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© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules
25: Integration by 25: Integration by PartsParts
Proof of Formulae Link
Integration by Parts
SUMMARY
dxdx
duvuvdx
dx
dvu
To integrate some products we can use the formula
Integration by Parts
This is in the formula book : you have to know how to use it
Integration by Parts
dx
duto get . . .
dxdx
duvuvdx
dx
dvu
Using this formula means that we differentiate
one factor, u
So,
Integration by Parts
So,
dxdx
duvuvdx
dx
dvu
and integrate the other , dx
dvto get v
Using this formula means that we differentiate
one factor, u dx
duto get . . .
Integration by Parts
So,
dxdx
duvuvdx
dx
dvu
e.g. 1 Find dxxx sin2
xu 2 xdx
dvsinand
2dx
duxv cos
differentiate
integrate
and integrate the other , dx
dvto get v
Using this formula means that we differentiate
one factor, u dx
duto get . . .
Having substituted in the formula, notice
that the 1st term, uv, is completed but the
2nd term still needs to be integrated.
( +C comes
later )
Integration by Parts
)cos(2 xx dxx 2)cos(
We can now substitute into the formula
So,xu 2
2dx
du xv cosdifferentiate
integrate
xdx
dvsinand
udx
dv u v vdx
du
dxdx
duvuvdx
dx
dvu
dxxx sin2
Integration by Parts
C
So,
dxdx
duvuvdx
dx
dvu
)cos(2 xx dxx 2)cos(
xsin2
xu 2
2dx
du xv cosdifferentiate
integrate
We can now substitute into the formula
xdx
dvsinand
dxxcos2xx cos2The 2nd term needs integrating
xx cos2
dxxx sin2
Integration by Parts
dxdx
duvuvdx
dx
dvu
2
)(2 xe
x
dx
e x
12
2
xu
1dx
dudifferentiate
integrate
xedx
dv 2and
dxexe xx
22
22
e.g. 2 Find dxxe x2
Solution:
v2
2 xe
Cexe xx
42
22
dxxe x2So,
This is a compound function, so we must be careful.
Integration by Parts
ExercisesFind dxxe x
1. dxxx 3cos2.
dxexe xx
Cexe xx
dxxx 3cos2. dxxx
x
3
3sin
3
3sin)(
Cxxx
9
3cos
3
3sin
1.
Solutions: dxxex
Integration by PartsDefinite Integration by PartsWith a definite integral it’s often easier to do the indefinite integral and insert the limits at the end.
1
0dxxe x
dxxe x dxexe xx
1
0xx exe
Cexe xx
0011 01 eeee
10 1
We’ll use the question in the exercise you have just done to illustrate.
Integration by Parts
Integration by parts cannot be used for every product.
Using Integration by Parts
It works if we can integrate one factor of the
product, the integral on the r.h.s. is easier* than
the one we started with.
* There is an exception but you need to learn the general rule.
Integration by Parts
The following exercises and examples are harder so you may want to practice more of the straightforward questions before you tackle them.
Integration by Parts
Solution:
What’s a possible problem?
Can you see what to do?
If we let and , we will need to
differentiate and integrate x.
xu ln xdx
dv
xln
ANS: We can’t integrate .
xln
xlnTip: Whenever appears in an integration by parts we choose to let it
equal u.
dxdx
duvuvdx
dx
dvue.g. 3 Find dxxx ln
Integration by Parts
dxdx
duvuvdx
dx
dvu
So, xu ln xdx
dv
xdx
du 1
2
2xv
dxxx ln xx
ln2
2
dxx
x 1
2
2
The r.h.s. integral still seems to be a product!BUT . . .
xx
dxxx ln2
ln2
4
2xC
x
cancels.
e.g. 3 Find dxxx ln
differentiate
integrate
So, xx
dxxx ln2
ln2
dxx
2
e.g. 3 Find xdxxln
Integration by Parts
e.g. 4 dxex x 2
Solution:
dxdx
duvuvdx
dx
dvu
Let 2xu xedx
dv
xdx
du2 xev
dxxeexdxex xxx 222
dxxeexdxex xxx 222
22
1 IexI x
The integral on the r.h.s. is still a product but using the method again will give us a simple function.We write
and
1I 2I
Integration by Parts
e.g. 4 dxex x 2
Solution:
2dx
duxev
22
1 IexI x
2ISo, xxe2 dxe x2
xxe2 dxe x2
Cexe xx 22Substitute in
( 1 )
. . . . . ( 1 )
xx exdxex 22
Let xu 2 xedx
dv and dxxeI x22
Cexe xx 22
Integration by Parts
dxdx
duvuvdx
dx
dvu
Solution:It doesn’t look as though integration by parts will help since neither function in the product gets easier when we differentiate it.
e.g. 5 Find dxxe x sin
However, there’s something special about the 2 functions that means the method does work.
The next example is interesting but is not essential. Click below if you want to miss it out.
Omit Example
Integration by Parts
dxdx
duvuvdx
dx
dvu
xeu xdx
dvsin
xedx
du xv cos
e.g. 5 Find dxxe x sin
dxxe x sin xe x cos dxxe x cos
xe x cos dxxe x cos
Solution:
We write this as: 21 cos IxeI x
Integration by Parts
dxdx
duvuvdx
dx
dvu
xeu xdx
dvcos
xedx
du xv sin
e.g. 5 Find dxxe x sin
xe x sin dxxe x sin
dxxeI x sin1where
So, 21 cos IxeI x
and dxxeI x cos2
We next use integration by parts for I2
dxxe x cos
12 sin IxeI x
Integration by Parts
dxdx
duvuvdx
dx
dvu
xeu xdx
dvcos
xedx
du xv sin
e.g. 5 Find dxxe x sin
dxxeI x sin1
xe x sin dxxe x sin
So,where
21 cos IxeI x
and dxxeI x cos2
We next use integration by parts for I2
dxxe x cos
12 sin IxeI x
Integration by Parts
dxdx
duvuvdx
dx
dvue.g. 5 Find dxxe x sin
So, 21 cos IxeI x
2 equations, 2 unknowns ( I1 and I2 )
!Substituting for I2 in
( 1 )
. . . . . ( 1 )
. . . . . ( 2 )
xeI x cos1
2I 1sin Ixe x
Integration by Parts
dxdx
duvuvdx
dx
dvue.g. 5 Find dxxe x sin
So, 21 cos IxeI x
2 equations, 2 unknowns ( I1 and I2 )
!Substituting for I2 in
( 1 )
. . . . . ( 1 )
. . . . . ( 2 )
xeI x cos1 1sin Ixe x
2I 1sin Ixe x
Integration by Parts
dxdx
duvuvdx
dx
dvue.g. 5 Find dxxe x sin
So, 21 cos IxeI x
2 equations, 2 unknowns ( I1 and I2 )
!Substituting for I2 in
( 1 )
. . . . . ( 1 )
. . . . . ( 2 )
xeI x cos1
xexeI xx sincos2 1
2
sincos1
xexeI
xx C
1sin Ixe x
2I 1sin Ixe x
Integration by PartsExercises
2.dxxx sin2( Hint: Although 2. is not a product it can be
turned into one by writing the function as . )
xln1
1. dxx2
1ln
Integration by PartsSolutions:
dxxx sin21.
xdx
du2 xv cos
2xu xdx
dvsinan
dLet
1I
dxxxxxI cos2cos21
dxxxxxI cos2cos21
2I
2dx
duxv sin
dxxxxI sin2sin22
Cxxx cos2sin2
Cxxxxxdxxx cos2sin2cossin 22
xu 2 xdx
dvcosan
dLetFor I2:
. . . . . ( 1 )
Subs. in ( 1 )
Integration by Parts
2.
xdx
du 1 xv
xu ln 1dx
dvand
Let
dxxln1 xx ln dxx
x 1
xx ln dx1
Cxxx ln
This is an important application of integration by parts
dxx2
1ln dxx
2
1ln1
dxx2
1lnSo, 2
1ln xxx
11ln122ln2 12ln2
Integration by Parts
There is a formula for integrating some products. I’m going to show you how we get the formula but it is tricky so if you want to go directly to the summary and examples click below.
Summary
Integration by Parts
dx
dvu
dx
duv
dx
dy
If ,uvy where u and v are
both functions of
x. Substituting for
y,
xxxdx
xxdcos1sin
)sin(
e.g. If , xxy sin
We develop the formula by considering how to differentiate products.
dx
dvu
dx
duv
dx
uvd
)(
Integration by Parts
dxxxdxxxx cossinsin
xxxdx
xxdcos1sin
)sin(So,
Integrating this equation, we get
dxxxdxxdxdx
xxd cossin)sin(
The l.h.s. is just the integral of a derivative, so, since integration is the reverse of differentiation, we get
Can you see what this has to do with integrating a product?
Integration by Parts
dxxxdxxxx cossinsin
Here’s the product . . .
if we rearrange, we get
dxxxxdxxx sinsincos
The function in the integral on the l.h.s. . . .. . . is a product, but the one on the r.h.s. . .
. is a simple function that we can integrate easily.
Integration by Parts
dxxxdxxxx cossinsin
Here’s the product . . .
if we rearrange, we get
dxxxxdxxx sinsincos
Cxxx )cos(sinCxxx cossin
We need to turn this method ( called integration by parts ) into a formula.
So, we’ve integrated ! xx cos
Integration by Parts
Integrating:
dxdx
dvudx
dx
duvdx
dx
uvd )(
dxdx
dvudx
dx
duvuv
dxdx
duvuvdx
dx
dvu
dx
dvu
dx
duv
dx
uvd
)(xxx
dx
xxdcossin
)sin(
dxxxdxxdxdx
xxd cossin)sin(
Rearranging:
Simplifying the l.h.s.:
dxxxdxxxx cossinsin
dxxxxdxxx sinsincos
Generalisation
Example
Integration by Parts
Integration by Parts
The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.
Integration by Parts
SUMMARY
dxdx
duvuvdx
dx
dvu
To integrate some products we can use the formula
Integration by Parts
Integration by Parts
dxdx
duvuvdx
dx
dvu
2
)(2 xe
x
dx
e x
12
2
xu
1dx
dudifferentiate
integrate
xedx
dv 2and
dxexe xx
22
22
e.g. Find dxxe x2
Solution:
v2
2 xe
Cexe xx
42
22
dxxe x2So,
Integration by Parts
Integration by parts can’t be used for every product.
Using Integration by Parts
It works if we can integrate one factor of the
product, the integral on the r.h.s. is easier* than
the one we started with.
* There is an exception but you need to learn the general rule.
Integration by Parts
differentiate
dxdx
duvuvdx
dx
dvu
We can’t integrate so,
xu ln xdx
dv
xdx
du 1
2
2xv
dxxx ln xx
ln2
2
dxx
x 1
2
2
The r.h.s. integral still seems to be a product!BUT . . .
xx
ln2
2
So, dxxx ln4
2x Cx cancels.
e.g. 3 Find dxxx ln
integrate
xln
Integration by Parts
2. dxxln dxx ln1
xdx
du 1 xv
xu ln 1dx
dvand
Let
dxxln1 xx ln dxx
x 1
xx ln dx1
Cxxx ln
This is an important application of integration by parts