M53 Lec4.1 Antidifferentiation.pdf

7/29/2019 M53 Lec4.1 Antidifferentiation.pdf

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Antiderivaties and Antidifferentiation

Mathematics 53

Institute of Mathematics (UP Diliman)

Lecture 4.1


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For today

1 Antiderivatives and Antidifferentiation

2 Integration by Substitution


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For today




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Antiderivatives

PROBLEM:

Suppose F(x) = 2x, what is F(x)?


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Antiderivatives

PROBLEM:

Suppose F(x) = 2x, what is F(x)?

ANSWER:

F(x) = x2


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Antiderivatives

Definition

A function F is an antiderivative of the function f on an interval I


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Antiderivatives

Definition

A function F is an antiderivative of the function f on an interval I if andonly if F(x)

=f(x) for every value of x in I.


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Antiderivatives

Definition

A function F is an antiderivative of the function f on an interval I if andonly if F(x)

=f(x) for every value of x in I.

Examples.


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Antiderivatives

Definition

A function F is an antiderivative of the function f on an interval I if andonly if F(x) = f(x) for every value of x in I.

Examples.

1 An antiderivative of f(x) = 12x2 + 2x is


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Antiderivatives

Definition


Examples.

1 An antiderivative of f(x) = 12x2 + 2x is F(x) = 4x3 + x2.


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Antiderivatives

Definition


Examples.

1 An antiderivative of f(x) = 12x2 + 2x is F(x) = 4x3 + x2.2 An antiderivative of g(x) = cos x is


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Antiderivatives

Definition


Examples.

1 An antiderivative of f(x) = 12x2 + 2x is F(x) = 4x3 + x2.2 An antiderivative of g(x) = cos x is G(x) = sin x.


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Antiderivatives

Remark

The antiderivative F of a function f is NOT unique.


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Antiderivatives

Remark


Examples.

1 Another antiderivative of f(x) = 12x2 + 2x is


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Antiderivatives

Remark


Examples.

1 Another antiderivative of f(x) = 12x2 + 2x is F1(x) = 4x3 + x2 1.


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d


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Antiderivatives

Remark


Examples.

1 Another antiderivative of f(x) = 12x2 + 2x is F1(x) = 4x3 + x2 1.2 Another antiderivative of g(x)

=cos x is G1(x)

=sin x

+.


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A tidiff ti ti


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Antidifferentiation

Terminologies and Notations:

A tidiff ti ti


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Antidifferentiation


Antidifferentation is the process of finding the antiderivative.

A tidiff ti ti


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Antidifferentiation



The symbol

, called the integral sign, denotes the operation of

antidifferentiation.

Antidifferentiation


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Antidifferentiation



The symbol



The function f is called the integrand.


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Antidifferentiation


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Antidifferentiation



The symbol



The function f is called the integrand.If F is an antiderivative of f, we write

f(x) d x= F(x) +C.

The expression F(x) +C is called the general antiderivative of f.


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Antidifferentiation


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Antidifferentiation

Example

1

12x2 + 2x

d x= 4x3 + x2 +C

2 cos x d x=sin x

+C

3

d x=

1 d x

Antidifferentiation


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Antidifferentiation

Example

1

12x2 + 2x

d x= 4x3 + x2 +C

2 cos x d x=sin x

+C

3

d x=

1 d x= x+C

Antidifferentiation


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Example

1

12x2 + 2x

d x= 4x3 + x2 +C

2 cos x d x=sin x

+C

3

d x=

1 d x= x+C

4

3x2 d x

Antidifferentiation


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Example

1

12x2 + 2x

d x= 4x3 + x2 +C

2 cos x d x=sin x

+C

3

d x=

1 d x= x+C

4

3x2 d x= x3 +C


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Antidifferentiation


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Example

1

12x2 + 2x

d x= 4x3 + x2 +C

2 cos x d x=sin x

+C

3

d x=

1 d x= x+C

4

3x2 d x= x3 +C

5

x2 d x= 1

3x3 +C

Theorems on Antidifferentiation


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Theorem



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Theorem

1

d x



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Theorem

1

d x= x+C


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Theorem

1

d x= x+C

2 If a is any constant,

a f(x) d x= a

f(x) d x

3 If f and g are defined on the same interval,

f(x) g(x)

d x


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Theorem

1

d x= x+C

2 If a is any constant,

a f(x) d x= a

f(x) d x

3 If f and g are defined on the same interval,

f(x) g(x)

d x=

f(x) d x

g(x) d x

4 If n is any rational number and n=

1,

xnd x= x

n+1

n+ 1 +C


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Example

Evaluate

x5 d x

x5 d x= x

6

6+C

Example

Evaluate

2 d x


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Example

Evaluate

x5 d x

x5 d x= x

6

6+C

Example

Evaluate

2 d x

2 d x= 2

d x



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Example

Evaluate

x5 d x

x5 d x= x

6

6+C

Example

Evaluate

2 d x

2 d x= 2

d x= 2x+C


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Example

Evaluate

3

x5d x

3x5 d x


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Example

Evaluate

3

x5d x

3x5 d x =

3x

5

d x

= 3

x5 d x

=3x4

4 +C



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Example

Evaluate

3

x5d x

3x5 d x =

3x

5

d x

= 3

x5 d x

=3x4

4 +C

= 34x4

+C



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Example

Evaluate

4u

3/4 d u



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Example

Evaluate

4u

3/4 d u

4u

3/4 d u



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Example

Evaluate

4u

3/4 d u

4u

3/4 d u = 4u7/4

7/4+C



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Example

Evaluate

4u

3/4 d u

4u

3/4 d u = 4u7/4

7/4+C

= 16u7/4

7+C



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ExampleEvaluate

12x2 + 2x

d x


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ExampleEvaluate

12x2 + 2x

d x

12x2 + 2x

d x =



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ExampleEvaluate

12x2 + 2x

d x

12x2 + 2x

d x =

12x2 d x



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ExampleEvaluate

12x2 + 2x

d x

12x2 + 2x

d x =

12x2 d x+

2x d x



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ExampleEvaluate

12x2 + 2x

d x

12x2 + 2x

d x =

12x2 d x+

2x d x

= 12x3

3



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ExampleEvaluate

12x2 + 2x

d x

12x2 + 2x

d x =

12x2 d x+

2x d x

= 12x3

3+ 2x

2

2



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ExampleEvaluate

12x2 + 2x

d x

12x2 + 2x

d x =

12x2 d x+

2x d x

= 12x3

3+ 2x

2

2+C



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ExampleEvaluate

12x2 + 2x

d x

12x2 + 2x

d x =

12x2 d x+

2x d x

= 12x3

3+ 2x

2

2+C

= 4x3

+ x2

+C


Example


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Example

Evaluate

t2

t3

t

d t


Example


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Example

Evaluate

t2

t3

t

d t

t2t 3

t d t


Example


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Example

Evaluate

t2

t3

t

d t

t2t 3

t d t =

2t

2 3t3/2 d t


Example


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Example

Evaluate

t2

t3

t

d t

t2t 3

t d t =

2t

2 3t3/2 d t=

2t2 d t

3t

3/2 d t


Example


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Example

Evaluate

t2

t3

t

d t

t2t 3

t d t =

2t

2 3t3/2 d t=

2t2 d t

3t

3/2 d t

= 2t3

3


Example


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p

Evaluate t2t 3

t d t

t2t 3

t d t =

2t

2 3t3/2 d t=

2t2 d t

3t

3/2 d t

= 2t3

3 3 2t

5/2

5


Example


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p

Evaluate t2t 3

t d t

t2t 3

t d t =

2t

2 3t3/2 d t=

2t2 d t

3t

3/2 d t

= 2t3

3 3 2t

5/2

5+C


Example


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p

Evaluate t2t 3

t d t

t2t 3

t d t =

2t

2 3t3/2 d t=

2t2 d t

3t

3/2 d t

= 2t3

3 3 2t

5/2

5+C

= 2t3

3 6t

5/2

5+C


Example


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Evaluate t2t 3

t d t

t2t 3

t d t =

2t

2 3t3/2 d t=

2t2 d t

3t

3/2 d t

= 2t3

3 3 2t

5/2

5+C

= 2t3

3 6t

5/2

5+C

Note:

t

2t 3

t

d t=

t d t

2t 3

t

d t


Example


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Evaluate x

2

+1

x2 d x


Example


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Evaluate x

2

+1

x2 d x

x2 + 1

x2

d x


Example


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Evaluate x

2

+1

x2 d x

x2 + 1

x2

d x

= 1

+x2 d x


Example


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Evaluate x

2

+1

x2 d x

x2 + 1

x2

d x

= 1

+x2 d x

= x


Example


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Evaluate x

2

+1

x2 d x

x2 + 1

x2

d x

= 1

+x2 d x

= x+ x1

1


Example


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Evaluate x

2

+1

x2 d x

x2 + 1

x2

d x

= 1

+x2 d x

= x+ x1

1 +C


Example


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Evaluate x

2

+1

x2 d x

x2 + 1

x2

d x

= 1

+x2 d x

= x+ x1

1 +C

=


Example


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Evaluate x

2

+1

x2 d x

x2 + 1

x2

d x

= 1

+x2 d x

= x+ x1

1 +C

= x 1x

+C


Example2


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Evaluate x

2 + 1x2 d x

x2 + 1

x2

d x

= 1

+x2 d x

= x+ x1

1 +C

= x 1x

+C

Note:

x2 + 1

x2d x=

x2 + 1 d x

x2 d x

Antiderivatives of Trigonometric Functions

Th


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Theorem

1

sin x d x


Th


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Theorem

1

sin x d x= cos x+C


Theorem


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Theorem

1

sin x d x= cos x+C

2

cos x d x


Theorem


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Theorem

1

sin x d x= cos x+C

2

cos x d x= sin x+C


Theorem


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Theorem

1

sin x d x= cos x+C

2

cos x d x= sin x+C

3 sec

2

x d x


Theorem


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Theorem

1

sin x d x= cos x+C

2

cos x d x= sin x+C

3 sec2x d x=

tanx+C


Theorem


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Theorem

1

sin x d x= cos x+C

2

cos x d x= sin x+C

3 sec2x d x=

tanx+C

4

csc

2 x d x


Theorem


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Theorem

1

sin x d x= cos x+C

2

cos x d x= sin x+C

3 sec2x d x=

tanx+C

4

csc

2 x d x= cot x+C


Theorem


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Theorem

1

sin x d x= cos x+C

2

cos x d x= sin x+C

3 sec2x d x=

tanx+C

4

csc

2 x d x= cot x+C

5

sec xtan x d x


Theorem


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Theorem

1

sin x d x= cos x+C

2

cos x d x= sin x+C

3 sec2x d x=

tanx+C

4

csc

2 x d x= cot x+C

5

sec xtan x d x= sec x+C


Theorem


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1

sin x d x= cos x+C

2

cos x d x= sin x+C

3 sec2x d x=

tanx+C

4

csc

2 x d x= cot x+C

5


6

csc xcot x d x


Theorem


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1

sin x d x= cos x+C

2

cos x d x= sin x+C

3 sec2 x d x= tan x+C4

csc

2 x d x= cot x+C

5


6

csc xcot x d x= csc x+C


Example

Evaluate (sin x cos x) d x


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Evaluate (sin x+cos x) d x


Example



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(sin x+ cos x) d x


Example



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(sin x+ cos x) d x = cos x


Example



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(sin x+ cos x) d x = cos x+ sin x


Example



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(sin x+ cos x) d x = cos x+ sin x+C


Example



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( +)



Example

Evaluate

(sin x cos x) d x


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(

+)


Example

Evaluate

tan

2 x d x


Example

Evaluate

(sin x+ cos x) d x


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Example

Evaluate

tan

2 x d x

tan

2 x d x


Example

Evaluate

(sin x+ cos x) d x


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Example

Evaluate

tan

2 x d x

tan

2 x d x =

sec2 x 1

d x


Example

Evaluate

(sin x+ cos x) d x


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Example

Evaluate

tan

2 x d x

tan

2 x d x =

sec2 x 1

d x

= tan x


Example

Evaluate

(sin x+ cos x) d x


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Example

Evaluate

tan

2 x d x

tan

2 x d x =

sec2 x 1

d x

= tan x x


Example

Evaluate

(sin x+ cos x) d x


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Example

Evaluate

tan

2 x d x

tan

2 x d x =

sec2 x 1

d x

= tan x x+C


Example


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p

Evaluate

cos x

sin2 xd x


Example


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Evaluate

cos x

sin2 xd x

cos xsin2 x

d x


Example


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Evaluate

cos x

sin2 xd x

cos xsin2 x

d x = 1sin x

cos xsin x

d x


Example


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Evaluate

cos x

sin2 xd x

cos xsin2 x

d x = 1sin x

cos xsin x

d x

=

csc xcot x d x


Example


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Evaluate

cos x

sin2 xd x

cos xsin2 x

d x = 1sin x

cos xsin x

d x

=

csc xcot x d x

= csc x+C


Example

Evaluate

d x

1 sin x


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+


Example

Evaluate

d x

1 sin x


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+

d x

1+ sin x


Example

Evaluate

d x

1 sin x


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+

d x

1+ sin x =

d x

1 + sin x


Example

Evaluate

d x

1 sin x


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+

d x

1+ sin x =

d x

1 + sin x 1 sin x1 sin x


Example

Evaluate

d x

1 sin x


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+

d x

1+ sin x =

d x


= 1

sin x

1 sin2 x d x


Example

Evaluate

d x

1 sin x


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+

d x

1+ sin x =

d x


= 1

sin x

1 sin2 x d x=

1 sin x

cos2 xd x


Example

Evaluate

d x

1 sin x


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+

d x

1+ sin x =

d x


= 1

sin x

1 sin2 x d x=

1 sin x

cos2 xd x

= 1

cos2

x

1

cosx

sin x

cosxd x


Example

Evaluate

d x

1 sin x


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+

d x

1+ sin x =

d x


= 1

sin x

1 sin2 x d x=

1 sin x

cos2 xd x

=

1

cos2

x

1

cosx

sin x

cosx

d x

=

sec2 x sec xtan x

d x


Example

Evaluate

d x

1

+sin x


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+

d x

1+ sin x =

d x


= 1

sin x

1 sin2 x d x=

1 sin x

cos2 xd x

=

1

cos2

x

1

cosx

sin x

cosx

d x

=

sec2 x sec xtan x

d x

= tan x


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Example

Evaluate

d x

1

+sin x


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+

d x

1+ sin x =

d x


= 1

sin x

1 sin2 x d x=

1 sin x

cos2 xd x

=

1

cos2

x

1

cosx

sin x

cosx

d x

=

sec2 x sec xtan x

d x

= tan x sec x+C

For today


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The Substitution RuleSuppose F(x) is an antiderivative of f(x).


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Dx[F(x)] = f(x)


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Dx[F(x)] = f(x)

R ll h h i l


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Recall the chain rule:

Dx

F(g(x))= F(g(x))g(x)


Dx[F(x)] = f(x)

R ll th h i l


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Dx

F(g(x))= F(g(x))g(x) = f(g(x))g(x)


Dx[F(x)] = f(x)

R ll th h i l


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Dx

F(g(x))= F(g(x))g(x) = f(g(x))g(x)

Therefore, F(g(x)) is an antiderivative of f(g(x))g(x)


Dx[F(x)] = f(x)



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Dx

F(g(x))= F(g(x))g(x) = f(g(x))g(x)


f(g(x))g(x) d x


Dx[F(x)] = f(x)



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Dx

F(g(x))= F(g(x))g(x) = f(g(x))g(x)


f(g(x))g(x) d x= F(g(x)) +C


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Dx[F(x)] = f(x)



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Dx

F(g(x))= F(g(x))g(x) = f(g(x))g(x)



Let u= g(x). Then d u= g(x) d x


Dx[F(x)] = f(x)



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Dx

F(g(x))= F(g(x))g(x) = f(g(x))g(x)



Let u= g(x). Then d u= g(x) d x

f(u) d u= F(u) +C

The Substitution Rule

Theorem

If u= g(x) is a differentiable function whose range is an interval I and f is


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g g fcontinuous on I, then


Theorem



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g g fcontinuous on I, then

f(g(x))g(x) d x=

f(u) d u


Theorem



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gcontinuous on I, then

f(g(x))g(x) d x=

f(u) d u

TIPS:

Assign u to be an "inner" function whose derivative (disregardingconstant factors) is a factor of the integrand.


Theorem

If u= g(x) is a differentiable function whose range is an interval I and f ish


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continuous on I, then

f(g(x))g(x) d x=

f(u) d u

TIPS:

Assign u to be an "inner" function whose derivative (disregardingconstant factors) is a factor of the integrand.

The entire integral must be expressed in terms of u.

Integration by Substitution

Example

Evaluate

(1 4x)1/2 d x


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Example

Evaluate

(1 4x)1/2 d x


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Let u= 1 4x


Example

Evaluate

(1 4x)1/2 d x


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Let u= 1 4xd u= 4 d x


Example

Evaluate

(1 4x)1/2 d x


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d u4

= d x


Example

Evaluate

(1 4x)1/2 d x


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d u4

= d x

(1 4x)1/2 d x


Example

Evaluate

(1 4x)1/2 d x


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d u4

= d x

(1 4x)1/2 d x =

u

1/2

d u

4


Example

Evaluate

(1 4x)1/2 d x


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d u4

= d x

(1 4x)1/2 d x =

u

1/2

d u

4

= 14

u

1/2 d u


Example

Evaluate

(1 4x)1/2 d x


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d u4

= d x

(1 4x)1/2 d x =

u

1/2

d u

4

= 14

u

1/2 d u

= 14

2u3/2

3+C


Example

Evaluate

(1 4x)1/2 d x


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d u4

= d x

(1 4x)1/2 d x =

u

1/2

d u

4

= 14

u

1/2 d u

= 14

2u3/2

3+C

= (1 4x)3/2

6+C


Example

Evaluate

x2(x3 1)10 d x


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Example

Evaluate

x2(x3 1)10 d x


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Let u = x3 1


Example

Evaluate

x2(x3 1)10 d x


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Let u = x3 1d u = 3x2 d x


Example

Evaluate

x2(x3 1)10 d x


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Let u = x3 1d u = 3x2 d xd u

3= x2 d x


Example

Evaluate

x2(x3 1)10 d x


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3= x2 d x

x2(x3 1)10 d x


Example

Evaluate

x2(x3 1)10 d x


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3= x2 d x

x2(x3 1)10 d x =

u10

du

3


Example

Evaluate

x2(x3 1)10 d x


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3= x2 d x

x2(x3 1)10 d x =

u10

du

3

= 13

u10 d u


Example

Evaluate

x2(x3 1)10 d x


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3= x2 d x

x2(x3 1)10 d x =

u10

du

3

= 13

u10 d u

= u11

33+C


Example

Evaluate

x2(x3 1)10 d x


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3= x2 d x

x2(x3 1)10 d x =

u10

du

3

= 13

u10 d u

= u11

33+C

= x3

13

33+C


Example

Evaluate

xx2 + 1

3 d x


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x + 1


Example

Evaluate

xx2 + 1

3 d x


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x + 1

Let u = x2 + 1


Example

Evaluate

xx2 + 1

3 d x


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Let u = x2 + 1d u = 2x d x


Example

Evaluate

xx2 + 1

3 d x


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Let u = x2 + 1d u = 2x d xd u

2= x d x


Example

Evaluate

xx2 + 1

3 d x


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2= x d x

x

x2 + 13

d x


Example

Evaluate

xx2 + 1

3 d x


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2= x d x

x

x2 + 13

d x=

1

2u3 d u


Example

Evaluate

xx2 + 1

3 d x


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2= x d x

x

x2 + 13

d x=

1

2u3 d u

= 12

u2

2 +C


Example

Evaluate

xx2 + 1

3 d x


169/274


2= x d x

x

x2 + 13

d x=

1

2u3 d u

= 12

u2

2 +C

= 14(

x

2

+1)2

+C


Example

Evaluate

tt 1 d t


170/274


Example

Evaluate

tt 1 d t


171/274

Let u = t 1


Example

Evaluate

tt 1 d t


172/274

Let u = t 1d u = d t


Example

Evaluate

tt 1 d t


173/274


Also t = u+ 1


Example

Evaluate

tt 1 d t


174/274


Also t = u+ 1

t

t 1 d t


Example

Evaluate

tt 1 d t


175/274


Also t = u+ 1

t

t 1 d t =

(u+ 1) u1/2 d u


Example

Evaluate

tt 1 d t


176/274


Also t = u+ 1

t

t 1 d t =

(u+ 1) u1/2 d u

= u3/2 +u1/2d u


Example

Evaluate

tt 1 d t


177/274


Also t = u+ 1

t

t 1 d t =

(u+ 1) u1/2 d u

= u3/2 +u1/2d u= 2u

5/2

5


Example

Evaluate

tt 1 d t


178/274


Also t = u+ 1

t

t 1 d t =

(u+ 1) u1/2 d u

= u3/2 +u1/2d u= 2u

5/2

5+ 2u

3/2

3


Example

Evaluate

tt 1 d t


179/274


Also t = u+ 1

t

t 1 d t =

(u+ 1) u1/2 d u

= u3/2 +u1/2d u= 2u

5/2

5+ 2u

3/2

3+C


Example

Evaluate

tt 1 d t


180/274


Also t = u+ 1

t

t 1 d t =

(u+ 1) u1/2 d u

= u3/2 +u1/2d u= 2u

5/2

5+ 2u

3/2

3+C

=2 (t

1)

5/2

5 +2 (t

1)

3/2

3 +C


Example

Evaluate

t3

t

2

+3

d t


181/274


Example

Evaluate

t3

t2

+3

d t


182/274

Let u = t2

+ 3


Example

Evaluate

t3

t2

+3

d t


183/274

Let u = t2

+ 3du = 2t d t


Example

Evaluate

t3

t2

+3

d t


184/274

Let u = t2

+ 3du = 2t d td u

2= t d t


Example

Evaluate

t3

t2

+3

d t


185/274

Let u = t2

+ 3du = 2t d td u

2= t d t

Also t2

=u

3


Example

Evaluate

t3

t2

+3

d t

t3

d t


186/274

Let u = t2

+ 3du = 2t d td u

2= t d t

Also t2

=u

3

t2 + 3d t


Example

Evaluate

t3

t2

+3

d t

t3 d t =

t2 t d t


187/274

Let u = t2

+ 3du = 2t d td u

2= t d t

Also t2

=u

3

t2 + 3d t

t2 + 3d t


Example

Evaluate

t3

t2

+3

d t

t3 d t =

t2 t d t


188/274

Let u = t2

+ 3du = 2t d td u

2= t d t

Also t2

=u

3

t2 + 3

t2 + 3

= u1/2

(u 3)d u

2


Example

Evaluate

t3

t2

+3

d t

t3 d t =

t2 t d t


189/274

Let u = t2

+ 3du = 2t d td u

2= t d t

Also t2

=u

3

t2 + 3

t2 + 3

= u1/2

(u 3)d u

2

= 12

u

1/2 3u1/2

d u


Example

Evaluate

t3

t2

+3

d t

t32

d t =

t2 t2

d t


190/274

Let u = t2

+ 3du = 2t d td u

2= t d t

Also t2

=u

3

t2 + 3

t2 + 3

= u1/2

(u3

)

d u

2

= 12

u

1/2 3u1/2

d u

= 12

2u

3/2

3


Example

Evaluate

t3

t2

+3

d t

t32

d t =

t2 t2

d t


191/274

Let u = t2

+ 3du = 2t d td u

2= t d t

Also t2

=u

3

t2 + 3

t2 + 3

= u1/2

(u

3)d u

2

= 12

u

1/2 3u1/2

d u

= 12

2u

3/2

3 6u1/2


Example

Evaluate

t3

t2

+3

d t

t32 3

d t =

t2 t2 3

d t


192/274

Let u = t2

+ 3du = 2t d td u

2= t d t

Also t2

=u

3

t2 + 3

t2 + 3

= u1/2

(u

3)d u

2

= 12

u

1/2 3u1/2

d u

= 12

2u

3/2

3 6u1/2

+C


Example

Evaluate

t3

t2

+3

d t

t3

t 2 3d t =

t2 tt 2 3

d t


193/274

Let u = t2

+ 3du = 2t d td u

2= t d t

Also t2

=u

3

t2 + 3

t2 + 3

= u1/2

(u

3)d u

2

= 12

u

1/2 3u1/2

d u

= 12

2u

3/2

3 6u1/2

+C

=

t2 + 33/2

3 3

t2 + 3

1/2 +C


Example

Evaluate

sec

3

xtan x d x


194/274


Example

Evaluate

sec

3

xtan x d x


195/274

Let u = sec x


Example

Evaluate

sec

3

xtan x d x


196/274

Let u = sec xd u = sec xtan x d x


Example

Evaluate

sec

3

xtan x d x

sec

3 x tan x d x


197/274


sec xtan x d x


Example

Evaluate

sec

3

xtan x d x

sec

3 x tan x d x =

sec2 x sec x tan x d x


198/274


sec xtan x d x =

sec x sec xtan x d x


Example

Evaluate

sec

3

xtan x d x

sec

3 x tan x d x =



199/274


sec xtan x d x


= u2 d u


Example

Evaluate

sec

3

xtan x d x

sec

3 x tan x d x =



200/274


sec xtan x d x


= u2 d u= u

3

3+C


Example

Evaluatesec3

xtan

x d x

sec

3 x tan x d x =



201/274


sec xtan x d x


= u2 d u= u

3

3+C

=sec3 x

3 +C


Example

tan1

s+ tan 1s sin 1ss2 cos 1s

d s


202/274


Example

tan1


d s

Let u= 1s


203/274


Example

tan1


d s

Let u= 1s

d u= 1s2

d s


204/274


Example

tan1


d s

Let u= 1s

d u= 1s2

d s d u= d ss2


205/274


Example

tan1


d s

Let u= 1s

d u= 1s2

d s d u= d ss2

tan1 + tan 1 sin 1


206/274

tan

s+ tan

ssin

s

s2

cos

1

s

d s


Example

tan1


d s

Let u= 1s

d u= 1s2

d s d u= d ss2

tan1 + tan 1 sin 1 tanu + tanu sinu


207/274

tan

s+ tan

ssin

s

s2

cos

1

s

d s

= tanu+ tanusinu

cosu

d u


Example

tan1


d s

Let u= 1s

d u= 1s2

d s d u= d ss2



208/274

tan

s+ tan

ssin

s

s2

cos

1

s

d s

= tanu+ tanusinu

cosu

d u

=

sec utanu+ tan2 u

d u


Example

tan1


d s

Let u= 1s

d u= 1s2

d s d u= d ss2



209/274

tan

s+ tan

ssin

s

s2

cos

1

s

d s

= tanu+ tanusinu

cosu

d u

=

sec utanu+ tan2 u

d u

=

sec utanu+ sec2 u 1

d u


Example

tan1


d s

Let u= 1s

d u= 1s2

d s d u= d ss2

tan1

s+ tan 1

ssin

1

s tanu + tanu sinu


210/274

s s s

s2

cos

1

s

d s

= tanu+ tanusinu

cosu

d u

=

sec utanu+ tan2 u

d u

=

sec utanu+ sec2 u 1

d u

= (sec u


Example

tan1


d s

Let u= 1s

d u= 1s2

d s d u= d ss2

tan1

s+ tan 1

ssin

1

s tanu + tanu sinu


211/274

s s s

s2

cos

1

s

d s

= tanu+ tanusinu

cosu

d u

=

sec utanu+ tan2 u

d u

=

sec utanu+ sec2 u 1

d u

= (sec u+ tan u


Example

tan1


d s

Let u= 1s

d u= 1s2

d s d u= d ss2

tan

1

s+ tan 1

ssin

1

s d

tanu+ tanusinu

d


212/274

s s s

s2

cos

1

s

d s =

cosu

d u

=

sec utanu+ tan2 u

d u

=

sec utanu+ sec2 u 1

d u

= (sec u+ tan u u)


Example

tan1


d s

Let u= 1s

d u= 1s2

d s d u= d ss2

tan

1

s+ tan 1

ssin

1

s d

tanu+ tanusinu

d


213/274

s s s

s2

cos

1

s

d s =

cosu

d u

=

sec utanu+ tan2 u

d u

=

sec utanu+ sec2 u 1

d u

= (sec u+ tan u u) +C


Example

tan1


d s

Let u= 1s

d u= 1s2

d s d u= d ss2

tan

1

s+ tan 1

ssin

1

s d

tanu+ tanusinu

d


214/274

s s s

s2 cos

1

s

d s =

cos ud u

=

sec utanu+ tan2 u

d u

=

sec utanu+ sec2 u 1

d u

= (sec u+ tan u u) +C= sec 1

s tan 1

s+ 1

s+C


Example

Evaluate

xcsc3

x2

cot

x2

d x


215/274


Example

Evaluate

xcsc3

x2

cot

x2

d x

Let u= cscx2


216/274


Example

Evaluate

xcsc3

x2

cot

x2

d x

Let u= cscx2 d u= cscx2cotx2 2x d x


217/274


Example

Evaluate

xcsc3

x2

cot

x2

d x

Let u= cscx2 d u= cscx2cotx2 2x d xd u

2= csc

x2

cot

x2 x d x


218/274


Example

Evaluate

xcsc3

x2

cot

x2

d x


2= csc

x2

cot

x2 x d x


219/274

xcsc

3 x

2cot

x

2d x


Example

Evaluate

xcsc3

x2

cot

x2

d x


2= csc

x2

cot

x2 x d x


220/274

xcsc

3 x

2cot

x

2d x = csc

2 x

2csc

x

2cot

x

2 x d x


Example

Evaluate

xcsc3

x2

cot

x2

d x


2= csc

x2

cot

x2 x d x


221/274

xcsc

3 x

2cot

x

2d x = csc

2 x

2csc

x

2cot

x

2 x d x= 1

2

u2 d u


Example

Evaluate

xcsc3

x2

cot

x2

d x


2= csc

x2

cot

x2 x d x


222/274

xcsc

3 x

2cot

x

2d x = csc

2 x

2csc

x

2cot

x

2 x d x= 1

2

u2 d u

=

1

2

u3

3+

C


Example

Evaluate

xcsc3

x2

cot

x2

d x


2= csc

x2

cot

x2 x d x


223/274

xcsc

3 x

2cot

x

2d x = csc

2 x

2csc

x

2cot

x

2 x d x= 1

2

u2 d u

=

1

2

u3

3+

C

= csc3

x2

6+C


Example

Evaluate

2x2 sin x3 d x


224/274


Example

Evaluate

2x2 sin x3 d x

Let u= x3


225/274


Example

Evaluate

2x2 sin x3 d x

Let u= x3 d u= 3x2 d x


226/274


Example

Evaluate

2x2 sin x3 d x

Let u= x3 d u= 3x2 d x d u3

= x2 d x


227/274


Example

Evaluate

2x2 sin x3 d x


= x2 d x

22

i3 d


228/274

2x2 sin x3 d x


Example

Evaluate

2x2 sin x3 d x


= x2 d x

2x2 sin x3 d x 2sin udu


229/274

2x2 sin x3 d x = 2sin u 3


Example

Evaluate

2x2 sin x3 d x


= x2 d x



230/274

2x sin x d x = 2sin u 3= 2

3

sinu d u


Example

Evaluate

2x2 sin x3 d x


= x2 d x



231/274


3

sinu d u

= 23

(cos u) +C


Example

Evaluate

2x2 sin x3 d x


= x2 d x



232/274


3

sinu d u

= 23

(cos u) +C

= 23

cos x3 +C


Example

Evaluate

4 +x d x


233/274


Example

Evaluate

4 +x d x

Let u= 4 +x


234/274


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2

xd x


235/274


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2

xd x 2 d u= d x

x


236/274


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2

xd x 2 d u= d x

x

4 +x d x


237/274


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2

xd x 2 d u=

d x

x

4 +x d x =

4 +x d x

xx


238/274


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2

xd x 2 d u=

d x

x

4 +x d x =

4 +x d x

xx


239/274

=

4 +xx d xx


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2

xd x 2 d u=

d x

x

4 +x d x =

4 +x d x

xx


240/274

=

4 +xx d xx

(

x= u 4)


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2

x d x2 d u=

d x

x

4 +x d x =

4 +x d x

xx


241/274

=

4 +xx d xx

(

x= u 4)

=

u1/2 (u 4) 2 d u


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2

x d x2 d u=

d x

x

4 +x d x =

4 +x d x

xx


242/274

=

4 +xx d xx

(

x= u 4)

=

u1/2 (u 4) 2 d u

= 2u3/2 8u1/2d u


Example

Evaluate

4 +x d x

Let u= 4 +x du= 12

xd x 2 d u= d x

x

3 1


243/274

4+x d x = 2u/2 8u/2d u


Example

Evaluate

4 +x d x

Let u= 4 +x du= 12

xd x 2 d u= d x

x

d

3/2

1

/2

d


244/274

4+x d x = 2u/2 8u/2d u

= 2 2u5/2

5


Example

Evaluate

4 +x d x

Let u= 4 +x du= 12

xd x 2 d u= d x

x

4 d

2

3

/2 81

/2

d


245/274

4+x d x = 2u/2 8u/2d u

= 2 2u5/2

5 2 8u

3/2

3


Example

Evaluate

4 +x d x

Let u= 4 +x du= 12

xd x 2 d u= d x

x

4 d

2

3

/2 81

/2

d


246/274

4+x d x = 2u/2 8u/2d u

= 2 2u5/2

5 2 8u

3/2

3+C


Example

Evaluate

4 +x d x

Let u= 4 +x du= 12

xd x 2 d u= d x

x

4+ d 2 3/2 8 1/2d


247/274

4+x d x = 2u/2 8u/2d u

= 2 2u5/2

5 2 8u

3/2

3+C

=44+

x

5/2

5 164+

x

3/2

3 +C


Example

Evaluate

4 +x d x


248/274


Example

Evaluate

4 +x d x

Let u= 4 +x


249/274


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x


250/274


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x 2x d u= d x


251/274


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x 2x d u= d x

x= u 4


252/274


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x 2x d u= d x

x= u 4 2 (u 4) d u= d x


253/274


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x 2x d u= d x

x= u 4 2 (u 4) d u= d x

4 +

x d x


254/274

+


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x 2x d u= d x

x= u 4 2 (u 4) d u= d x

4 +

x d x

= u

1/2

2 (u

4) d u


255/274

+


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x 2x d u= d x

x= u 4 2 (u 4) d u= d x

4 +

x d x

= u

1/2

2 (u

4) d u


256/274

+ =

2u

3/2 8u1/2

d u


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x 2x d u= d x

x= u 4 2 (u 4) d u= d x

4 +

x d x

= u

1/2

2 (u

4) d u


257/274

=

2u

3/2 8u1/2

d u

= 2 2u5/2

5


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x 2x d u= d x

x= u 4 2 (u 4) d u= d x

4 +

x d x

= u

1/2

2 (u

4) d u


258/274

=

2u

3/2 8u1/2

d u

= 2 2u5/2

5 2 8u

3/2

3


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x 2x d u= d x

x= u 4 2 (u 4) d u= d x

4 +

x d x

= u

1/2

2 (u

4) d u

3/ 1/


259/274

=

2u

3/2 8u1/2

d u

= 2 2u5/2

5 2 8u

3/2

3+C


Example

Evaluate

4 +x d x

Let u= 4 +x du=1

2x d x 2x d u= d x

x= u 4 2 (u 4) d u= d x

4 +

x d x

= u

1/2

2 (u

4) d u

3/ 1/


260/274

=

2u

3/2 8u1/2

d u

= 2 2u5/2

5 2 8u

3/2

3+C

= 4

4 +x5/25

16

4 +x3/23

+C

End-of-Class Exercise

Evaluate

t

t+ 3d t


261/274


Evaluate

t

t+ 3d t

Solution:


262/274


Evaluate

t

t+ 3d t

Solution:

Let u= t+ 3


263/274


Evaluate

t

t+ 3d t

Solution:

Let u= t+ 3 d u= d t


264/274


Evaluate

t

t+ 3d t

Solution:

Let u= t+ 3 d u= d t t= u 3


265/274


Evaluate

t

t+ 3d t

Solution:


tt+ 3

d t


266/274


Evaluate

t

t+ 3d t

Solution:


tt+ 3

d t =

u 3u1/2

d u


267/274


Evaluate

t

t+ 3d t

Solution:


tt+ 3

d t =

u 3u1/2

d u

=


268/274


Evaluate

t

t+ 3d t

Solution:


tt+ 3

d t =

u 3u1/2

d u

= u1/2 3u1/2d u


269/274


Evaluate

t

t+ 3d t

Solution:


tt+ 3

d t =

u 3u1/2

d u

= u1/2 3u1/2d u


270/274

= 2u3/2

3


Evaluate

t

t+ 3d t

Solution:


tt+ 3

d t =

u 3u1/2

d u

= u1/2 3u1/2d u


271/274

= 2u3/2

3 3 2u

1/2

1


Evaluate

t

t+ 3d t

Solution:


tt+ 3

d t =

u 3u1/2

d u

= u1/2 3u1/2d u3/ 1/


272/274

= 2u3/2

3 3 2u

1/2

1+C


Evaluate

t

t+ 3d t

Solution:


tt+ 3

d t =

u 3u1/2

d u

= u1/2 3u1/2d u3/ 1/


273/274

= 2u3/2

3 3 2u

1/2

1+C

=2 (t+ 3)3/2

3 6 (t

+3)

1/2

+C

The End.


274/274

Documents

M53 Lec4.1 Antidifferentiation.pdf