4.6 Integration by parts

198

CHAPTER 4. INTEGRALS

4.6 Integration by parts

4.6.1 General Concept

If f and g are two di?erentiable functions, then the product rule gives us: (f (x) g (x))0 = f (x) g0 (x) + f 0 (x) g (x)

If we integrate both sides, we get

Z

Z

Z

(f (x) g (x))0 dx = f (x) g0 (x) dx + f 0 (x) g (x) dx

From the fundamental theorem of calculus, we know that the integral of the

derivative of a function is the function itself, therefore we have:

Z

Z

f (x) g (x) = f (x) g0 (x) dx + f 0 (x) g (x) dx

Or, solving for the ...rst integral

Z

Z

f (x) g0 (x) dx = f (x) g (x) f 0 (x) g (x) dx

This is the integration by parts formula. However, we usually do not remember

it this way. If we let u = f (x), then du = f 0 (x) dx. Similarly, if v = g (x), then

dv = g0 (x) dx. Doing the two substitutions gives us:

Z

Z

udv = uv vdu

(4.4)

The goal whenRusing this formula is to pretend that the integral we are given is of the form udv. We ...nd u and dv that accomplish this. Once we have u

and dv, we ...nd du by di?erentiatiRng u and v by integrating dv. Then, we can rewrite the given integral as uv vdu. This will work if the new integral we

obtained can be evaluated or is easier to evaluate than the one we started with.

Also, once we have selected dv, v is an antiderivative of dv. therefore, we must

be able to ...nd an antiderivative for dv.

For de...nite integrals, the integration by parts formula becomes

Zb f (x) g0 (x) dx = f (x) g (x)jba

a

Zb f 0 (x) g (x) dx

a

We illustrate this technique with several examples. For clarity, the quantities

we select from the integral (u and dv) will be in bold characters. The quantities

we deduce from our selection (du and v) will be in normal characters.

R Example 293 Find x sin xdx

If we select

u=x

du = dx

v = cos x dv = sin xdx

4.6. INTEGRATION BY PARTS

199

Then, applying formula 4.4 gives us:

Z

Z

x sin xdx = x cos x

cos xdx

Z

= x cos x + cos xdx

= x cos x + sin x + C

Remark 294 There is usually more than one way to select u and dv. However, we should do it so that when we apply the integration by parts formula, we obtain an integral we can solve, or at least one that does not seem more complicated. In the above example, if we had selected

u = sin x du = cos xdx

x2 v=

2

dv = xdx

then the integration by parts formula would have given us

Z

x2

x sin xdx = sin x

Z 1 x2 cos xdx

2

2

which is more complicated than the integral we started with.

R Example 295 Find xexdx Following the previous example, we select

u = x du = dx v = ex dv = exdx

Then, applying formula 4.4 gives

Z xexdx = xex

Z exdx

= xex ex + C

R Remark 296 In the past two examples, we had an integral of the form xf (x) dx where we knew an antiderivative of f (x). We did integrationR by parts with u = x. This will be the case for most integrals of this form. x ln xdx is an exception.

Example

297

Find

Re

1

ln

xdx

If we select

u = ln x v=x

du = 1 dx x

dv = dx

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CHAPTER 4. INTEGRALS

Then, applying formula 4.4 gives

Ze ln xdx = x ln xje1

1

= x ln xje1

Ze

dx

1

xje1

= (e ln e ln 1) (e 1)

=1

Remark 298 You will note that in the problem we just did, we found that Z

ln xdx = x ln x x + C

R Example 299 Find tan 1 xdx If we select

u = tan 1x

v=x

1 du = 1 + x2 dx

dv = dx

Then, applying formula 4.4 gives Z

tan 1 xdx = x tan 1 x

Z x

1 + x2 dx

(4.5)

We

will

do

R

x 1 + x2 dx

separately,

using

the

substitution

u

=

1+

x2.

Then,

du = 2xdx therefore

Z

Z

x

1 du

1 + x2 dx = 2 u

=

1 2

ln juj + C

1 = ln

1 + x2

+C

2

1 = ln

1 + x2

+C

2

Using what we just found in equation 4.5, gives us

Z

tan 1 xdx = x tan 1 x

1 ln

1 + x2

+C

2

Remark 300 In the previous two examples, the integrand consisted of just one function. We used integration by parts with u set to that function. This technique should be remembered.

4.6.2 Repeated Integration by Parts

In some cases, applying the integration by parts formula one time will not be enough. You may need to apply it twice, or more. We look at some example to illustrate the various cases which can occur.

4.6. INTEGRATION BY PARTS

201

R Example 301 Find x2exdx

If we select

u = x2

v = ex

du = 2xdx dv = exdx

Then, applying formula 4.4 gives Z x2exdx = x2ex

Z 2 xexdx

At this point, we are left with an integral that we still cannot do. However, we see that it looks simpler than the one we started with. So, we try the integration by parts formula again. In fact, we already did this integral in an example above, so we will simply use the result.

Z x2exdx = x2ex 2 (xex ex + C)

= x2ex 2xex + 2ex + C

Note that we replaced 2C by C. We only wish to denote that there is some

constant.

R Example 302 Find x2 sin xdx

If we select

u = x2

du = 2xdx

v = cos x dv = sin xdx

Then, applying formula 4.4 gives

Z

Z

x2 sin xdx = x2 cos x + 2 x cos xdx

(4.6)

R We apply the integration by parts formula to x cos xdx. If we select

u=x

du = dx

v = sin x dv = cos xdx

Then, applying formula 4.4 gives

Z

Z

x cos xdx = x sin x sin xdx

= x sin x + cos x + C

If we use what we just found in equation 4.6, we obtain Z x2 sin xdx = x2 cos x + 2 (x sin x + cos x + C)

= x2 cos x + 2x sin x + 2 cos x + C

Here again, we replaced 2C by C.

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CHAPTER 4. INTEGRALS

In these two examples, we applied the integration by parts formula several

time. We noticed that every time, the integral was getting easier. So, our hope

was that we would eventually be able to ...nd it, which we did. In some cases, a

di?erent situation will present itself, as illustrated in the next example.

Example 303 Find R ex sin xdx

If we select

u = ex

du = exdx

v = cos x dv = sin xdx

Then, applying formula 4.4 gives

Z

Z

ex sin xdx = ex cos x + ex cos xdx

(4.7)

The integral we obtain is not simpler, but it is not more di? cult either. Furthermore, it looks like the original, sin x having been replaced by cos x. We apply the integration by parts formula to it. If we select

u = ex du = exdx v = sin x dv = cos xdx

Then, applying formula 4.4 gives Z ex cos xdx = ex sin x

Z ex sin xdx

If we replace what we just found in equation 4.7, we obtain

Z

Z

ex sin xdx = ex cos x + ex sin x

ex sin xdx

R Therefore, we can solve for ex sin xdx. We obtain

Z

Z

ex sin xdx + ex sin xdx = ex cos x + ex sin x

Z

2 ex sin xdx = ex (sin x cos x)

Z

ex sin xdx =

ex (sin x

cos x) + C

2

Note, we inserted the constant C in the last step.

Using integration by parts, we can prove important formulae, called reduction formulae. The most important ones are listed as a theorem.

Theorem 304 Let n be an integer such that n 2.

1. R sinn xdx = 1 cos x sinn 1 x + n 1 R sinn 2 xdx

n

n

4.6. INTEGRATION BY PARTS

203

2.

R cosn xdx =

1 sin x cosn

1x+ n

1 R cosn 2 xdx

n

n

3. R (ln x)n dx = x (ln x)n n R (ln x)n 1 dx

R

R

4. xnexdx = xnex n xn 1exdx

Proof. We only prove part 1. The other parts are left as an exercise.

1. Prove that R sinn xdx = 1 cos x sinn 1 x + n 1 R sinn 2 xdx.

n

n

select

u = sinn 1x du = (n 1) sinn 2 x cos xdx

v = cos x

dv = sin xdx

If we

Then, applying formula 4.4 gives

Z

Z

sinn xdx = sinn 1 x cos x + (n 1) cos2 x sinn 2 xdx

Z

= sinn 1 x cos x + (n 1) 1 sin2 x sinn 2 xdx

Z

Z

= sinn 1 x cos x + (n 1) sinn 2 xdx (n 1) sinn xdx

Z

Z

Z

sinn xdx + (n 1) sinn xdx = sinn 1 x cos x + (n 1) sinn 2 xdx

Z

Z

n sinn xdx = sinn 1 x cos x + (n 1) sinn 2 xdx

Z

Z

sinn xdx = 1 cos x sinn 1 x + n 1 sinn 2 xdx

n

n

2. see exercise 34 3. see exercise 37 4. do as an exercise

We illustrate with an example how these formulas might be used.

Example 305 Find R (ln x)3 dx = 6x 1 ln3 x 1 ln2 x + ln x 1

6

2

Using part 3 of theorem 304, we obtain

Z

Z

(ln x)3 dx = x (ln x)3 3 (ln x)2 dx

To ...nd R (ln x)2 dx, we apply the same formula again.

Z

Z

(ln x)2 dx = x (ln x)2 2 ln xdx

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CHAPTER 4. INTEGRALS

The last integral was found in an example above (otherwise, we would ...nd it using integration by parts). It was found to be

Z ln xdx = x ln x x

Combining the three gives us Z (ln x)3 dx = x (ln x)3

= x (ln x)3

= x (ln x)3

Z 3 (ln x)2 dx

Z 3 x (ln x)2 2 ln xdx

Z 3x (ln x)2 + 6 ln xdx

= x (ln x)3 3x (ln x)2 + 6 [x ln x x] = x (ln x)3 3x (ln x)2 + 6x ln x 6x

Remark 306 As this example shows, the reduction formula may have to be applied several times before we ...nd the answer.

4.6.3 Things to know

Know and be able to apply the integration by parts formula. Remember that this formula may have to be applied more than once. In particular, remember the following techniques we learned in the examples:

R ? To integrate xnf (x) dx where f is a function for which we know

an antiderivative (such as sin x, cos x, ex, :::) we let u = xn. When n > 1, we will have to do repeated integration by parts (n times).

R ? When the integrand contains only one function as in f (x) dx, one

thing to try is to let u = f (x).

Using integration by parts, be able to prove and use the following reduction

formulas

Z

Z

sinn xdx = 1 cos x sinn 1 x + n 1 sinn 2 xdx

Z

n

nZ

cosn xdx = 1 sin x cosn 1 x + n 1 cosn 2 xdx

Z

n

Z

n

(ln x)n dx = x (ln x)n n (ln x)n 1 dx

Z

Z

xnexdx = xnex n xn 1exdx

4.6. INTEGRATION BY PARTS

205

4.6.4 Problems

R 1. Using integration by parts, evaluate x ln xdx with u = ln x and dv = xdx

R 2. Evaluate x cos 2xdx

3.

Evaluate

R

te

t 3

dt

4. Evaluate R x2 sin xdx

R 5. Evaluate ln (5x + 4) dx

R 6. Evaluate arctan 3tdt

7. Evaluate R e2x sin 5xdx

R 8. Evaluate 0 x sin 3xdx

9.

Evaluate

R2

1

ln x x2 dx

10.

Evaluate

R1

0

x e2x dx

11.

Evaluate

R2

1

(ln

x)2

dx

12.

R Evaluate sin

1 xdx

Rp 13. Use a substitution then integration by parts to evaluate sin xdx

14.

R Using the formulas in theorem 304, prove that

sin2 xdx =

x

sin 2x +C

24

15.

Using

the

formulas

in

theorem

304,

prove

that

R 2 0

sinn

xdx

=

n

n

1

R 2 0

sinn

2 xdx

16. Using the formulas in theorem 304, ...nd R (ln x)3 dx

17. Finish proving theorem 304

18.

If f

f (0) (a) g0

=g (a)

(0) f

=0 0 (a)

and f 00 g (a) +

aR0nadfg0000(xa)regc(oxn)tdinxuous,

show

that

Ra

0

f

(x)

g00

(x)

dx

=

4.6.5 Answers

R 1. Using integration by parts, evaluate x ln xdx with u = ln x and dv = xdx

R

x2 ln x x2

x ln xdx =

+C

2

4

R

1

1

2. Evaluate x cos 2xdx = 4 cos 2x + 2 x sin 2x + C

3.

Evaluate

R

te

t 3

dt

=

3e

1 3

t

(t

3) + C

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