE to make our choice of $u$ , we end up with $u={t}^{3}$ and $dv={e}^{{t}^{2}}dt$ . Unfortunately, this choice won’t work because we are unable to evaluate ${\displaystyle\int }^{\text{ }}{e}^{{t}^{2}}dt$ . However, since we can evaluate ${\displaystyle\int }^{\text{ }}t{e}^{{t}^{2}}dx$ , we can try choosing $u={t}^{2}$ and $dv=t{e}^{{t}^{2}}dt$ . With these choices we have

\begin{matrix} u & = & t^{2} & d v & = & t e^{t^{2}} d t d u & = & 2 t d t & v & = & \int t e^{t^{2}} d t = \frac{1}{2} e^{t^{2}} . \end{matrix}

$\begin{array}{ccccccc}\hfill u& =\hfill & {t}^{2}\hfill & & \hfill dv& =\hfill & t{e}^{{t}^{2}}dt\hfill \\ \hfill du& =\hfill & 2tdt\hfill & & \hfill v& =\hfill & {\displaystyle\int }^{\text{ }}t{e}^{{t}^{2}}dt=\frac{1}{2}{e}^{{t}^{2}}.\hfill \end{array}$

Thus, we obtain

\begin{matrix} \int t^{3} e^{t^{2}} d t & = \frac{1}{2} t^{2} e^{t^{2}} - \int \frac{1}{2} e^{t^{2}} 2 t d t = \frac{1}{2} t^{2} e^{t^{2}} - \frac{1}{2} e^{t^{2}} + C . \end{matrix}

$\begin{array}{cc}{\displaystyle\int {t}^{3}{e}^{{t}^{2}}dt}\hfill & =\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-{\displaystyle\int \frac{1}{2}{e}^{{t}^{2}}2tdt}\hfill \\ \hfill & =\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-\frac{1}{2}{e}^{{t}^{2}}+C.\end{array}$

Example: Applying Integration by Parts More Than Once

Evaluate ${\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx$ .

Show Solution

Solution

This integral appears to have only one function—namely, $\sin\left(\text{ln}x\right)$ —however, we can always use the constant function 1 as the other function. In this example, let’s choose $u=\sin\left(\text{ln}x\right)$ and $dv=1dx$ . (The decision to use $u=\sin\left(\text{ln}x\right)$ is easy. We can’t choose $dv=\sin\left(\text{ln}x\right)dx$ because if we could integrate it, we wouldn’t be using integration by parts in the first place!) Consequently, $du=\frac{1}{x}\cos\left(\text{ln}x\right)dx$ and $v={\displaystyle\int }^{\text{ }}1dx=x$ . After applying integration by parts to the integral and simplifying, we have

\int sin (ln x) d x = x sin (ln x) - \int cos (ln x) d x

${\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx=x\sin\left(\text{ln}x\right)-{\displaystyle\int }^{\text{ }}\cos\left(\text{ln}x\right)dx$ .

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. This time let’s choose $u=\cos\left(\text{ln}x\right)$ and $dv=1dx$ , making $du=\text{-}\frac{1}{x}\sin\left(\text{ln}x\right)dx$ and $v={\displaystyle\int }^{\text{ }}1dx=x$ . Substituting, we have

\int sin (ln x) d x = x sin (ln x) - (x cos (ln x) - \int - sin (ln x) d x)

${\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx=x\sin\left(\text{ln}x\right)-\left(x\cos\left(\text{ln}x\right)-{\displaystyle\int }^{\text{ }}-\sin\left(\text{ln}x\right)dx\right)$ .

After simplifying, we obtain

\int sin (ln x) d x = x sin (ln x) - x cos (ln x) - \int sin (ln x) d x

${\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx=x\sin\left(\text{ln}x\right)-x\cos\left(\text{ln}x\right)-{\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx$ .

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 $I={\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx$ . Thus, the equation becomes

I = x sin (ln x) - x cos (ln x) - I

$I=x\sin\left(\text{ln}x\right)-x\cos\left(\text{ln}x\right)-I$ .

First, add $I$ to both sides of the equation to obtain

2 I = x sin (ln x) - x cos (ln x)

$2I=x\sin\left(\text{ln}x\right)-x\cos\left(\text{ln}x\right)$ .

Next, divide by 2:

I = \frac{1}{2} x sin (ln x) - \frac{1}{2} x cos (ln x)

$I=\frac{1}{2}x\sin\left(\text{ln}x\right)-\frac{1}{2}x\cos\left(\text{ln}x\right)$ .

Substituting $I={\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx$ again, we have

\int sin (ln x) d x = \frac{1}{2} x sin (ln x) - \frac{1}{2} x cos (ln x)

${\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx=\frac{1}{2}x\sin\left(\text{ln}x\right)-\frac{1}{2}x\cos\left(\text{ln}x\right)$ .

From this we see that $\frac{1}{2}x\sin\left(\text{ln}x\right)-\frac{1}{2}x\cos\left(\text{ln}x\right)$ is an antiderivative of $\sin\left(\text{ln}x\right)dx$ . For the most general antiderivative, add $+C\text{:}$

${\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx=\frac{1}{2}x\sin\left(\text{ln}x\right)-\frac{1}{2}x\cos\left(\text{ln}x\right)+C$ .

Analysis

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

$\begin{array}{c}\frac{d}{dx}\left(\frac{1}{2}x\sin\left(\text{ln}x\right)-\frac{1}{2}x\cos\left(\text{ln}x\right)\right)\hfill \\ \\ =\frac{1}{2}\left(\sin\left(\text{ln}x\right)\right)+\cos\left(\text{ln}x\right)\cdot \frac{1}{x}\cdot \frac{1}{2}x-\left(\frac{1}{2}\cos\left(\text{ln}x\right)-\sin\left(\text{ln}x\right)\cdot \frac{1}{x}\cdot \frac{1}{2}x\right)\hfill \\ =\sin\left(\text{ln}x\right).\hfill \end{array}$

try it

Evaluate ${\displaystyle\int }^{\text{ }}{x}^{2}\sin{x}dx$ .

Hint

This is similar to the previous example: Evaluate ${\displaystyle\int }^{\text{ }}{x}^{2}{e}^{3x}dx$ .

Show Solution

$\text{-}{x}^{2}\cos{x}+2x\sin{x}+2\cos{x}+C$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for this segmented clip of “3.1 Integration by Parts” here (opens in new window).

Try It

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.

Integration by Parts for Definite Integrals

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

${\displaystyle\int }_{a}^{b}udv={uv|}_{a}^{b}-{\displaystyle\int }_{a}^{b}vdu$ .

Example: 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 $\left[0,1\right]$ .

Show Solution

This region is shown in Figure 1. To find the area, we must evaluate $\underset{0}{\overset{1}{\displaystyle\int }}{\tan}^{-1}xdx$ .

This figure is the graph of the inverse tangent function. It is an increasing function that passes through the origin. In the first quadrant there is a shaded region under the graph, above the x-axis. The shaded area is bounded to the right at x = 1.

For this integral, let’s choose $u={\tan}^{-1}x$ and $dv=dx$ , thereby making $du=\frac{1}{{x}^{2}+1}dx$ and $v=x$ . After applying the integration-by-parts formula we obtain

$\text{Area}=x{\tan}^{-1}{x|}_{0}^{1}-\underset{0}{\overset{1}{\displaystyle\int }}\frac{x}{{x}^{2}+1}dx$ .

Use u-substitution to obtain

$\underset{0}{\overset{1}{\displaystyle\int }}\frac{x}{{x}^{2}+1}dx=\frac{1}{2}\text{ln}|{x}^{2}+1{|}_{0}^{1}$ .

Thus,

$\text{Area}=x{\tan }^{-1}x{|}_{0}^{1}-\frac{1}{2}\mathrm{ln}|{x}^{2}+1|{|}_{0}^{1}=\frac{\pi }{4}-\frac{1}{2}\mathrm{ln}2$ .

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

Example: Finding a volume of revolution

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

Show Solution

Solution

The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).

This figure is the graph of the function e^-x. It is an increasing function on the left side of the y-axis and decreasing on the right side of the y-axis. The curve also comes to a point on the y-axis at y=1. Under the curve there is a shaded rectangle in the first quadrant. There is also a cylinder under the graph, formed by revolving the rectangle around the y-axis.

To find the volume using shells, we must evaluate $2\pi {\displaystyle\int }_{0}^{1}x{e}^{\text{-}x}dx$ . To do this, let $u=x$ and $dv={e}^{\text{-}x}$ . These choices lead to $du=dx$ and $v={\displaystyle\int }^{\text{ }}{e}^{\text{-}x}=\text{-}{e}^{\text{-}x}$ . Substituting into the integration-by-parts for definite integrals formula, we obtain

$\begin{array}{cccc}\hfill \text{Volume}& =2\pi \underset{0}{\overset{1}{\displaystyle\int }}x{e}^{\text{-}x}dx=2\pi \left(\text{-}x{e}^{\text{-}x}{|}_{0}^{1}+\underset{0}{\overset{1}{\displaystyle\int }}{e}^{\text{-}x}dx\right)\hfill & & \text{Use integration by parts}.\hfill \\ & =-2\pi x{e}^{\text{-}x}{|}_{0}^{1}-2\pi {e}^{\text{-}x}{|}_{0}^{1}\hfill & & \text{Evaluate}\underset{0}{\overset{1}{\displaystyle\int }}{e}^{\text{-}x}dx=\text{-}{e}^{\text{-}x}{|}_{0}^{1}.\hfill \\ & =2\pi -\frac{4\pi }{e}.\hfill & & \text{Evaluate and simplify}.\hfill \end{array}$

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 $\frac{1}{e}$ added to the volume of a cone of base radius $1$ and height of $1-\frac{1}{3}$ . Consequently, the solid should have a volume a bit less than

$\pi {\left(1\right)}^{2}\frac{1}{e}+\left(\frac{\pi }{3}\right){\left(1\right)}^{2}\left(1-\frac{1}{e}\right)=\frac{2\pi }{3e}-\frac{\pi }{3}\approx 1.8177$ .

Since $2\pi -\frac{4\pi }{e}\approx 1.6603$ , we see that our calculated volume is reasonable.

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Evaluate ${\displaystyle\int }_{0}^{\pi \text{/}2}x\cos{x}dx$ .

Hint

Use the integration-by-parts formula with $u=x$ and $dv=\cos{x}dx$ .

Show Solution

$\frac{\pi }{2}-1$

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Module 3: Techniques of Integration

Integration-by-Parts Formula

Learning Outcomes

Integration by Parts

Example: Using Integration by Parts

Interactive

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Recall: Derivative rules for transcendental functions

example: Using integration by parts

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example: Applying integration by parts more than once

Example: Applying Integration by Parts When LIATE Doesn’t Quite Work

Example: Applying Integration by Parts More Than Once

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Integration by Parts for Definite Integrals

Integration by Parts for Definite Integrals

Example: finding the area of a region

Example: Finding a volume of revolution

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