CHAPTER 6: DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELING SECTION 6.3: ANTIDIFFERENTIATION BY...

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