chapter 8 integration techniques. 8.1 integration by parts

Chapter 8

Integration Techniques

8.1

Integration by Parts

4

Integrate by parts: Practice!

.

http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/intbypartsdirectory/IntByParts.html

8.2

Trigonometric Integrals

8.3

Trigonometric Substitutions

15

Trigonometric Substitution

In finding the area of a circle or an ellipse, an integral of the form dx arises, where a > 0.

If it were the substitution

u = a2 – x2 would be effective but, as it stands,

dx is more difficult.

16

Trigonometric Substitution

If we change the variable from x to by the substitutionx = a sin , then the identity 1 – sin2 = cos2 allows us to get rid of the root sign because

17

Example 1

Evaluate

Solution:

Let x = 3 sin , where – /2 /2. Then dx = 3 cos d and

(Note that cos 0 because – /2 /2.)

18

Example 1 – Solution

By Inverse Substitution we get:

cont’d

19

Example 1 – Solution Since this is an indefinite integral, we must return to the

original variable x. This can be done either by using trigonometric identities to express cot in terms of sin = x/3 or by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle.

cont’d

sin =

Figure 1

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Since sin = x/3, we label the opposite side and the hypotenuse as having lengths x and 3.

Then the Pythagorean Theorem gives the length of the adjacent side as so we can simply read the value of cot from the figure:

cont’d

21


Since sin = x/3, we have = sin–1(x/3) and so

cont’d

24

Example 2

Find

Solution:Let x = 2 tan , – /2 < < /2. Then dx = 2 sec2 d and

=

= 2| sec |

= 2 sec

25


Thus we have

To evaluate this trigonometric integral we put everything in terms of sin and cos :

cont’d

26


=

Therefore, making the substitution u = sin , we have

cont’d

27

Example 2 – Solutioncont’d

28


We use the figure below to determine that csc = and so

cont’d

Figure 3

29

Example 3

Find

Solution:First we note that (4x2 + 9)3/2 = so trigonometric substitution is appropriate.

Although is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitution u = 2x.

30


When we combine this with the tangent substitution, we have x = which gives and

When x = 0, tan = 0, so = 0; when x = tan = so = /3.

cont’d

31


Now we substitute u = cos so that du = –sin d.When = 0, u = 1; when = /3, u =

cont’d

32


Thereforecont’d

8.4

Partial Fractions

35

Integration of Rational Functions by Partial Fractions

To see how the method of partial fractions works in general, let’s consider a rational function

where P and Q are polynomials.

It’s possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree of Q. Such a rational function is called proper.

36


If f is improper, that is, deg(P) deg(Q), then we must take the preliminary step of dividing Q into P (by long division) until a remainder R (x) is obtained such that deg(R) < deg(Q).

where S and R are also polynomials.

37

Example 1

Find

Solution:

Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write:

38


If f(x) = R (x)/Q (x) is a proper rational function:

factor the denominator Q (x) as far as possible.

Ex: if Q (x) = x4 – 16, we could factor it as

Q (x) = (x2 – 4)(x2 + 4) = (x – 2)(x + 2)(x2 + 4)

39


Next: express the proper rational function as a sum of partial fractions of the form

or

A theorem in algebra guarantees that it is always possible to do this.

Four cases can occur.

40


Case I The denominator Q (x) is a product of distinct linear factors.

This means that we can write

Q (x) = (a1x + b1)(a2x + b2) . . . (akx + bk)

where no factor is repeated (and no factor is a constant multiple of another).

41


In this case the partial fraction theorem states that there exist constants A1, A2, . . . , Ak such that

These constants can be determined as in the next example.

42

Example 2

Evaluate

Solution:

Since the degree of the numerator is less than the degree of the denominator, we don’t need to divide.

We factor the denominator as

2x3 + 3x2 – 2x = x(2x2 + 3x – 2)

= x(2x – 1)(x + 2)

43


Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand has the form

To determine the values of A, B, and C, we multiply both sides of this equation by the product of the denominators, x(2x – 1)(x + 2), obtaining

x2 + 2x – 1 = A(2x – 1)(x + 2) + Bx(x + 2) + Cx(2x – 1)

44


Expanding the right side and writing it in the standard form for polynomials, we get

x2 + 2x – 1 = (2A + B + 2C)x2 + (3A + 2B – C)x – 2A

These polynomials are identical, so their coefficients must be equal. The coefficient of x2 on the right side, 2A + B + 2C, must equal the coefficient of x2 on the left side—namely, 1.

Likewise, the coefficients of x are equal and the constant terms are equal.

cont’d

45


This gives the following system of equations for A, B, and C:

2A + B + 2C = 1

3A + 2B – C = 2

–2A = –1

Solving, we get, A = B = and C = and so

cont’d

46


In integrating the middle term we have made the mental substitution u = 2x – 1, which gives du = 2 dx and dx = du.

cont’d

47

Note:We can use an alternative method to find the coefficients A, B and C. We can choose values of x that simplify the equation:

x2 + 2x – 1 = A(2x – 1)(x + 2) + Bx(x + 2) + Cx(2x – 1)

If we put x = 0, then the second and third terms on the right side vanish and the equation then becomes –2A = –1, or A = .

Likewise, x = gives 5B/4 = and x = –2 gives 10C = –1, so B = and C =

48


Case II: Q (x) is a product of linear factors, some of which are repeated.

Suppose the first linear factor (a1x + b1) is repeated r times; that is, (a1x + b1)r occurs in the factorization of Q (x). Then instead of the single term A1/(a1x + b1) in the equation:

we use

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Example, we could write

50

Example 3

Find

Solution:

The first step is to divide. The result of long division is

51


The second step is to factor the denominator Q (x) = x3 – x2 – x + 1.

Since Q (1) = 0, we know that x – 1 is a factor and we obtain

x3 – x2 – x + 1 = (x – 1)(x2 – 1)

= (x – 1)(x – 1)(x + 1)

= (x – 1)2(x + 1)

cont’d

52


Since the linear factor x – 1 occurs twice, the partial fraction decomposition is

Multiplying by the least common denominator, (x – 1)2(x + 1), we get

4x = A (x – 1)(x + 1) + B (x + 1) + C (x – 1)2

cont’d

53


= (A + C)x2 + (B – 2C)x + (–A + B + C)

Now we equate coefficients:

A + C = 0

B – 2C = 4

–A + B + C = 0

cont’d

54


Solving, we obtain A = 1, B = 2, and C = –1, so

cont’d

55


Case III: Q (x) contains irreducible quadratic factors, none of which is repeated.

If Q (x) has the factor ax2 + bx + c, where b2 – 4ac < 0, then, in addition to the partial fractions, the expression for R (x)/Q (x) will have a term of the form

where A and B are constants to be determined.

56


Example:f (x) = x/[(x – 2)(x2 + 1)(x2 + 4)] has the partial fraction decomposition:

Any term of the form: can be integrated by completing the square (if necessary) and using the formula

57

Example 4

Evaluate

Solution:

Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain

58


Notice that the quadratic 4x2 – 4x + 3 is irreducible because its discriminant is b2 – 4ac = –32 < 0. This means it can’t be factored, so we don’t need to use the partial fraction technique.

To integrate the given function we complete the square in the denominator:

4x2 – 4x + 3 = (2x – 1)2 + 2

This suggests that we make the substitution u = 2x – 1.

cont’d

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Then du = 2 dx and x = (u + 1), so

cont’d

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Example 4 – Solutioncont’d

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Note:Example 6 illustrates the general procedure for integrating a partial fraction of the form

We complete the square in the denominator and then make a substitution that brings the integral into the form

Then the first integral is a logarithm and the second is expressed in terms of

where b2 – 4ac < 0

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Case IV: Q (x) contains a repeated irreducible quadratic factor.

If Q (x) has the factor (ax2 + bx + c)r, where b2 – 4ac < 0, then instead of the single partial fraction , the sum:

occurs in the partial fraction decomposition of R (x)/Q (x).

Each of the terms can be integrated by using a substitution or by first completing the square if necessary.

63

Example 5

Evaluate

Solution:

The form of the partial fraction decomposition is

Multiplying by x(x2 + 1)2, we have

–x3 + 2x2 – x + 1 = A(x2 +1)2 + (Bx + C)x(x2 + 1) + (Dx + E)x

64


= A(x4 + 2x2 +1) + B(x4 + x2) + C(x3 + x) + Dx2 + Ex

= (A + B)x4 + Cx3 + (2A + B + D)x2 + (C + E)x + A

If we equate coefficients, we get the system

A + B = 0 C = –1 2A + B + D = 2 C + E = –1 A = 1

which has the solution A = 1, B = –1, C = –1, D = 1 and E = 0.

cont’d

65


Thus

cont’d

8.7

Improper Integrals

68

Type 1: Infinite Intervals

70

Examples:

71

Practice Example:

Determine whether the integral is convergent or divergent.

Solution:

According to part (a) of Definition 1, we have

The limit does not exist as a finite number and so the

Improper integral is divergent.

74

Examples:

75

Type 2: Discontinuous Integrands

Suppose that f is a positive continuous function defined on a finite interval [a, b) but has a vertical asymptote at b.

Let S be the unbounded region under the graph of f and above the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely in a horizontal direction. Here the region is infinite in a vertical direction.)

The area of the part of S between a and t is

Figure 7

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Type 2: Discontinuous Integrands

77

Practice Example:

Find Solution:

We note first that the given integral is improper because has the vertical asymptote x = 2.

Since the infinite discontinuity occurs at the left endpoint of [2, 5], we use part (b) of Definition 3:

78

Example – Solution

Thus the given improper integral is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region.

Figure 10

cont’d

79

Gabriel’s Horn:

8.8

Introduction to Differential Equations

81

Definition

A differential equation is an equation containing an unknown function and its derivatives.

32 xdx

dy

032

2

aydx

dy

dx

yd

364

3

3

y

dx

dy

dx

yd

Examples:.

y is the dependent variable and x is independent variable.

1.

2.

3.

Ordinary Differential Equations

82

Partial Differential Equation

Examples:0

2

2

2

2

y

u

x

u

04

4

4

4

t

u

x

u

t

u

t

u

x

u

2

2

2

2

u is the dependent variable and x and y are independent variables.

u is dependent variable and x and t are independent variables

1.

2.

3.

83

Order of a Differential Equation

The order of the differential equation is the order of the highest derivative in the differential equation.

Differential Equation ORDER

32 x

dx

dy

0932

2

ydx

dy

dx

yd

364

3

3

y

dx

dy

dx

yd

1

2

3

84

Degree of Differential Equation

Differential Equation Degree

032

2

aydx

dy

dx

yd

364

3

3

y

dx

dy

dx

yd

0353

2

2

dx

dy

dx

yd

1

1

3

The degree of a differential equation is the power of the highest order derivative term in the differential equation.

85

Linear Differential Equation A differential equation is linear, if:

1. The dependent variable and its derivatives are of degree one,2. The coefficient of any term does not contain the dependent variable, y.

364

3

3

y

dx

dy

dx

yd is non-linear because the 2nd term is not of degree one.

.0932

2

ydx

dy

dx

ydExamples:is linear.

.0932

2

ydx

dy

dx

yd is non-linear because 3rd term contains y

86

32

22 x

dx

dyy

dx

ydx

is non-linear because in the 2nd term the coefficient contains y.

3.

is non–linear because the coefficient on the left hand side contains y

ydx

dysin4.

8787

Solving Differential Equations

88

The most general first order differential equation can be written as:

There is no general formula for the solution. We will look at two types of these and how to solve them:Linear EquationsSeparable Equations

First Order Differential Equations

89

Linear Differential Equations: Integrating Factor Method

If not already in the following form, re-express the equation in the form:

(1) where both p(t) and g(t) are continuous functions

Assume there is a function, , called an integrating factor.

Multiply each term in (1) by . This will give:

90

NOW: assume that whatever is, it will satisfy the following :

Do not worry about how we can find a that will satisfy the above. As long as p(t) is continuous we will find it!

The equation becomes:

We recognize that the left side is nothing more than the following product rule:

91

The equation becomes:

integrate both sides :

Finally, the solution y(t) is:

92

What is for any given equation ?

We started with assuming:

So:

Finally:

93

Solve with y(0) = 1.Practice Example:

Step1: Compare the equation with the standard form:

We identify: and .

Step2: Find the integrating factor

Step3: Multiplying through by the integrating factor, we get .

Step4: Rewrite as the derivative .

Step5: Integrate both sides with respect to x and get

Step6: Use the initial condition to find c. , gives: . So the solution to the problem is and finally:

94

Linear Differential Equations: Separable Equations

A separable differential equation is any differential equation that we can write in the following form

Now rewrite the differential equation as:

Integrate both sides. Use the initial condition to find the constant of integration.

95

Solve with y(0) = 1.Practice Example:

Step1: Divide through by y. We get:

Step2: Integrate both sides:

Step3: Solving for y gives: where

Step4: Use the initial condition: to get: A = 1 So the solution to the problem is:

96

Examples from textbook:

97

Example 2 (textbook) : Initial Value Problem Drug Dosing

98

Example 3 (textbook) : Separable Equation

99

Example 4 (textbook) : Separable Equation Logistic Population Growth

100

Directional Fields: Reading assignment (for your general Math knowledge).

chapter 8 integration techniques. 8.1 integration by parts

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