Presentation on theme: "BY PARTS. Integration by Parts Although integration by parts is used most of the time on products of the form described above, it is sometimes effective."— Presentation transcript:
Integration by Parts Although integration by parts is used most of the time on products of the form described above, it is sometimes effective on single functions. Looking at the following example. If is a product of a power of x (or a polynomial) and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing according to the type of function.
6.3 Integration By Parts Start with the product rule: This is the Integration by Parts formula.
The Integration by Parts formula is a “product rule” for integration. u differentiates to zero (usually). dv is easy to integrate. Choose u in this order: LIPET Logs, Inverse trig, Polynomial, Exponential, Trig
20 Recall Basic Identities Pythagorean Identities Half-Angle Formulas These will be used to integrate powers of sin and cos
21 Integral of sin n x, n Odd Split into product of an even and sin x Make the even power a power of sin 2 x Use the Pythagorean identity Let u = cos x, du = -sin x dx
22 Integral of sin n x, n Odd Integrate and un-substitute Similar strategy with cos n x, n odd
23 Integral of sin n x, n Even Use half-angle formulas Try Change to power of cos 2 x Expand the binomial, then integrate
24 Combinations of sin, cos General form If either n or m is odd, use techniques as before –Split the odd power into an even power and power of one –Use Pythagorean identity –Specify u and du, substitute –Usually reduces to a polynomial –Integrate, un-substitute Try with
25 Combinations of sin, cos Consider Use Pythagorean identity Separate and use sin n x strategy for n odd
26 Combinations of tan m, sec n When n is even –Factor out sec 2 x –Rewrite remainder of integrand in terms of Pythagorean identity sec 2 x = 1 + tan 2 x –Then u = tan x, du = sec 2 x dx Try
27 Combinations of tan m, sec n When m is odd –Factor out tan x sec x (for the du) –Use identity sec 2 x – 1 = tan 2 x for even powers of tan x –Let u = sec x, du = sec x tan x Try the same integral with this strategy Note similar strategies for integrals involving combinations of cot m x and csc n x
28 Integrals of Even Powers of sec, csc Use the identity sec 2 x – 1 = tan 2 x Try
Here are a couple of shortcuts that are result from Trigonometric Substitution: These are on your list of formulas. They are not really new.
39 New Patterns for the Integrand Now we will look for a different set of patterns And we will use them in the context of a right triangle Draw and label the other two triangles which show the relationships of a and x a x
40 Example Given Consider the labeled triangle –Let x = 3 tan θ(Why?) –And dx = 3 sec 2 θ dθ Then we have 3 x θ Use identity tan 2 x + 1 = sec 2 x
41 Finishing Up Our results are in terms of θ –We must un-substitute back into x –Use the triangle relationships 3 x θ
This would be a lot easier if we could re-write it as two separate terms. 1 These are called non- repeating linear factors. You may already know a short-cut for this type of problem. We will get to that in a few minutes.
This would be a lot easier if we could re-write it as two separate terms. Multiply by the common denominator. Set like-terms equal to each other. Solve two equations with two unknowns. 1
The short-cut for this type of problem is called the Heaviside Method, after English engineer Oliver Heaviside. Multiply by the common denominator. 1 Let x = - 1 Let x = 3
The short-cut for this type of problem is called the Heaviside Method, after English engineer Oliver Heaviside. 1
Good News! The AP Exam only requires non-repeating linear factors! The more complicated methods of partial fractions are good to know, and you might see them in college, but they will not be on the AP exam or on my exam.
Repeated roots: we must use two terms for partial fractions. 2
If the degree of the numerator is higher than the degree of the denominator, use long division first. 4 (from example one)
irreducible quadratic factor repeated root first degree numerator A challenging example:
We can do this problem on the TI-89: expand ((-2x+4)/((x^2+1) * (x-1)^2)) Of course with the TI-89, we could just integrate and wouldn’t need partial fractions! 3F2
1. Using Table of Integration Formulas 2. Simplify the Integrand if Possible Sometimes the use of algebraic manipulation or trigonometric identities will simplify the integrand and make the method of integration obvious. 3. Look for an Obvious Substitution Try to find some function in the integrand whose differential also occurs, apart from a constant factor. 3. Classify the Integrand According to Its Form Trigonometric functions, Rational functions, Radicals, Integration by parts. 4. Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the denominator or using trigonometric identities) may be useful in transforming the integral into an easier form. 5. Relate the problem to previous problems When you have built up some experience in integration, you may be able to use a method on a given integral that is similar to a method you have already used on a previous integral. Or you may even be able to express the given integral in terms of a previous one. 6. Use several methods Sometimes two or three methods are required to evaluate an integral. The evaluation could involve several successive substitutions of different types, or it might combine integration by parts with one or more substitutions. Strategy for Integration
Until now we have been finding integrals of continuous functions over closed intervals. Sometimes we can find integrals for functions where the function or the limits are infinite. These are called improper integrals.
Example 1: The function is undefined at x = 1. Since x = 1 is an asymptote, the function has no maximum. Can we find the area under an infinitely high curve? We could define this integral as: (left hand limit) We must approach the limit from inside the interval.
Idea of the Comparison Theorem The improper integral converges if the area under the red curve is finite. We show that this is true by showing that the area under the blue curve is finite. Since the area under the red curve is smaller than the area under the blue curve, it must then also be finite. This means that the complicated improper integral converges.
Examples To show that the area under the blue curve in the previous figure is finite, compute as follows:previous figure