Integration by Parts

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}) d x

defies us. Many students want to know whether there is a product rule for integration. There isn’t, 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.

The Integration-by-Parts Formula

then by using the product rule, we obtain

h^{'} (x) = f^{'} (x) g (x) + g^{'} (x) f (x) .

Although at first it may seem counterproductive, let’s now integrate both sides of this equation:

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

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.

Using Integration by Parts

Use integration by parts with $u = x$

and $d v = sin x d x$

to evaluate $\int x sin x d x .$

By choosing $u = x,$

we have $d u = 1 d x .$

Since $d v = sin x d x,$

we get $v = \int sin x d x = − cos x .$

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}

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}

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 $d v = x .$

If we had done so, then we would have $d u = cos x$

and $v = \frac{1}{2} x^{2} .$

Thus, after applying integration by parts, we have $\int^{} x sin x d x = \frac{1}{2} x^{2} sin x - \int^{} \frac{1}{2} x^{2} cos x d x .$

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 $d v$

before finding a choice that works.

Second, you may wonder why, when we find $v = \int^{} sin x d x = − cos x,$

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{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}

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

Therefore, the antiderivative checks out.

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 logarithmic function and an algebraic function, we should choose

u

to be the logarithmic function, because L comes before A in LIATE. The integral in [link] has a trigonometric function

(sin x)

is selected to be the remaining part of the function to be integrated, together with

d x .

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

d v .

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

d v .

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

Algebraic functions are generally easy both to integrate and to differentiate, and they come in the middle of the mnemonic.

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

Applying Integration by Parts More Than Once

Evaluate $\int^{} x^{2} e^{3 x} d x .$

Using LIATE, choose $u = x^{2}$

and $d v = e^{3 x} d x .$

Thus, $d u = 2 x d x$

and $v = \int e^{3 x} d x = (\frac{1}{3}) e^{3 x} .$

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}

Substituting into [link] 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 .

We still cannot integrate $\int \frac{2}{3} x e^{3 x} d x$

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 $d v = \frac{2}{3} e^{3 x} d x .$

Thus, $d u = d x$

and $v = \int (\frac{2}{3}) e^{3 x} d x = (\frac{2}{9}) e^{3 x} .$

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}

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) .

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 .

Applying Integration by Parts More Than Once

Evaluate $\int^{} sin (ln x) d x .$

This integral appears to have only one function—namely, $sin (ln x)$

—however, we can always use the constant function 1 as the other function. In this example, let’s choose $u = sin (ln x)$

and $d v = 1 d x .$

(The decision to use $u = sin (ln x)$

is easy. We can’t choose $d v = sin (ln x) d x$

because if we could integrate it, we wouldn’t be using integration by parts in the first place!) Consequently, $d u = (1 / x) cos (ln x) d x$

and $v = \int^{} 1 d x = x .$

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 .

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 $u = cos (ln x)$

and $d v = 1 d x,$

making $d u = − (1 / x) sin (ln x) d x$

and $v = \int^{} 1 d x = x .$

Substituting, we have

\int^{​} sin (ln x) d x = x sin (ln x) - (x cos (ln x) — \int^{​} - sin (ln x) d x) .

After simplifying, we obtain

\int^{​} sin (ln x) d x = x sin (ln x) - x cos (ln x) - \int^{​} sin (ln x) d x .

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 $I = \int^{} sin (ln x) d x .$

Thus, the equation becomes

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

First, add $I$

to both sides of the equation to obtain

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

Next, divide by 2:

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

Substituting $I = \int^{} sin (ln x) d x$

again, we have

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

From this we see that $(1 / 2) x sin (ln x) - (1 / 2) x cos (ln x)$

is an antiderivative of $sin (ln x) d x .$

For the most general antiderivative, add $+ C :$

\int^{​} sin (ln x) d x = \frac{1}{2} x sin (ln x) - \frac{1}{2} x cos (ln x) + C .

Analysis

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

\begin{matrix} \frac{d}{d x} (\frac{1}{2} x sin (ln x) - \frac{1}{2} x cos (ln x)) \\ = \frac{1}{2} (sin (ln x)) + cos (ln x) \cdot \frac{1}{x} \cdot \frac{1}{2} x - (\frac{1}{2} cos (ln x) - sin (ln x) \cdot \frac{1}{x} \cdot \frac{1}{2} x) \\ = sin (ln x) . \end{matrix}

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.

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.

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 $2 π \int_{0}^{1} x e^{− x} d x .$

To do this, let $u = x$

and $d v = e^{− x} .$

These choices lead to $d u = d x$

and $v = \int^{} e^{− x} = − e^{− x} .$

Substituting into [link], we obtain

\begin{matrix} Volume & = 2 π \int_{0}^{1} x e^{− x} d x = 2 π (− x e^{− x} {\|}_{0}^{1} + \int_{0}^{1} e^{− x} d x) & Use integration by parts . \\ = −2 π x e^{− x} {\|}_{0}^{1} - 2 π e^{− x} {\|}_{0}^{1} & Evaluate \int_{0}^{1} e^{− x} d x = − e^{− x} {\|}_{0}^{1} . \\ = 2 π - \frac{4 π}{e} . & Evaluate and simplify . \end{matrix}

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 $1 - \frac{1}{3} .$

Consequently, the solid should have a volume a bit less than

π {(1)}^{2} \frac{1}{e} + (\frac{π}{3}) {(1)}^{2} (1 - \frac{1}{e}) = \frac{2 π}{3 e} - \frac{π}{3} \approx 1.8177 .

Since $2 π - \frac{4 π}{e} \approx 1.6603,$

we see that our calculated volume is reasonable.

Key Concepts

Key Equations

In using the technique of integration by parts, you must carefully choose which expression is u. For each of the following problems, use the guidelines in this section to choose u. Do not evaluate the integrals.

\int x^{3} e^{2 x} d x

u = x^{3}

\int x^{3} ln (x) d x

\int y^{3} cos y d x

u = y^{3}

\int x^{2} arctan x d x

\int e^{3 x} sin (2 x) d x

u = sin (2 x)

Find the integral by using the simplest method. Not all problems require integration by parts.

\int v sin v d v

\int ln x d x

(Hint: $\int ln x d x$

is equivalent to $\int 1 \cdot ln (x) d x .)$

− x + x ln x + C

\int x cos x d x

\int {tan}^{−1} x d x

x {tan}^{−1} x - \frac{1}{2} ln (1 + x^{2}) + C

\int x^{2} e^{x} d x

\int x sin (2 x) d x

- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x) + C

\int x e^{4 x} d x

\int x e^{− x} d x

e^{− x} (−1 - x) + C

\int x cos 3 x d x

\int x^{2} cos x d x

2 x cos x + (−2 + x^{2}) sin x + C

\int x ln x d x

\int ln (2 x + 1) d x

\frac{1}{2} (1 + 2 x) (−1 + ln (1 + 2 x)) + C

\int x^{2} e^{4 x} d x

\int e^{x} sin x d x

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

\int e^{x} cos x d x

\int x e^{− x^{2}} d x

- \frac{e^{− x^{2}}}{2} + C

\int x^{2} e^{− x} d x

\int sin (ln (2 x)) d x

- \frac{1}{2} x cos [ln (2 x)] + \frac{1}{2} x sin [ln (2 x)] + C

\int c o s (ln x) d x

\int {(ln x)}^{2} d x

2 x - 2 x ln x + x {(ln x)}^{2} + C

\int ln (x^{2}) d x

\int x^{2} ln x d x

(− \frac{x^{3}}{9} + \frac{1}{3} x^{3} ln x) + C

\int {sin}^{−1} x d x

\int {cos}^{−1} (2 x) d x

- \frac{1}{2} \sqrt{1 - 4 x^{2}} + x {cos}^{−1} (2 x) + C

\int x arctan x d x

\int x^{2} sin x d x

− (−2 + x^{2}) cos x + 2 x sin x + C

\int x^{3} cos x d x

\int x^{3} sin x d x

− x (−6 + x^{2}) cos x + 3 (−2 + x^{2}) sin x + C

\int x^{3} e^{x} d x

\int x {sec}^{−1} x d x

\frac{1}{2} x (− \sqrt{1 - \frac{1}{x^{2}}} + x \cdot {sec}^{−1} x) + C

\int x {sec}^{2} x d x

\int x cosh x d x

− cosh x + x sinh x + C

Compute the definite integrals. Use a graphing utility to confirm your answers.

\int_{1 / e}^{1} ln x d x

\int_{0}^{1} x e^{−2 x} d x

(Express the answer in exact form.)

\frac{1}{4} - \frac{3}{4 e^{2}}

\int_{0}^{1} e^{\sqrt{x}} d x (let u = \sqrt{x})

\int_{1}^{e} ln (x^{2}) d x

\int_{0}^{π} x cos x d x

\int_{− π}^{π} x sin x d x

(Express the answer in exact form.)

2 π

\int_{0}^{3} ln (x^{2} + 1) d x

(Express the answer in exact form.)

\int_{0}^{π / 2} x^{2} sin x d x

(Express the answer in exact form.)

−2 + π

\int_{0}^{1} x 5^{x} d x

(Express the answer using five significant digits.)

Evaluate $\int cos x ln (sin x) d x$

− sin (x) + ln [sin (x)] sin x + C

Derive the following formulas using the technique of integration by parts. Assume that n is a positive integer. These formulas are called reduction formulas because the exponent in the x term has been reduced by one in each case. The second integral is simpler than the original integral.

\int x^{n} e^{x} d x = x^{n} e^{x} - n \int x^{n - 1} e^{x} d x

\int x^{n} cos x d x = x^{n} sin x - n \int x^{n - 1} sin x d x

Answers vary

\int x^{n} sin x d x = \_\_\_\_\_\_

Integrate $\int 2 x \sqrt{2 x - 3} d x$

using two methods:

Using parts, letting $d v = \sqrt{2 x - 3} d x$
Substitution, letting $u = 2 x - 3$

a. $\frac{2}{5} (1 + x) {(−3 + 2 x)}^{3 / 2} + C$

b. $\frac{2}{5} (1 + x) {(−3 + 2 x)}^{3 / 2} + C$

State whether you would use integration by parts to evaluate the integral. If so, identify u and dv. If not, describe the technique used to perform the integration without actually doing the problem.

\int x ln x d x

\int \frac{{ln}^{2} x}{x} d x

Do not use integration by parts. Choose u to be $ln x,$

and the integral is of the form $\int u^{2} d u .$

\int x e^{x} d x

\int x e^{x^{2} - 3} d x

Do not use integration by parts. Let $u = x^{2} - 3,$

and the integral can be put into the form $\int e^{u} d u .$

\int x^{2} sin x d x

\int x^{2} sin (3 x^{3} + 2) d x

Do not use integration by parts. Choose u to be $u = 3 x^{3} + 2$

and the integral can be put into the form $\int sin (u) d u .$

Sketch the region bounded above by the curve, the x-axis, and $x = 1,$

and find the area of the region. Provide the exact form or round answers to the number of places indicated.

y = 2 x e^{− x}

(Approximate answer to four decimal places.)

y = e^{− x} sin (π x)

(Approximate answer to five decimal places.)

The area under graph is 0.39535.* * *

This figure is the graph of y=e^-x sin(pi*x). The curve begins in the third quadrant at x=0.5, increases through the origin, reaches a high point between 0.5 and 0.75, then decreases, passing through x=1.

Find the volume generated by rotating the region bounded by the given curves about the specified line. Express the answers in exact form or approximate to the number of decimal places indicated.

y = sin x, y = 0, x = 2 π, x = 3 π

about the y-axis (Express the answer in exact form.)

y = e^{− x}

y = 0, x = −1 x = 0;

about $x = 1$

(Express the answer in exact form.)

2 π e

A particle moving along a straight line has a velocity of $v (t) = t^{2} e^{− t}$

after t sec. How far does it travel in the first 2 sec? (Assume the units are in feet and express the answer in exact form.)

Find the area under the graph of $y = {sec}^{3} x$

from $x = 0 to x = 1 .$

(Round the answer to two significant digits.)

2.05

Find the area between $y = (x - 2) e^{x}$

and the x-axis from $x = 2$

to $x = 5 .$

(Express the answer in exact form.)

Find the area of the region enclosed by the curve $y = x cos x$

and the x-axis for

\frac{11 π}{2} \leq x \leq \frac{13 π}{2} .

(Express the answer in exact form.)

12 π

Find the volume of the solid generated by revolving the region bounded by the curve $y = ln x,$

the x-axis, and the vertical line $x = e^{2}$

about the x-axis. (Express the answer in exact form.)

Find the volume of the solid generated by revolving the region bounded by the curve $y = 4 cos x$

and the x-axis, $\frac{π}{2} \leq x \leq \frac{3 π}{2},$

about the x-axis. (Express the answer in exact form.)

8 π^{2}

Find the volume of the solid generated by revolving the region in the first quadrant bounded by $y = e^{x}$

and the x-axis, from $x = 0$

to $x = ln (7),$

about the y-axis. (Express the answer in exact form.)

Integration by Parts

The Integration-by-Parts Formula