8.2 integration by parts. formula for integration by parts the idea is to use the above formula to...
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8.2 Integration by parts
Formula for Integration by parts
( ) '( ) ( ) ( ) ( ) '( )
f x g x dx f x g x g x f x dx
udv uv vdu
• The idea is to use the above formula to simplify an integration task.• One wants to find a representation for the function to be integrated in the form udv so that the function vdu is easier to integrate than the original function.
• The rule is proved using the Product Rule for differentiation.
Deriving the Formula
Start with the product rule:
d dv duuv u v
dx dx dx
d uv u dv v du
d uv v du u dv
u dv d uv v du
u dv d uv v du
u dv d uv v du
u dv uv v du This is the Integration by Parts formula.
u dv uv v du
u differentiates to zero (usually).
dv is easy to integrate.
Choose u in this order: LIPET
Logs, Inverse trig, Polynomial, Exponential, Trig
Choosing u and v
Example 1:
cos x x dxpolynomial factor u x
du dx
cos dv x dx
sinv x
u dv uv v du LIPET
sin cosx x x C
u v v du
sin sin x x x dx
Example 2:
ln x dxlogarithmic factor lnu x
1du dx
x
dv dx
v x
u dv uv v du LIPET
lnx x x C
1ln x x x dx
x
u v v du
This is still a product, so we need to use integration by parts again.
Example 3:2 xx e dx
u dv uv v du LIPET
2u x xdv e dx
2 du x dx xv e u v v du
2 2 x xx e e x dx 2 2 x xx e xe dx u x xdv e dx
du dx xv e 2 2x x xx e xe e dx
2 2 2x x xx e xe e C
Example 4:
cos xe x dxLIPET
xu e sin dv x dx xdu e dx cosv x
u v v du sin sinx xe x x e dx
sin cos cos x x xe x e x x e dx
xu e cos dv x dx xdu e dx sinv x
sin cos cos x x xe x e x e x dx This is the expression we started with!
uv v du
Example 4(cont.):
cos xe x dxLIPET
u v v du
cos xe x dx 2 cos sin cosx x xe x dx e x e x
sin coscos
2
x xx e x e x
e x dx C
sin sinx xe x x e dx xu e sin dv x dx
xdu e dx cosv x
xu e cos dv x dx xdu e dx sinv x
sin cos cos x x xe x e x e x dx
sin cos cos x x xe x e x x e dx
This is called “solving for the unknown integral.”
It works when both factors integrate and differentiate forever.
Integration by Parts for Definite Integrals
12
0
arcsin x dx
b b b
b
aa a a
udv uv vdu u b v b u a v a vdu
2
Choose arcsin . Then
, and .1
u x
dxdv dx v x du
x
Formula
Integration by Parts Formula and the Fundamental Theorem of Calculus imply the above Integration by Parts Formula for Definite Integrals. Here we must assume that the functions u and v and their derivatives are all continuous.
Example
1 12 21
2
0 20 0
arcsin arcsin1
xdxx dx x x
x
Integration by Parts for Definite Integrals
12
2
20
Compute by the substitution 1 , 2 .1
1 30 1 and
2 4
xdxt x dt xdx
x
x t x t
331 42 4
20 1 1
3 2 31
2 221
xdx dtt
tx
Example (cont’d)
1 1 12 2 21
212
0 2 20 0 0
arcsinarcsin arcsin
21 1
xdx xdxx dx x x
x x
By the computations on the previous slide we now have
12 1
2
0
arcsin 2 3 2 3arcsin
2 2 12 2x dx
Combining these results we get the answer