# Techniques of Integration

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Techniques of Integration
Chapter 7 Techniques of Integration

7.1 Integration by parts Question: How to integrate
, where the integrands are the product of two kinds of functions? Every differentiation rule has a corresponding integration rule: Differentiation Integration the Chain Rule the Substitution Rule the Product Rule the Rule for Integration by Parts

The formula for integration by parts (1)
Let u = f (x) and v = g(x) are both differentiable, then du= f’(x) dx and dv= g’(x) dx (2)

Example 1. Find Example 2. Evaluate Example 3. Find Example 4. Evaluate

The formula for definite integration by parts (3)
Example 5. Calculate Example 6. Prove the reduction formula where is an integer.

Summarize When the integrands are the product of two kinds of functions and neither of them is derivative of the other, we use the integration by parts. We can compare:

First we recognize u and v, then confirm to be more easily integrated than . For example

7.2 Trigonometric integrals
Question 1: How to evaluate ？ (a) If the power of cosine is odd ( n=2k+1 ). Save one cosine factor to Use to express the remaining factors in terms of sine: Then substitute u=sinx.

(b) If the power of sine is odd ( m=2k+1 ).
Save one sine factor to Use to express the remaining factors in terms of sine: Then substitute u=cosx. (c) If the powers of both sine and cosine are even, use the half-angle identities

It is sometimes helpful to use the identity
Example 1. Evaluate Example 2. Evaluate Example 3. Find

Question 2: How to evaluate ？
(a) If the power of secant is even ( n=2k). Save a factor of Use to express the remaining factors in terms of tanx: Then substitute u=tanx. Example 4. Find

(b) If the power of tangent is odd ( m=2k+1).
Save a factor of Use to express the remaining factors in terms of secx: Then substitute u=secx. Example 5. Find

(c) If n = 0, only tanx occurs. Use and, if necessary, the formula
Example 6. Find (d) If n is odd and m is even, we express the integrand completely in term of secx. Power of secx may require integration by parts. Example 7. Find Example 8. Find

Question 3: How to evaluate ？
To evaluate the integrals (a) (b) use the corresponding identity: Example 9. Evaluate

7.3 Trigonometric substitution
How to find the area of a circle or an ellipse? How to integrate In general we can make a substitution of the form x=g(t) by using the Substitution Rule in reverse(called inverse substitution): Assume that g has an inverse function, that is, g is one-to-one, we obtain

Table of trigonometric substitution
One kind of inverse substitution is trigonometric substitution. Table of trigonometric substitution Expression Substitution Identity

Example 1. Evaluate Example 2. Find Example 3. Evaluate Example 4. Evaluate

Inverse Substitution Formula For Definite Integral
Let x=g( t ) and g has an inverse function, we have where Example 5. Find Example 6. Find the area enclosed by the ellipse Example 7. Find

7.4 Integration of rational functions
by partial fraction In this section we show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fraction(called partial fraction). Consider a rational function where P and Q are polynomials. If where then the degree of P is n and we write deg(P) = n

Step 1. First express f as a sum of simpler fractions.
We can integrate rational functions according to 3 steps: Step 1. First express f as a sum of simpler fractions. Provide that the degree of P is less than the degree of Q. Such a rational function is called proper. If f is improper, that is, we can divide Q into P by long division until a remainder R(x) is obtained such that deg(R)<deg(Q). The division statement is (1) where S and R are also polynomials.

Step 2. Second factor the denominator Q(x) as far as possible
Step 2. Second factor the denominator Q(x) as far as possible. It can be shown that any polynomial Q can be factored as a product of linear factors(of the form ax+b) and irreducible quadratic factors (of the form Step 3. Finally express the proper rational function R(x)/Q(x) as a sum of partial fractions of the form A theorem in algebra guarantees that it is always possible to do it.

Case 1. The denominator Q(x) is a product of distinct linear factors.
We explain the details for the 4 cases that occur. Case 1. The denominator Q(x) is a product of distinct linear factors. The partial fraction theorem states there exist constants such that (2) These constants need to be determined. Example 1. Evaluate

Example 2. Find Case 2. Q(x) is a product of linear factors, some of which are repeated. Suppose the first linear factor is repeated r times; that is, occurs in the factorization of Q(x). Then instead of the single term in Equation 2, we would use (3) Example 3. Find

Case 3. Q(x) contains irreducible quadratic factors, none of which is repeated.
If Q(x) has the factor then, in addition to the partial fraction in Equation 2 and 3, the expression for R(x)/Q(x) will have a term of the form (4) where A and B are constants to be determined. We can integrate (4) by completing the square and using the formula (5) Example 3. Find

Case 4. Q(x) contains a repeated irreducible quadratic factors.
Example 4. Evaluate Case 4. Q(x) contains a repeated irreducible quadratic factors. If Q(x) has the factor then instead of the single partial fraction(4), the sum (6) occurs in the partial fraction decomposition of R(x)/Q(x). Each of the term in (6) can be integrated by completed the square and making a tangent substitution. Example 5. Evaluate

7.5 Rationalizing substitutions
By means of appropriate substitutions, some functions can be changed into rational functions. In particular, when an integrand contains an expression of the form , then the substitution u = may be effective. Example 1. Evaluate Let Example 2. Find Let

The substitution t = tan(x/2) will convert any rational function of sinx and cosx into an ordinary rational function. This is called Weierstrass substitution. Let Then Therefore

(1) Since t = tan(x/2), we have , so Thus if we make the substitution t = tan(x/2), then we have Example 3. Find

7.6 Strategy For Integration
Integration is more challenging than differentiation. In finding the derivative of a function it is obvious which differentiation formula we should apply. But when integrating a given function, it may not be obvious which techniques we should use. First it is useful to be familiar with the basic integration formulas. Table of integration formulas Constants of integration have been omitted.

1. Simplify the integrand if possible.
Secondly if you do not immediately see how to attack a given integral, you might try the following four-step strategy. 1. Simplify the integrand if possible. 2. Look for an obvious substitution. 3. Classify the integrand according to its form. 4. Try again. (a) Try substitution (b) Try parts. (c) Manipulate the integrand. (d) Relate the problem to previous problems. (e) Use several methods.

Example 1. Example 2. Example 3. Example 4. Example 5.

Computer Algebra Systems
7.7 Using Tables of Integrals and Computer Algebra Systems

7.8 Approximation Integration
How to integrate It is difficult, or even impossible, to find an antiderivative. When the function is determined from a scientific experiment through instrument readings, how to integrate such discrete function? In both cases we need to find approximate values of definite integrals.

The left endpoint approximation
Using Riemann sums The left endpoint approximation (1) The right endpoint approximation (2) (3) Midpoint rule

Example 1. Use (a) the Trapezoidal Rule (b) the Midpoint Rule with n=5
to approximate the integral

(5) Error bounds Suppose If
Notice The error in using an approximation is defined to be the amount that needs to be added to the approximation to make it exact. . In general, we have (5) Error bounds Suppose If and are the errors in the Trapezoidal and Midpoint Rules, then

Example 2. (a) Use the Midpoint Rule with n=10 to approximate the integral
(b) Give an upper bound for the error involved in this approximation. (6) Simpson’s Rule where n is even and Example 3. Use Simpson’s Rule with n=10 to approximation

(7) Error bound for Simpson’s Rule Suppose that
If is the error involved in using Simpson’s Rule, then Example 4. (a) Use Simpson’s Rule with n=10 to approximate the integral (b) Estimate the error involved in this approximation.

Type 1 Infinite Intervals
7.9 Improper Integrals In defining a definite integral we deal with (1) the function f defined on a finite interval [a, b]; (2) f is a bounded function. Question: How to integrate a definite integral when the interval is infinite or when f is unbounded ? Type 1 Infinite Intervals Consider the infinite region that lies under the curve , above the x-axis and from the line x=1 to infinite, can this area A be infinite?

Notice and So the area of the infinite region is equal to 1 and we write

(1)Definition of An Improper Integral of Type 1
(a) If exists for every number , then provide this limit exists(as a finite number). (b) If exists for every number , then The improper integrals in (a) and (b) are called convergent if the limit exists and divergent if the limit does not exist. (c) If both and are convergent, then we define

Example 1. Determine whether the integral
is convergent or divergent. Example 2. Evaluate Example 3. Evaluate Example 4. For what value of p is the integral convergent? (2) is convergent if p >1 and divergent if

Type 2 Discontinuous Integrands
(3) Definition of An Improper Integral of Type 2 (a) If f is continuous on [a,b) and , then if this limit exists(as a finite number). (b) If f is continuous on (a,b] and , then The improper integrals in (a) and (b) are called convergent if the limit exists and divergent if the limit does not exist.

(c) If , where a<c<b, and both
and are convergent, then we define Example 5. Find Example 6. Determine whether converges or diverges. Example 7. Evaluate if possible. Example 8. Find

A Comparison Test For Improper Integrals
(4)Comparison Theorem Suppose that f and g are continuous functions with (a) If is convergent, then is convergent. (b) If is divergent, then is divergent. Example 9. Show that (a) is convergent. (b) is divergent.