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Drill: Find dy/dx y = x 3 sin 2x y = e 2x ln (3x + 1) y = tan -1 2x Product rule: •x 3 (2cos 2x) + 3x 2 sin (2x) • 2x 3 cos 2x + 3x 2 sin (2x) Product Rule •e 2x (3/(3x +1)) + 2e 2x ln (3x + 1) 3e 2x /(3x +1) + 2e 2x ln (3x + 1) 2 2 4 1 2 2 1 2 x x

# Drill: Find dy / dx

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Drill: Find dy / dx. y = x 3 sin 2x y = e 2x ln (3x + 1) y = tan -1 2x Product rule: x 3 (2cos 2x) + 3x 2 sin (2x) 2x 3 cos 2x + 3x 2 sin (2x). Product Rule e 2x (3/(3x +1)) + 2e 2x ln (3x + 1) 3e 2x /(3x +1) + 2e 2x ln (3x + 1). Antidifferentiation by Parts. Lesson 6.3. - PowerPoint PPT Presentation

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Drill: Find dy/dx

• y = x3 sin 2x

• y = e2x ln (3x + 1)

• y = tan-1 2x

• Product rule:• x3 (2cos 2x) + 3x2 sin (2x)• 2x3 cos 2x + 3x2 sin (2x)

• Product Rule• e2x (3/(3x +1)) + 2e2x ln (3x + 1)• 3e2x/(3x +1) + 2e2x ln (3x + 1)

22 41

2

21

2

xx

Antidifferentiation by Parts

Lesson 6.3

Objectives

• Students will be able to:– use integration by parts to evaluate indefinite and definite

integrals.– use rapid repeated integration or tabular method to evaluate

indefinite integrals.

Integration by Parts Formula

A way to integrate a product is to write it in the form

If u and v are differentiable function of x, then

.functionanother of aldifferentifunction one

. duvuvdvu

Example 1 Using Integration by Parts

Evaluate .cos dxxx xdvxu cos,Let ,dxdu xdxv cos

xv sin duvuvdvu

xdxxxxdxx sinsincos

Cxxx cossin

Example 1 Using Integration by Parts

Evaluate .5 3 dxxe x dxedvxu x3,5Let ,5dxdu dxev x3

xev 3

3

1

duvuvdvu

dxeexdxxe xxx 5

3

1

3

155 333

dxexe xx 33

3

15

3

5C

exe

xx

)

33(5

3

5 33

Ce

xex

x 9

5

3

5 33

Example 1 Using Integration by PartsEvaluate .15cos dxxx

dxxdvxu 15cos,Let

dxxvdxdu 15cos, 15sin5

1 x

duvuvdvu

dxxxxdxxx 15sin

5

115sin

5

115cos

Cxxx 15cos

5

1

5

115sin

5

1

Cxxx )15cos(25

115sin

5

1

Example 2 Repeated Use ofIntegration by Parts

Evaluate .2 dxex xdxedvxu x ,Let 2

dxevdxxdu x,2xev duvuvdvu

dxxeexdxex xxx 222

dxxeex xx 22

dxxeex xx 22

dxedvxu x ,Let dxevdxdu x, xev

)(22 dxexeex xxx

)(22 Cexeex xxx

Cexeex xxx 222

Example 2 Repeated Use ofIntegration by Parts

Evaluate .sin2 dxxx dxxdvxu sin,Let 2

dxxvdxxdu sin,2

xv cos duvuvdvu

dxxxxxdxxx 2coscossin 22

dxxxxx cos2cos2

xdxxxx cos2cos2

dxxdvxu cos,Let

dxxvdxdu cos, xv sin

)sinsin(2cos2 xdxxxxx

))cos(sin(2cos2 Cxxxxx

Cxxxxx cos2sin2cos2

Example 3 Solving an Initial Value Problem

• Solve the differential equation dy/dx = xlnx subject to the initial condition y = -1 when x = 1

.ln xdxx xdxdvxu ,lnLet

dxxvdxx

du ,1

2

2xv

It is typically better to let u = lnx

dxx

xxxdvu

1

2)

2(ln

22

dxxxx

2

1ln)

2(

2

Cx

xx

22

1ln)

2(

22

Cx

xx

4

ln)2

(22

C4

11ln)

2

1(1

22

C4

101

C4

3

4

3

4ln)

2(

22

xx

xy

DrillSolve the differential equation: dy/dx = x2e4x (This means you will need to find the anti-derivative of dy/dx = x2e4x )

.42 dxex xxedvxu 42 ,Let

dxevdxxdu x4,2

4

4xev

xdxee

xdvuxx

244

442

dxxeex x

x4

42

2

1

4

dxxeex x

x4

42

2

1

4

xedvxu 4,Let

dxevdxdu x4,

4

4xev

dx

eex

ex xxx

442

1

4

4442

C

exeex xxx

1642

1

4

4442

C

exeex xxx

3284

4442

Cexeex xxx

3284

4442

Example 4Solving for the unknown integral

xdxe x cos xdxdveu x cos,Let

dxxvdxedu x cos,

xv sin xdxexedvu xx sinsin

xdxexexe xxx sinsincosxdxdveu x sin,Let

dxxvedu x sin,xv cos

xdxexexexe xxxx coscossincos

xdxexexexe xxxx coscossincos

xexexexe xxxx coscossincos

xdxexexexe xxxx coscossincos

xexexe xxx cossincos2 C

xexexe

xxx

2

cossincos

Rapid Repeated Integration by PartsAKA: The Tabular Method

• Choose parts for u and dv.• Differentiate the u’s until you have 0.• Integrate the dv’s the same number of times.• Multiply down diagonals.• Alternate signs along the diagonals.

Example 5 Rapid Repeated Integration by Parts

Evaluate

u and its derivatives dv and its integrals

.4sin2 dxxx

2x x4sin

x2

2

0

4

4cos x

16

4sin x

64

4cos x

Cxx

xx

x

64

4cos2

16

4sin2

4

4cos2

Example 5 Rapid Repeated Integration by Parts

Evaluate .4sin2 dxxx Cxx

xx

x

64

4cos2

16

4sin2

4

4cos2

Cxxxxx 4cos32

14sin

8

14cos

4

1 2

Example 5 Rapid Repeated Integration by Parts

Evaluate

u and its derivatives dv and its integrals

.2 dxex x

2x xe

x2

2

0

xe

xe

xe

Ceexex xxx 222

Example 5 Antidifferentiating ln x

xdxln dxdvxu ,ln

xvdxx

du ,1

dxx

xxxxdx 1

lnln

dxxxxdx 1lnln

Cxxxxdx lnln

Example 6 Antidifferentiating sin-1 x

xdx1sin dxdvxu ,sin 1

xvdxx

du

,1

12

dxx

xxxxdx

2

11

1

1sinsin

dxx

xxx

2

1

1sin

dxx

xxx

2

1

1sin

dxx

duxdxduxu

2,2,1 2

x

du

u

xxx

2sin 1

uduxxduu

xx

2

111

2

1sin

1

2

1sin

CuxxCu

xx

2

11

2

1

1 sin

212

1sin

Cxxx 2

121 )1(sin

Homework

• Page 346/7: Day #1: 1-15 odd• Page 347: 17-24