17 formulas from integration by parts

Formulas From Integration by Parts

Formulas From Integration by PartsIntegration by parts:Let u = u(x), v = v(x), then

where dv = v'dx and du = u'dx. ∫ udv = uv – ∫ vdu



The integrals of some functions require using the integration by parts many times.




This leads to special methods and formulas that summarize the results.




This leads to special methods and formulas that summarize the results.

These formulas usually reduce the complexity of the integrands in stages until the final answers are obtained.

Formulas From Integration by PartsThe Tableau Method


Given

+u

∫ udv, set up the table as shown.

dv


Given

+u


differentiatewith alternate signs

–u'

+u'' •–u''' dv


Given

+u



integrate

v –u'

+u''∫ v dx ∫[∫ v dx]dx

••

–u''' dv


Given

+u



integrate

v –u'


••

–u'''

The integral ∫ u dv is the sum of the diagonal products.

dv


Given

+u



integrate

v –u'


••

–u'''


dv

That is, ∫ u dv = uv


Given

+u



integrate

v –u'


••

–u'''


dv

That is, ∫ u dv = uv – u' ∫vdx


Given

+u



integrate

v –u'


••

–u'''


dv

That is, ∫ u dv = uv – u' ∫vdx + u'' ∫[∫ v dx]dx ..


Example: Find ∫(x3 + 2x) ex/2dx



x3+2x

ex/2

Set u = x3 + 2x, dv = ex/2dx



x3+2x


ex/2




x3+2x


ex/2


–(3x2+2)



x3+2x


6x

ex/2


–(3x2+2)



x3+2x


6x –6 ex/2


–(3x2+2) 0



x3+2x


integrate

6x –6 ex/2


–(3x2+2) 0



x3+2x


integrate

6x –6 ex/2


–(3x2+2) 0

2ex/2



x3+2x


integrate

6x –6 ex/2


–(3x2+2) 0

2ex/2 4ex/2



x3+2x


integrate

6x –6 ex/2


–(3x2+2) 0

2ex/2 4ex/2 8ex/2 16ex/2



x3+2x


integrate

6x –6 ex/2


–(3x2+2) 0

2ex/2 4ex/2 8ex/2 16ex/2

Hence ∫(x3 + 2x) ex/2dx

=



x3+2x


integrate

6x –6 ex/2


–(3x2+2) 0

2ex/2 4ex/2 8ex/2 16ex/2


= 2ex/2(x3+2x)



x3+2x


integrate

6x –6 ex/2


–(3x2+2) 0

2ex/2 4ex/2 8ex/2 16ex/2


= 2ex/2(x3+2x) – 4ex/2(3x2+2)



x3+2x


integrate

6x –6 ex/2


–(3x2+2) 0

2ex/2 4ex/2 8ex/2 16ex/2


= 2ex/2(x3+2x) – 4ex/2(3x2+2) + 48xex/2 – 96ex/2 + c



x3+2x


integrate

6x –6 ex/2


–(3x2+2) 0

2ex/2 4ex/2 8ex/2 16ex/2


= 2ex/2(x3+2x) – 4ex/2(3x2+2) + 48xex/2 – 96ex/2 + c

= ex/2 [2(x3+2x) – 4(3x2+2) + 48x – 96] + c



x3+2x


integrate

6x –6 ex/2


–(3x2+2) 0

2ex/2 4ex/2 8ex/2 16ex/2


= 2ex/2(x3+2x) – 4ex/2(3x2+2) + 48xex/2 – 96ex/2 + c

= ex/2 [2(x3+2x) – 4(3x2+2) + 48x – 96] + c

= ex/2 [2x3 – 12x2 + 52x – 104] + c


The tableau method is used when the integrand is a product of polynomial with a functions that can be integrated repeatedly such as ex, sin(x), and cos(x).



Antiderivatives of trig-power



Antiderivatives of trig-powerFrom here on, we use s for sin(x), c for cos(x), and t for tan(x).



Antiderivatives of trig-powerFrom here on, we use s for sin(x), c for cos(x), and t for tan(x). Also we assume that n > 2.




∫ cosn(x) dx

From here on, we use s for sin(x), c for cos(x), and t for tan(x). Also we assume that n > 2.




Write ∫cndx = ∫cn–1 cdx, use integration by parts.

∫ cosn(x) dx





set u = cn–1 dv = cdx


∫ cosn(x) dx






du = -(n – 1)cn–2sdx


∫ cosn(x) dx






du = -(n – 1)cn–2sdx v = s


∫ cosn(x) dx



∫cndx =


du = -(n – 1)cn–2sdx v = s


∫cndx = cn–1s


du = -(n – 1)cn–2sdx v = s


∫cndx = cn–1s + (n – 1)∫cn–2s2dx


du = -(n – 1)cn–2sdx v = s




du = -(n – 1)cn–2sdx v = s

∫cndx = cn–1s + (n – 1)∫cn–2(1 – c2)dx




du = -(n – 1)cn–2sdx v = s

∫cndx = cn–1s + (n – 1)∫cn–2(1 – c2)dx

∫cndx = cn–1s + (n – 1)∫cn–2dx – (n – 1) ∫cndx




du = -(n – 1)cn–2sdx v = s

∫cndx = cn–1s + (n – 1)∫cn–2(1 – c2)dx


combine like–terms




du = -(n – 1)cn–2sdx v = s

∫cndx = cn–1s + (n – 1)∫cn–2(1 – c2)dx


n∫cndx = cn–1s + (n – 1)∫cn–2dx




du = -(n – 1)cn–2sdx v = s

∫cndx = cn–1s + (n – 1)∫cn–2(1 – c2)dx


n∫cndx = cn–1s + (n – 1)∫cn–2dx

Hence, ∫cndx = + ∫cn–2dx n

cn–1sn

n–1


∫cndx = + ∫cn–2dx n

cn–1sn

n–1

Using this formula repeatedly eventually reduces the problem to integrating cosine to the power 0 or 1, which terminates the process.



cn–1sn

n–1

Example: Find ∫ c5 dx




cn–1sn

n–1


n = 5,




cn–1sn

n–1


n = 5, ∫ c5 dx = + 5c4s




cn–1sn

n–1


n = 5, ∫ c5 dx = + ∫c3dx 5c4s

54




cn–1sn

n–1


n = 5, ∫ c5 dx = + ∫c3dx 5c4s

54

n = 3,




cn–1sn

n–1


n = 5, ∫ c5 dx = + ∫c3dx 5c4s

54

n = 3, (5c4s

54 = + 3

c2s +

)




cn–1sn

n–1


n = 5, ∫ c5 dx = + ∫c3dx 5c4s

54

n = 3, (5c4s

54 = + 3

c2s +

32 ∫ c dx )




cn–1sn

n–1


n = 5, ∫ c5 dx = + ∫c3dx 5c4s

54

n = 3, (5c4s

54 = + 3

c2s +

32 ∫ c dx )

(5c4s

54 = + 3

c2s +

32 s) + k


Formulas From Integration by PartsFor ∫sndx

u = sn–1Set


u = sn–1 dv = sdxSet


u = sn–1 dv = sdxdu = (n – 1)sn–2cdx

Set


u = sn–1 dv = sdxdu = (n – 1)sn–2cdx v = –c

Set


∫sndx = + ∫sn–2dx n

–sn–1cn

n–1

For ∫sndxu = sn–1 dv = sdx

du = (n – 1)sn–2cdx v = –c Use integration by parts we get the formula:

Set



–sn–1cn

n–1



For use the fact that ∫secn(x)dx, ∫sec2(x)dx = t

Set



–sn–1cn

n–1



For use the fact that ∫secn(x)dx, ∫sec2(x)dx = tSet u = sec(x)n–2

Set



–sn–1cn

n–1



For use the fact that ∫secn(x)dx, ∫sec2(x)dx = tSet u = sec(x)n–2 dv = sec2(x)dx

Set



–sn–1cn

n–1




du = (n – 2)sn–2tdx

Set



–sn–1cn

n–1




du = (n – 2)sn–2tdx v = t

Set



–sn–1cn

n–1



For use the fact that ∫secn(x)dx,

∫secn(x)dx = + ∫secn–2(x)dx n – 1 secn–2(x)tan(x)

n–1 n–2

∫sec2(x)dx = tSet u = sec(x)n–2 dv = sec2(x)dx

du = (n – 2)sn–2tdx v = t Use integration by parts we get the formula:

Set

Formulas From Integration by PartsFor the antiderivatives of tn, use change of variable via the formula t2 = sec2(x) – 1.


= ∫tn–2 t2 dx

∫tndx


= ∫tn–2 t2 dx= ∫tn–2(sec2(x) –1)dx

∫tndx


= ∫tn–2 t2 dx= ∫tn–2(sec2(x) –1)dx

∫tndx

= ∫ tn–2sec2(x) – tn–2dx


= ∫tn–2 t2 dx= ∫tn–2(sec2(x) –1)dx

∫tndx


Use change of variable


= ∫tn–2 t2 dx= ∫tn–2(sec2(x) –1)dx

∫tndx


Use change of variableSet u = tan(x)


= ∫tn–2 t2 dx= ∫tn–2(sec2(x) –1)dx

∫tndx


Use change of variableSet u = tan(x) then du = sec2(x)dx


= ∫tn–2 t2 dx= ∫tn–2(sec2(x) –1)dx

∫tndx



∫ tn–2 sec2(x) dx


= ∫tn–2 t2 dx= ∫tn–2(sec2(x) –1)dx

∫tndx



∫ tn–2 sec2(x) dx = ∫ un–2 du =


= ∫tn–2 t2 dx= ∫tn–2(sec2(x) –1)dx

∫tndx



∫ tn–2 sec2(x) dx = ∫ un–2 du = n – 1 un–1


= ∫tn–2 t2 dx= ∫tn–2(sec2(x) –1)dx

∫tndx




= n – 1 tn–1


= ∫tn–2 t2 dx= ∫tn–2(sec2(x) –1)dx

∫tndx




= n – 1 tn–1

So ∫tndx = – ∫tn–2dx n – 1 tn–1



–sn–1cn

n–1

We summarize the formulas below.

∫secn(x)dx = + ∫secn–2(x)dx n – 1

secn–2(x)tan(x)n–1 n–2


cn–1sn

n–1

From integration by parts, we get:



–sn–1cn

n–1

We summarize the formulas below.

∫tndx = – ∫tn–2dx n – 1 tn–1

∫secn(x)dx = + ∫secn–2(x)dx n – 1

secn–2(x)tan(x)n–1 n–2


cn–1sn

n–1

From integration by parts, we get:

From change of variable, we get:

Formulas From Integration by PartsExercise.

∫cscn(x)dx2. Find by integration by parts.

1. Finish the algebra for the antiderivatives of sn, secn(x).

∫cotn(x)dx3. Find by change of variable.

17 formulas from integration by parts

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