ntegration by Parts, Reduction Formulae

Remembertheproductrule:
(uv)? = u?v + uv?
Wecanrewritethatas
uv? =(uv)? ? u?v
Integratethistogettheformulaforintegrationbyparts:
uv? dx = uv ? u?v dx
Example 1. tan?1 xdx.
At first, it’s not clear how integration by parts helps. Write
tan?1 xdx = tan?1 x(1 dx)= uv? dx
·
with u = tan?1 x and v? =1.
Therefore,
1
v = x and u? =
1+ x2
Plugalloftheseintotheformulaforintegrationbypartstoget:
1
tan?1 xdx = uv? dx = (tan?1 x)x ? 1+ x2 (x)dx
= x tan?1 x ? 12 ln |1+ x2| + c
Alternative Approach to Integration by Parts
Asabove,theproductrule:
(uv)? = u?v + uv?
canberewrittenas
uv? =(uv)? ? u?v
Thistime,let’stakethe definite integral:
? b ? b ? b
uv? dx =(uv)? dx ? u?v dx
aa a
1
? ?
?
?
Lecture 30 18.01 Fall 2006
Bythefundamentaltheoremofcalculus,wecansay
? b ?? b
?b
uv? dx = uv? u?v dx
a ?
aa
Anothernotationintheindefinitecaseis
udv = uv ? v du
Thisisthesamebecause
dv = v? dx = uv? dx = udv and du = u? dx = u?v dx = vu? dx = v du
??
Example 2. (ln x)dx 1
u = ln x; du = dx and dv = dx; v = x
x
? ??? ?
1
(ln x)dx = x ln x ? x dx = x ln x ? dx = x ln x ? x + c
x
We can also use “advanced guessing” to solve this problem. We know that the derivative of something equals ln x:
d
(??) = ln x
dx
Let’stry
d 1
(x ln x) = ln x + x = ln x +1
dx · x
That’s almost it, but not quite. Let’s repair this guess to get:
d
(x ln x ? x) = ln x +1 ? 1 = ln x
dx
Reduction Formulas (Recurrence Formulas)
Example 3. (ln x)n dx
Let’stry: ??
1
u = (ln x)n = u? = n(ln x)n?1
? x v? = dx; v = x
Pluggingtheseintotheformulaforintegrationbypartsgivesus:
? ??? 1
1??
(ln x)ndx = x(ln x)n n(ln x)n?1x ?dx
? ?x Keeprepeatingintegrationbypartstogetthefullformula: n (n?1) (n?2) (n?3) etc
? ????
Example 4. xnex dx Let’stry:
u = xn = u? = nxn?1; v? = ex = v = ex
??
2
? ?
?
?
?
Lecture 30 18.01 Fall 2006
Puttingtheseintotheintegrationbypartsformulagivesus:
n nx
xex dx = xe nxn?1ex dx
?
Repeat,goingfrom n ? (n ? 1) ? (n ? 2) ? etc.
Bad news: Ifyouchangetheintegralsjustalittlebit,theybecomeimpossibletoevaluate:
??2
tan?1 x dx = impossible
x
e
dx = alsoimpossible
x
Good news: Whenyoucan’tevaluateanintegral,then
? 2 x
e
dx
1 x
isan answer, not a question. This is thesolution–youdon’thavetointegrateit!
The most important thing is setting up the integral! (Once you’ve done that, you can always evaluate it numerically on a computer.) So, why bother to evaluate integrals by hand, then? Because youoftengetfamiliesofrelatedintegrals,suchas
x
? e
F(a)= dx
xa
1
whereyouwanttofindhowtheanswerdependson,say, a.
3