Calculus 2 Chapter 6(4)

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INTEGRATION TECHNIQUES

6.2: INTEGRATION BY PARTS

Let u = f (x) and v = g(x), are two differentiable functions, the

integration by parts rule given as follow:

To apply integration by parts, you need to make a careful

choice of u and dv so that the integral on the right-hand side

is one that you know how to evaluate.

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Example 1:

Solution:


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Realize that the choice of u and dv is critical.



A Poor Choice of u and dv, Let us choose

So the integration

The last integral is one that we do not know how to

calculate any better than the original one. Prepared by Dr. F.G.A

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Ex 2:

Ex 3:

Ex 4:

Ex 5:

Ex 6:

Ex 7:


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cosx xdx2xxe dx

2 lnx xdx

1tan xdx



Remark: For any positive integer n, the integral will require integration by parts.

At this point, we take

Ex 8: Evaluate the integral

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2 xx e dx

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Definite Integrals: For the definite integral we have the following formula: Ex 9: Solution:


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Ex 10:

Exercises: evaluate the following:

1-

2-

3-

4- 5-


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1

0

sin 2x x dx

cosxe xdx

sin x dxxe dx

sin ln x dx1secx xdx


6.3: TRIGONOMETRIC TECHNIQUES OF INTEGRATION

Consider integrals of the form

where m and n are positive integers. We have three cases:

Case 1: m Odd Positive Integer

If the power of sine is odd, save one sine factor and use

to express the remaining in term of cosine. Ex 1: Sol: re write the integral to be


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2 2sin 1 cosx x

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Let


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4 2cos sin sinx x xdx



Ex 2:

Ex 3:


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2 5cos sinx x dx

22

0

cos sinx x dx

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Case 2: n Odd Positive Integer

If the power of cosine is odd, save one cosine factor and use

to express the remaining in term of sine:

Ex 4:

Sol:


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2 2cos 1 sinx x




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Ex 5:

Case 3: Both m and n are even

If the power of both sine and cosine are even, use the half-

angle identities

It is sometimes helpful to use the identity


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21

cos 1 cos 22

x x 21

sin 1 cos 22

x x

1sin cos sin 2

2x x x

4cos sinx xdx



Ex 6:


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Ex 7:

Ex 8:

Ex 9:


1- 2-

3-


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2 2cos sinx xdx2 4cos sinx xdx

4cos xdx

3 4cos sinx x dx

3cos sinx x dx

3 2sin cosx x dx




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Solution Method Power of

Cos

Power of

Sin

Case 1: save (sine) factor. Use

and

even odd

Case 2: save (cosine) factor. Use

and

odd even

Case 3: Use the half angle identities

and

even

even

Any method of Case I or Case 2 odd

odd

2 2sin 1 cosx x

cosu x

sindu xdx

sinu x

cosdu xdx 2 2cos 1 sinx x

21

cos 1 cos 22

x x 21

sin 1 cos 22

x x

1sin cos sin 2

2x x x

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To evaluate the integrals

1-

2-

3-

Use the corresponding identity


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sin cosmx nxdx

sin sinmx nxdxcos cosmx nxdx

11 sin cos sin sin

2

12 sin sin cos cos

2

13 cos cos cos cos

2

A B A B A B

A B A B A B

A B A B A B



Ex 10

Solution:

Ex11:

Ex 12:


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sin 4 cos5x xdx

1 1

sin 4 cos5 sin sin 9 sin sin 92 2

1 1cos cos9

2 9

x x dx x x dx x x dx

x x c

sin 4 cos 2x xdx

cos3 cos 2x xdx

1

sin cos sin sin2

A B A B A B

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To evaluate the integral

m and n are integers

Case 1: m Is an Odd Positive Integer

If the power of tangent is odd, save a factor of

and use to express the remaining factors

in terms of then substitute

Ex 13:

Sol:


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sec tanx x2 2tan sec 1x x sec x sec .u x




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Case 2: n Is an Even Positive Integer

If the power of the secant is even, save a factor of

and use to express the remaining factors

in term of then substitute

Ex 14:

Sol:


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tan x

2 2sec 1 tanx x

2sec x

tan .u x




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Case 3: m Is an Even Positive Integer and n Is an Odd

Positive Integer

Replace any factors of with and then use

a special reduction formula to evaluate integrals of the

form

Ex 15:

Sol: Notice that if we multiply the integrand by the fraction


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2tan x 2sec 1x



we get

The numerator is exactly the derivative of the denominator.

Let


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sec

Power of

tan

Case 1: save (sec x tan x) factor. Use

and

odd

odd

Case 2: save ( ) factor. Use

and

even even

Case 3: Replace any factors of with

and then use a special reduction

formula

odd

even

Use Case 1 or Case 2

Note: Use case 2 only when you have

otherwise its better to use case 1

even

odd

2 2tan sec 1x x

secu x

sec tandu x xdx

tanu x2secdu xdx

2sec x2 2sec 1 tanx x

2tan x2sec 1x

2sec x



To evaluate we can use the same techniques

whish were used to evaluate we just have to

replace cot x wit tan x and csc x with sec x.

Ex: Evaluate

1-

2-

3-


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cot cscm nx x dxtan secm nx x dx

3 3cot cscx x dx

2 2cot cscx x dx2cot cscx x dx

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csc

Power of

cot

Case 1: save (csc x cot x) factor. Use

and

odd

odd

Case 2: save ( ) factor. Use

and

even even

Case 3: Replace any factors of with

and then use a special reduction

formula

odd

even

Use Case 1 or Case 2

Note: Use case 2 only when you have

otherwise its better to use case 1

even

odd

2 2cot csc 1x x

cscu x

csc cotdu x xdx

cotu x2cscdu xdx

2csc x2 2csc 1 cotx x

2cot x2csc 1x

2csc x


6.3: Trigonometric Substitution

If an integral contains a term of the form

for some a > 0, you can often evaluate the integral by making

a substitution involving a trig function (hence, the name

trigonometric substitution).

The substitution are:

1- For use


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2 2a x sinx a

2 2a x

ax

2 2

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Ex 16:

Sol:


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2sinx

2

2

cos 1 1csc cot

4sin cos 4 4d d c

2 2 2 22cos cos

4sin 4 4sin 2sin 2 1 sind d




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2- For use

Ex 17:

Sol:


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a

x2 2a x tanx a

, sec 02 2

3tanx




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2 2

2 2

3sec 3sec

9 9 tan 3 1 tand d

2secsec ln sec tan

secd d c

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3- For use

Ex 18:

Sol:


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2 2x a

a

x2 2x a secx a

5secx




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Ex: Evaluate

1- 2-


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2

2

6dx

x x

3

21

xdx

x




1- 2-

3- 4-

5- 6-

7-


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2 29

dx

x x

2

29

xdx

x

24 9xdx

x

2 4

xdx

x x

210 2

xdx

x x

3

21

xdx

x

2

2

4dx

x x x

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6.4: INTEGRATION OF RATIONAL FUNCTIONS USING

PARTIAL FRACTIONS

To evaluate integrals of the form where p(x) and

q(x) are polynomial.

In partial fraction method the degree of p(x) should be less

than the degree of q(x) if not we use the long division.

1- If a rational expression has a denominator that factors into

n distinct linear factors, then we can write

Where are constant. Prepared by Dr. F.G.A

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( )

( )

p xdx

q x

1 1 2 2

1 2

1 1 2 2

( ) ( )

( ) ...

...

n n

n

n n

p x p x

q x a x b a x b a x b

A A A

a x b a x b a x b

1 2, ,..., nA A A



PARTIAL FRACTIONS

Ex:

Sol:

Likewise, taking x = 1, we find B = 3 and taking x = 1, we find C = 4.


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PARTIAL FRACTIONS

Repeated Linear factor: If the denominator of a rational

expression contains repeated linear factors, the decomposition

looks like the following.

Where are constant. Prepared by Dr. F.G.A Sharjah University

1 2

2

( ) ( )...

( )

n

n n

A A Ap x p x

q x ax bax b ax b ax b

1 2, ,..., nA A A



PARTIAL FRACTIONS

Ex: Evaluate

Sol:


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2

3 2

5 20 6

2

x xdx

x x x

repeated linear factor

Taking x = 0, we find A = 6. Likewise, taking x = 1, we find that C = 9.

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PARTIAL FRACTIONS

You should find that B = 1. Now we have

Partial Fractions Where Long Division Is Required: If the

numerator p(x) of a rational expression has the same or higher

degree than the denominator q(x) , you must first perform a

long division and follow this with a partial fractions

decomposition of the remaining fraction. Prepared by Dr. F.G.A

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PARTIAL FRACTIONS

Ex: Evaluate

Sol:


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3 2

2

2 4 15 5

2 8

x x xdx

x x

Since the degree of the numerator exceeds that of

the denominator, first divide:

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PARTIAL FRACTIONS


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It is a simple matter to solve for the constants:



PARTIAL FRACTIONS


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PARTIAL FRACTIONS

Denominators Containing Irreducible Quadratic Factors:

If the degree of p(x) is less than the degree of the denominator

q(x) and all of the factors in the denominator are distinct, then

we can write


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PARTIAL FRACTIONS

Ex: Evaluate

Sol:


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2

3

2 5 2x xdx

x x

2, 5, 2

0

A B C A

B

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PARTIAL FRACTIONS

When an integral contains a quadratic expression

You can sometime simplify the integral by completing the

square.

Ex: Evaluate


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2ax bx c

2

2

5 6 2

2 2 5

x xdx

x x x



PARTIAL FRACTIONS

Sol:


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Solve this system of linear equations using

elimination to get:

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PARTIAL FRACTIONS


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PARTIAL FRACTIONS

Ex: Evaluate

1- 2-

3- 4- 5-


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2 4 1

1 1 3

x xdx

x x x

2

2 1

dx

x x

3

2

2

2 8

x xdx

x x

2

6 7

2

xdx

x

2

2 1

7 10

xdx

x x

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PARTIAL FRACTIONS


1- 2-

3- 4-

5-


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2

2 1

dx

x x

3 2

4

3 2

xdx

x x x

3 4dx

x x

3

3 2

4

2 2

xdx

x x x

4

3 2

2

1

xdx

x x x


6.6: IMPROPER INTEGRALS

Definition:

If is continuous on the interval [a, b) and as,

, we define the improper integral of on [a, b] by

In either case, if the limit exists and equals some value L,

we say that the improper integral converges (to L).

If the limit does not exist, we say that the

improper integral diverges. Prepared by Dr. F.G.A

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f x x b

f xf

Similarly, if is continuous on (a, b] and as x a+, we define the improper integral of on [a, b] by

f x f xf

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EX1: Determine whether the improper integral

converges or diverges.

Sol:

Since the defining limit does not exist, the improper integral

diverges.




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0

2

1

1dx

x

1

0

1dx

x



Sol:

The improper integral converges to 2.

Definition:

Suppose that f is continuous on the interval [a, b], except at

some c (a, b), and as x c. The integral is improper and we write


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f x

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If both and converge (to L1 and L2,

respectively), we say that the improper integral

converges.

If either of the improper integrals or

diverges,

then we say that the improper integral diverges.



Sol:


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2

2

1

1dx

x



In example 1, we showed that diverges.

Thus, also diverges.

EX: Determine whether the following improper integral


1- 2- 3-

4- 5- 6-

7-


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0

2

1

1dx

x

2

2

1

1dx

x

1

30

1dx

x

1

3 40

1dx

x

5

1

2

5dx

x

2

0

secxdx

1

0

ln xdx2

0

sec xdx

3

2

0

2

1dx

x

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Definition: If f is continuous on the interval , we define

the improper integral to be

Similarly, if f is continuous on , we define

In either case, if the limit exists and equals some value L,

we say that the improper integral converges (to L).


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( )a

f x dx

,a

,a



If the limit does not exist, we say that the improper integral

diverges.

Ex 4: Determine whether the improper integral


Sol:

diverges.

Remark:

This integral converges whenever p > 1 and diverges

for p 1. Check!!!!! Prepared by Dr. F.G.A

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1

1dx

x

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Sol:


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0

xxe dx

indeterminate form 0



Resolve with lHpitals Rule:

So the integral converges Prepared by Dr. F.G.A

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Definition: If f is continuous on (,), we write

for any constant a, where converges if and only

if both and converge.

If either one diverges, the original improper integral also

diverges.


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Sol:

Similarly,


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2xxe dx

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Hence,

The integral converges. Prepared by Dr. F.G.A

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Sol:

Since diverges diverges. Prepared by Dr. F.G.A

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xe dx

0

xe dx

xe dx

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Theorem (Comparison Test) :Suppose that f (x) and g (x) are

continuous on and 0 f (x) g(x), for all



Sol: Using the comparison test for an improper integral

Its an easy to show that Prepared by Dr. F.G.A

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,a ,x a

0

1xdx

x e



Converges to 1.

From the above theorem, it follows that

converges.



Sol: First, recall that

Then,


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00 0

1 1lim lim lim 1 1

RR

x x

x RR R Rdx e dx e

e e

0

1xdx

x e

1

2 sin xdx

x

1 2 sin0

x

x x

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We showed in example 4 that

diverges. The Comparison Test now tells us that

must diverge, also.



Exercises: Determine whether the following improper

integral converges or diverges.

1- 2-


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3

01

xdx

x

2

2

3dx

x

2

1 1

dx

x



3- 4- 5 -

6- 7- 8-

9- 10- 11-

12- Prepared by Dr. F.G.A

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0

23

dx

x

32 1

xdx

x

2

4

1

2

3

xdx

x

2

1

2 sec xdx

x

21dx

x

1dx

x

xe dx

4

3

1

x dx

2

0

sin

1 xxdx

e

0

ln

1 xx

dxe

Calculus 2 Chapter 6(4)

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Transcript of Calculus 2 Chapter 6(4)