Lecture note (last updated on July 6, 2010) Dr. Firoz

Week 1 (July 6-10) Chapter 5 and 6

Section 5.5 The Substitution Rule (Page # 293).

Complete the integrals with the given hints.

1. 2 3 2 31 , 1x x dx u x

2. sin 3 2cos , 3 2cosx x dx u x

3. 2sin,

xdx u x

x

4. sin cos , sinxe xdx u x

5. 20(3 2) , 3 2x dx u x

6. 1

, 5 35 3

dx u xx

7. sin ,xdx u x

8. 1

(1 )x

x x

x

edx e dx x e c

e

9. , 11

xx

x

edx u e

e

10. 2 2 2

1 1

1 1 1

x xdx dx dx

x x x

11. 2

4,

1

xdx u x

x

12. 2 2

0

cos( ) ,x x dx u x

Mat 266

Summer 2010

Section 6.1 Integration by parts (Page # 304)

In this section we need the following formula for integration by parts:

vduuvudv

Following formula is called the Fundamental theorem of calculus.

)()()( aFbFdxxf

b

a

Formulas to remember for this chapter:

xxdxx

xxdxxxdxxxdxxxdx

xxdxbababa

babababababa

xxxxxx

xxxxxx

xxxxxxx

csccotcsc

cotcsctansecsincoscossin

cossin))cos()(cos(2/1sinsin

))cos()(cos(2/1coscos))sin()(sin(2/1cossin

)3coscos3(4/1cos)3sinsin3(4/1sin

cot1csctan1seccos1sin

2sin2/1cossin)2cos1(2/1cos)2cos1(2/1sin

22

33

222222

22

Examples.

1. vduuvudvdxxxcos , when dxduxu

Cxxx

xdxxx

cossin

sinsin and vxdvxdx sincos

2. vduuvudvdxex x2 , when xdxduxu 22 , vedvdxe xx

Cexeex

dxexeex

vduvuex

dxxeex

xxx

xxx

x

xx

22

22

][

2

2

2

11

2

2

and also we consider dxduxu 22 11

3. vduuvudvdxxe x cos , when dxedueu xx

dxexexe

duvuvxe

dxxexe

xxx

x

xx

coscossin

][sin

sinsin

11 and vxdvxdx sincos also

xdxe x cos2 = xx xexe cossin + C consider xvxdxdv cossin 11

4. vduuvudvdxxe x sin , when dxedueu xx

Cxxe x )cos(sin2

1 and vxdvxdx cossin also

5. vduuvudvdppp ln5 , when pdpdupu /1ln

dpppp 56 6/1ln6/1 and vpdvdpp 65 6/1 also

Cppp 66 36/1ln6/1

6. dxx

x2

1

2

ln. We consider dx

x

x2

ln and dxxdmxm /1ln and also mex .

We have now dmmedme

mdx

x

x m

m2

ln. Now try to use integration by parts to find

that Cxxxdmmedxx

x m /1ln/1ln

2

2

12ln

2

11

2

12ln

2

11ln

1ln2

1

2

1

2 xx

xdx

x

x

OR: We consider dxx

x2

ln and v

xdvdx

xdxxdxxu

11,/1ln

2

Then we have x

xx

dxx

xx

vduuvudvdxx

x 1ln

11ln

1ln22

Now 2

12ln

2

11

2

12ln

2

11ln

1ln2

1

2

1

2 xx

xdx

x

x

7. vduuvudvdxx)2ln( dxxduxu /12ln

Cxxxdxxx 2ln2ln xvdxdv

8. xdxdxxarc 1tantan dxx

duxu2

1

1

1tan

udv xvdxdv

Cxxx

x

xxx

vduuv

)1ln(2/1tan

1tan

21

2

1

9. xdxdxxarc 1coscos dxx

duxu2

1

1

1cos

udv xvdxdv

Cxxx

x

xxx

vduuv

21

2

1

1cos

1cos

10. Prove the reduction formula 2,sin1

sincos1

sin 21 nxdxn

nxx

nxdx nnn

Let us consider xdxxnduxu nn cossin)1(sin 21 and

xvxdxdv cossin

Now

xdxn

nxx

nxdx

xdxnxxxdxn

dxxxnxx

xdxxnxx

xdxxnxx

vduuvudvxdxxxdx

nnn

nnn

nnn

nn

nn

nn

21

21

21

221

221

1

sin1

sincos1

sin

sin)1(sincossin

)sin(sin)1(sincos

sin)sin1()1(sincos

sincos)1(sincos

sinsinsin

11. Use the reduction formula in example 6 to prove that

22642

)12(531sin

2/

0

2

n

nxdxn

We have xdxn

nxx

nxdx nnn 21 sin

1sincos

1sin

Plug n2 for n: to get

xdxn

nxx

nxdx nnn 22122 sin

2

12sincos

2

1sin … … (1)

2/

0

22

222/

0

2/

0

1222/

0

sin2

12

sin2

12sincos

2

1sin

xdxn

n

xdxn

nxx

nxdx

n

nnn

… …. (2)

Now let us plug 22n , 42n , 62n , …3, 2 successively for 2n in (2): to get the

following results: 2/

0

42222/

0

sin22

32sin xdx

n

nxdx nn

2/

0

62422/

0

sin42

52sin xdx

n

nxdx nn

2/

0

82622/

0

sin22

32sin xdx

n

nxdx nn

2/

0

242/

0

sin4

3sin xdxxdx

22

1

2

1sin

2

1sin

2/

0

2/

0

022/

0

xxdxxdx

Finally from (2) we have the following result:

224)42)(22(2

13)52)(32)(12(sin

2/

0

2

nnn

nnnxdxn

12 . Cttttdtttvduuvudvtdtt 2sin4

12cos

2

12cos

2

12cos

2

12sin

where tvdvtdtdvtdtdtdutu 2cos2

12sin2sinand

13. 2 sinx xdx

Cxxx

xx

xdxxx

xx

xdxxxx

vduuvudvxdxx

cos2

sin2

cossin1

sin2

cos

cos2

cossin

32

22

22

where xvdvxdxdvxdxxdxduxu cos1

sinsinand22

14. 2/5)1/1()]12([][)1( 1

0

2

1

0

1

0

1

0

22 eexxeedxexdxex xxxx

15. dxenxexvduuvdxex xnxnxn 1

where xxxnn evdvdxedvdxedxnxduxu and1

Section 6.2 Trigonometric Integrals and Trigonometric Substitution (Page # 310)

Following trigonometric identities are useful:

)2cos1(2/1sinsin21

)2cos1(2/1cos1cos2

sincos2cos

2sin2/1cossincossin22sin

22

22

22

xxx

xxx

xxx

xxxxxx

)3coscos3(4/1coscos3cos43cos

)3sinsin3(4/1sinsin4sin33sin

33

33

xxxxxx

xxxxxx

Example 1 Evaluate Cxxdxxxxdx )3sin3/1sin3(4/1)3coscos3(4/1cos3

Or we could solve by substitution as

Cuuduuxdxxxdxxxdx 32223 3/1)1(cos)sin1(coscoscos

where xdxduxu cossin

Thus we have Cxxxdx 33 sin3/1sincos .

Our results are equivalent.

Example 2 Evaluate Cuuuduuuuxdxx 9/7/25/)2(sincos 97586445

Where xdxduxu cossin and 22224 )sin1()(coscos xxx

Example 3 Evaluate dxx

2/

0

5cos

2/

0

2/

0

224

2/

0

5 cos)sin1(coscoscos xdxxxdxxdxx

We have

duuuduuxdx )21()1(cos 42225 , where uxsin . Do the rest!

Example 4

3/

6/

3csc xdx

7825.1csccotcotcsc(ln2/1csc

cotcsclncsccot

cotcsc

)cot(csccsccsccotcsccsccotcsc2

csccsccsccot)1(csccsccsccot

cotcsccsccotcsccsccsc

3/

6/

3/

6/

3

3

32

223

xxxxdxx

xxxx

dxxx

xxxxxdxxxdxx

xdxdxxxdxxxx

dxxxxxvduuvudvxdxxxdx

where

xvxdxdv

xdxxduxu

cotcsc

cotcsccsc

2

Example 5 xdxx 6sin3sin . Use formula )cos()cos(sinsin2 BABABA

Thus we have sin3 sin 6 1/ 2(cos3 cos9 ) 1/ 6sin3 1/18sin9x xdx x x dx x x C

Example 6

Cuumduum

dxmxmxdxmxmxdxmx )3/(/1)1(1

)sin())(cos1()sin()(sin)(sin 32223

where dxmxmdumxu )sin()cos(

Example 7 )4(sin8/12

)4cos1(2/1)2(sin)(sin 22 dddxmx

2/

0

2

42sin d , because

where 2/or0when04sin

Example 8. 097.01

12

223

dttt

xdxdxxx

xxdt

tt

2

2323cos

1secsec

tansec

1

1 , where tx sec

3/then,2,4/then2hen xtxtW

Example 9 dxxx

2

0

23 4

uduuuduudxxx 3322323 sectan2)sec2(4tan4tan4 , where ux tan

Next you plug um sec and compete the integral

Example 10 dxx

ax4

22

uduua

uduuaua

uadx

x

axcossin

1tansec

sec

tan 2

2444

22

, where uax sec

Example 11 duuu 25

1

xdxxdxxx

duuu

csc5

1cos5

sin55.sin5

1

5

1

22, where xu sin5

Example 12. dxxx

3/2

0

23 94

uduuuduuudxxx 232323 cossin81/32)cos3/2(sin44)sin3/2(94

where ux sin3/2

OR: 294 xu , duuudxxx )4(164/194 2/32/123

Example 13. dxxx

x

2

2

4

dxx

xdx

xx

xdx

xx

x

2

2

2

2

2

2

)2(4)44(44

Next put 2xu

Example 14. dxax

x2/322

2

)(

a) put ,sec,tan 2 uduadxuax

duu

udx

ax

x

sec

tan

)(

2

2/322

2

b) put ,cosh,sinh tdtadxtax

Cttdtthtdtdxax

xtanh)sec1(tanh

)(

22

2/322

2

, now substitute back what is t.

Section 6.3 Integration of rational functions by partial fraction (Page# 320)

Review the following rules

1. a) 13

4)1)(3(

12134

342

3

x

B

x

Ax

xx

xx

xx

x

b) 4)4(1)1()1()4()1(

1222223223 x

GFx

x

EDx

x

C

x

B

x

A

xx

x

2. )()())((

1

bx

B

ax

A

bxax

baB

baAaxBbxA

1,

1)()(1

Cax

bx

badx

bxbadx

axbadx

bxaxln

1

)(

11

)(

11

))((

1

3. 11

123

2

x

C

x

B

x

A

xx

xx, find that 1,1,1 CBA

4. 15)5()1()5(

122 x

C

x

B

x

A

xx, find that 36/1,36/1,6/1 CBA

5. 2 2 2 2

1

( 1) 1 ( 1)

A B C D

s s s s s s, find that 1,2,1,2 DCBA

6. 33

623

2

x

CBx

x

A

xx

xx, find that 1,1,2 CBA

7. 1)1(1)1()1(

122222

2

x

DCx

x

B

x

A

xx

xx, find that 1,1,1,1 DCBA

8 xdxx 1tan , proceed with integration by parts dvxdxxu ,tan 1

9 duu

duu

udu

u

udx

x

xdx

xx

x

16

1

162

1

2

1

16

2

16)32(

12

7124

1222222

which is performed by first completing square and then substitution method for the first part and apply

formula (6) for the second part.

This problem also could be solved (if method is not mentioned):

dxxx

xdx

xx

x

)12)(72(

12

7124

122

, and then use partial fraction.

10. duu

dxx

x

24 3

1

2

1

3, where

2xu

Now apply formula a

xdx

xa

1

22sin

1

11. xdxxx cotcsc

Precede using integration by parts xdxxdvxu cotcsc,

12. rdrr ln4

Proceed using integration by parts drrdvru 4,ln . Do not forget to find definite integral result.

13.

4

0

2 54

1dx

xx

x, Factor the denominator and apply integration by partial fraction.

14. dxxx

x

ln

ln1, proceed by substitution xu ln12

.

15.

4

0

2 54

1dx

xx

x, Factor the denominator and apply integration by partial fraction.

16.

2/2

02

2

1dx

x

x, substitute ux sin

17. dxxx

xdx

xx

x

)2)(4(

2263

82

232

2

by long division. Now use partial fraction.

18. dxxdxxxdxxx 222 )1(4)2(323 , use substitution ux sin21

19.

2/

4/

2/

4/cossin4

cos4sin

cot4

cot41dx

xx

xxdx

x

x, substitution xxu cossin4

Section 6.4 Integration Using Tables and computer Algebra Systems (Page# 328)

Tables of indefinite integrals are very useful when we are confronted by an integral that is difficult by hand

or very lengthy in calculation and we do not have access to a computer algebra system. We have a list of

few such integrals on the reference pages at the end of our text book. Those formulas we do not need to

memorize, rather those will be provided on the test. In this section we will learn how to use those given

formulas.

Example 1. dzz

z2

3 )/1(tan

This problem we could solve without using formula 69. Just substitute dzz

dxzx2

1/1

33 2 2

2

tan (1/ ) 1tan tan (sec 1) tan ln sec

2

zdz xdx x x dx x x C

z,

for detail work look at example 7, page number 314.

Example 2.

3

222 74

1dx

xx, substitute dxduxu 22

u

udu

uudx

xx 7

7

7

12

74

1 245

2222, 45 above equal sign means we are using formula

45 from our textbook. You need to evaluate the result at the given limits.

Example 3. dxxxx )3cos()sin( 22, substitute xdxduxu 22

duuuuduudxxxx )2sin4(sin4/13cossin2/1)3cos()sin( 22, using the

trigonometric formula xxyxyxyx sin)sin()],sin()[(sin(2/1cossin . Or it could

be solved just using formula 81.

Example 4.

2

0

423 4 dxxxx , substitute xdxduxu 22

2

2cos

4

84

12

1222)2(22/14 12

21142423 u

uuuu

duuuudxxxx

Example 5. Cxduuudxxx 534432 )1(15/1)(3/1)1( , but if you do it in Maple you

may have result like Cxxxxxdxxx 3691215432

3

1

3

2

3

2

3

1

15

1)1( . The results are

equivalent one can check using binomial expansion.

Example 6. Cx

x

x

xxdxx

3

3

5

342

cos15

sin2

cos5

sinsectan , by Maple

By hand one can substitute xdxduxu 2sectan , then

Cxxduuuxdxx 532242 tan5

1tan

3

1)1(sectan . The results are equivalent if you

substitute )cos1(sinsin 235 xxx

Section 6.5 Approximate Integration (Page# 333)

There are two situations in which we can not find exact value of a definite integral or it is impossible to

evaluate. The fisr situation arises from the fact that in order to evaluate the definite integral

dxxf

b

a

)( using the fundamental theorem of calculus we do not know the antiderivative of f(x). We have

seen some examples in section 7.5, where our integrals are rather difficult. Also we have some cases where

the integrals are impossible like dxxdxe x

1

1

5

1

0

1,2

. Second we may have some practical problem

which does not fit with any formulas or simplifications. In this section we will use only three methods.

Mid Point Rule: )()(1

n

i

in

b

a

xfxMdxxf , where step size =n

abx and

2

1 iii

xxx , also bxxxxxxxax n,,,, 12010

The error estimate is 2

3

24

)(

n

abkEM

where kxf )( for bxa

Trapezoidal Rule: )()()(22

)(1

1

0

n

i

nin

b

a

xfxfxfx

Tdxxf , where

step size =n

abx , bxxxxxxxax n,,,, 12010

The error estimate is 2

3

12

)(

n

abkET

where kxf )( for bxa

Simpson’s Rule:

)()(2)(4)(2)(4)(3

)( 43210 nn

b

a

xfxfxfxfxfxfx

Sdxxf , where

step size =n

abx , bxxxxxxxax n,,,, 12010

The error estimate is 4

5

180

)(

n

abkES

where kxf )()4(for bxa

Note: nnn MTS3

2

3

12

Example 1. Approximate the integral dxx

2

1

1 for 5n using

a) Mid point rule ( 5n ) b) Trapezoidal rule ( 5n ) and c) Simpson’s rule (n = 10)

d) Find error estimations in each rule. e) Verify that nnn MTS3

2

3

12

a) Mid point rule )()(1

n

i

in

b

a

xfxMdxxf , 2,1,2.05

1250 xx

n

abx

691908.0))9.1()7.1()5.1()3.1()1.1()1((2.01

2

1

ffffffdxx

Midpoint rule:

b) Trapezoidal rule )()()(22

)(1

1

0

n

i

nin

b

a

xfxfxfx

Tdxxf

695635.0))2()8.1(2)6.1(2)4.1(2)2.1(2)1((2

2.012

1

ffffffdxx

c) Simpson’s rule

)()(2)(4)(2)(4)(3

)( 43210 nn

b

a

xfxfxfxfxfxfx

Sdxxf

693150.0))2()4.1(2)3.1(4)2.1(2)1.1(4)1((3

1.012

1

ffffffdxx

d) Error bounds for 5n ; 5)4(3 /24)(,/2)( xxfxxf

0033.0)5(24

)12(2

24

)(2

3

2

3

n

abkEM where 2/2)( 3 kkxkxf for 21 x

0067.0)5(12

)12(2

12

)(2

3

2

3

n

abkET

00002.0)10(108

)12(24

180

)(4

5

4

5

n

abkES where 24/24)( 5)4( kkxkxf for

21 x

e)

)6919.0(3/2)6956.0(3/16931.03/23/13

2

3

155102 MTSMTS nnn which

is true.

Example 2. How large should we take n in order to guarantee that the trapezoidal rule midpoint rule

approximations of dxx

2

1

1 are accurate to within 0.0001?

From example 1 we have k = 2.

299.280001.0)(24

)12(2

24

)(2

3

2

3

nnnn

abkEM and

418.400001.0)(12

)12(2

12

)(2

3

2

3

nnnn

abkET

89.60001.0)(108

)12(24

180

)(4

5

4

5

nnnn

abkES (because n must be even)