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Section 6.3 – Antidifferentiation by Parts Remember the Product Rule for Derivatives: By reversing this derivative formula, we obtain the integral formula
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CHAPTER 6:DIFFERENTIAL EQUATIONS AND
MATHEMATICAL MODELINGSECTION 6.3:
ANTIDIFFERENTIATION BY PARTS
AP CALCULUS AB
What you’ll learn about
Product Rule in Integral FormSolving for the Unknown IntegralTabular IntegrationInverse Trigonometric and Logarithmic
Functions
… and whyThe Product Rule relates to derivatives as the
technique of parts relates to antiderivatives.
Section 6.3 – Antidifferentiation by Parts
Remember the Product Rule for Derivatives:
By reversing this derivative formula, we obtain the integral formula
dv d duu dx uv v dxdx dx dx
duuv v dxdx
d dv duuv u vdx dx dx
Section 6.3 – Antidifferentiation by Parts
Integration by Parts Formula
udv uv vdu
Section 6.3 – Antidifferentiation by Parts
Example:
cos
Let and costhen and sin
So, cos sin sin
sin cos sin cos
x x dx
u x dv xdxdu dx v x
x x dx x x xdx
x x x C
x x x C
Example Using Integration by Parts
Evaluate sin .x xdx
Use - with sin -cos
sin - cos cos - cos sin
udv uv vduu x dv xdxdu dx v x
x xdx x x xdxx x x C
Section 6.3 – Antidifferentiation by Parts
Try to pick for your u something that will get smaller, or more simple when you take the derivative.
Sometimes, you may have to integrate by parts more than one time to get to something you can integrate.
Sometimes, integration by parts does not work.
Section 6.3 – Antidifferentiation by Parts
Repeated use2
2
2 2
2
cos
Let and costhen 2 sin
So, cos sin 2 sin
Repeating the processlet 2 and sinthen 2 cos
And cos
x xdx
u x dv xdxdu xdx v x
x xdx x x x xdx
u x dv xdxdu dx v x
x xdx x
2
2
2
sin 2 cos 2cos
sin 2 cos 2 cos
sin 2 cos 2sin
x x x xdx
x x x x xdx
x x x x x C
Example Repeated Use of Integration by Parts
2Evaluate 2 .xx e dx
2
2 2
2 2
2
Let 2 4 2 2 4
Now let 2 2 4
2 4 4
x
x
x x x
x
x
x x x x
x x x
u x dv e dxdu xdx v ex e dx x e xe dx
u x dv e dxdu dx v e
x e dx x e xe e dxx e xe e C
Example Antidifferentiating ln x
Find ln .xdx
Let ln 1
1ln ln -
ln - ln -
u x dv dx
du dx v xx
xdx x x x dxx
x x dxx x x C
Section 6.3 – Antidifferentiation by Parts
Solving for the Unknown IntegralSometimes, when performing this process, we seem to get a circular argument going. When this occurs, collect your like terms, and you can solve for the unknown integral.
Example: cos
Let and cos
then sin
So, cos sin sin
Repeating the process
let and sin
then cos
And cos
x
x
x
x x x
x
x
x
e xdx
u e dv xdx
du e dx v x
e xdx e x e xdx
u e dv xdx
du e dx v x
e x
sin cos cos
sin cos cos
Adding the like integrals together we get
2 cos sin cos sin cos
So, cos sin cos2
x x x
x x x
x x x x
xx
dx e x e x e xdx
e x e x e dx
e xdx e x e x e x x
ee dx x x C
Example Solving for the Unknown Integral
Evaluate sin .xe xdx
Let sin -cos
sin cos cos Now let cos sin
sin cos sin sin 2 sin cos sin
sin
x
x
x x x
x
x
x x x x
x x x
x
u e dv xdxdu e dx v xe xdx e x e xdx
u e dv xdxdu e dx v x
e xdx e x e x e xdxe xdx e x e x
e xdx
cos sin 2
x xe x e x C
Section 6.3 – Antidifferentiation by Parts
Tabular IntegrationWhen many repetitions of integration by parts is needed, there is a way to organize the calculations that saves a great deal of work and confusion. It is called tabular integration.
Section 6.3 – Antidifferentiation by Parts
Example: 3 xx e dx3
2
and its derivatives and its integrals
3
6
x
x
u dv
x e
x e
x
6
0
x
x
x
e
e
e
(+)
(-)
(+)
(-)
3 3 2So 3 6 6x x x x xx e dx x e x e xe e C
