# Techniques of Integration

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 CHPTER 7 Techniques of Integration 7.. Substitution Integration, unlike differentiation, is more of an art-form than a collection of algorithms. Many problems in applied mathematics involve the integration of functions given by complicated formulae, and practitioners consult a Table of Integrals in order to complete the integration. There are certain methods of integration which are essential to be able to use the Tables effectively. These are: substitution, integration by parts and partial fractions. In this chapter we will survey these methods as well as some of the ideas which lead to the tables. fter the examination on this material, students will be free to use the Tables to integrate. The idea of substitution was introduced in section 4. (recall Proposition 4.4). To integrate a differential f x which is not in the table, we first seek a function u u x so that the given differential can be rewritten as a differential g u du which does appear in the table. Then, if g u du G u C, we know that f x G u x C. Finding and employing the function u often requires some experience and ingenuity as the following examples show. Example 7. x x? (7.) Let u x, so that du and x u. Then x x u u du 4 u3 u du 4 5 u5 3 u3 C (7.) 30 u3 3u 5 C where at the end we have replaced u by x. 30 x 3 6x C 5 x 3 3x C Example 7. tan x? 07

2 Chapter 7 Techniques of Integration 08 Since this isn t on our tables, we revert to the definition of the tangent: tanx letting u cos x du sinx we obtain (7.3) tan x sinx cosx du u lnu C lncosx C lnsecx C sinx cosx. Then, Example 7.3 sec x?. This is tricky, and there are several ways to find the integral. However, if we are guided by the principle of rewriting in terms of sines and cosines, we are led to the following: (7.4) secx cosx cosx cos x Now we can try the substitution u sin x du cos x. Then (7.5) sec x cosx sin x du u This looks like a dead end, but a little algebra pulls us through. The identity (7.6) leads to (7.7) Using u du u sinx, we finally end up with (7.8) sec x u u du u u u ln sinx ln sinx C ln ln u ln u C sinx C sinx Example 7.4 s a circle rolls along a horizontal line, a point on the circle traverses a curve called the cycloid. loop of the cycloid is the trajectory of a point as the circle goes through one full rotation. Let us find the length of one loop of the cycloid traversed by a circle of radius. Let the variable t represent the angle of rotation of the circle, in radians, and start (at t 0) with the point of intersection P of the circle and the line on which it is rolling. fter the circle has rotated through t radians, the position of the point is as given as in figure 7.. The point of contact of the circle with the line is now t units to the right of the original point of contact (assuming no slippage), so (7.9) x t t sint y t cost To find arc length, we use ds dy, where cost dt dy sintdt. Thus (7.0) ds cost sin t dt cost dt so ds costdt, and the arc length is given by the integral (7.) L 0 π costdt

3 7. Integration by Parts 09 Figure 7. PSfrag replacements cost P t sint t t To evaluate this integral by substitution, we need a factor of sint. We can get this by multiplying and dividing by cost: cos (7.) cost t sint cost cost By symmetry around the line t π, the integral will be twice the integral from 0 to π. In that interval, sint is positive, so we can drop the absolute value signs. Now, the substitution u cost du sintdt will work. When t 0 u, and when t π u. Thus (7.3) L u du u du u Integration by Parts Sometimes we can recognize the differential to be integrated as a product of a function which is easily differentiated and a differential which is easily integrated. For example, if the problem is to find (7.4) x cos x then we can easily differentiate f x x, and integrate cos x separately. When this happens, the integral version of the product rule, called integration by parts, may be useful, because it interchanges the roles of the two factors. Recall the product rule: d uv udv vdu, and rewrite it as (7.5) udv d uv vdu In the case of 7.4, taking u x dv cosx, we have du v sinx. Putting this all in 7.5: (7.6) xcosx d xsinx sinx

4 Chapter 7 Techniques of Integration 0 and we can easily integrate the right hand side to obtain (7.7) xcosx xsinx sinx xsinx cosx C Proposition 7. (Integration by Parts) For any two differentiable functions u and v: (7.8) udv uv vdu To integrate by parts:. First identify the parts by reading the differential to be integrated as the product of a function u easily differentiated, and a differential dv easily integrated.. Write down the expressions for u dv and du v. 3. Substitute these expressions in Integrate the new differential vdu. Example 7.5 Find xe x. Let u x dv e x. Then du v e x. 7.8 gives us (7.9) xe x xe x e x xe x e x C Example 7.6 Find x e x. The substitution u x dv e x du x v e x doesn t immediately solve the problem, but reduces us to example 3: (7.0) x e x x e x xe x x e x xe x e x C x e x xe x e x C Example 7.7 To find lnx, we let u lnx dv, so that du x v x, and (7.) lnx xlnx x x xlnx xlnx x C This same idea works for arctanx: Let (7.) u arctanx dv du x v x and thus (7.3) arctanx x arctanx x x xarctanx ln x C where the last integration is accomplished by the new substitution u x du x.

5 7. Integration by Parts Example 7.8 These ideas lead to some clever strategies. Suppose we have to integrate e x cosx. We see that an integration by parts leads us to integrate e x sinx, which is just as hard. But suppose we integrate by parts again? See what happens: Letting u e x dv cosx du e x v sinx, we get (7.4) e x cosx e x sinx e x sinx Now integrate by parts again: letting u e x dv sinx du e x v cosx, we get (7.5) e x sinx e x cosx e x cosx Inserting this in 7.4 leads to (7.6) e x cosx e x sinx e x cosx e x cosx Bringing the last term over to the left hand side and dividing by gives us the answer: (7.7) e x cosx ex sinx e x cosx C Example 7.9 If a calculation of a definite integral involves integration by parts, it is a good idea to evaluate as soon as integrated terms appear. We illustrate with the calculation of (7.8) Let u lnx dv so that du x v x, and 4 lnx (7.9) 4 4 lnx xlnx 4 4 4ln4 x 4ln4 3 Example 7.0 (7.30) 0 arcsinx? We make the substitution u arcsinx dv du x v x. Then (7.3) 0 arcsinx xarcsinx 0 0 x x Now, to complete the last integral, let u x du x, leading us to (7.3) 0 arcsinx π u du π u 3 4 π 3

6 Chapter 7 Techniques of Integration 7.3. Partial Fractions The point of the partial fractions expansion is that integration of a rational function can be reduced to the following formulae, once we have determined the roots of the polynomial in the denominator. Proposition 7. a) x a ln x a C du b) u b b arctan u C b udu c) u b ln u b C These are easily verified by differentiating the right hand sides (or by using previous techniques). Example 7. Let us illustrate with an example we ve already seen. To find the integral (7.33) we check that (7.34) so that (7.35) x a x b x a x b x a x b a b x a x b a b ln x a ln x b C x a a b ln x b The trick 7.34 can be applied to any rational function. ny polynomial can be written as a product of factors of the form x r or x a b, where r is a real root and the quadratic terms correspond to the conjugate pairs of complex roots. The partial fraction expansion allows us to write the quotient of polynomials as a sum of terms whose denominators are of these forms, and thus the integration is reduced to Proposition 7.. Here is the partial fractions procedure.. Given a rational function R x, if the degree of the numerator is not less than the degree of the denominator, by long division, we can write (7.36) R x Q x p x q x where now deg p deg q.. Find the roots of q x 0. If the roots are all distinct (there are no multiple roots), write p q as a sum of terms of the form C (7.37) x r B x a b Cx x a b 3. Find the values of B C. 4. Integrate term by term using Proposition 7..

7 7.3 Partial Fractions 3 If the roots are not distinct, the expansion is more complicated; we shall resume this discussion later. For the present let us concentrate on the case of distinct roots, and how to find the coefficients B C in x Example 7. Integrate x x First we write (7.38) x x x x Now multiply this equation by x x, getting (7.39) x x B x B x If we substitute x, we get, so ; now letting x, we get B, so B, and 7.38 becomes (7.40) x x x x x Integrating, we get (7.4) x x x ln x ln x C ln x x C So, this is the procedure for finding the coefficients of the partial fractions expansion when the roots are all real and distinct:. Write down the expansion with unknown coefficients.. Multiply through by the product of all the terms x r. 3. Substitute each root in the above equation; each substitution determines one of the coefficients. x Example 7.3 Integrate 3 x x 3 Here the roots are 3, so we have the expansion (7.4) leading to x 3 x x 3 x B x C x 3 (7.43) x 3 x x 3 B x x 3 C x x Substitute x : 3 4, so 4. Substitute x : 3 B, so B. Substitute x 3 : 9 3 C 4, so C 3 4, and 7.4 becomes (7.44) and the integral is (7.45) x 3 x x 3 x 3 x x 3 4 x 4 ln x x ln x 3 4 x ln x 3 C

8 Chapter 7 Techniques of Integration Quadratic Factors Example 7.4 x 4x 5? Here we can factor: x 4x 5 x (7.46) x 5, so we can write x 4x 5 x B x 5 and solve for and B as above: 6 B 6, so we have (7.47) and the integral is (7.48) x 4x 5 6 x 5 x x 4x 5 6 ln x 5 C x Example 7.5 x 4x 5? Here we can t find real factors, because the roots are complex. But we can complete the square: x 4x 5 x, and now use Proposition 7.b: (7.49) x 4x 5 x arctan x C x 3 Example 7.6 x 4x 5? Here we have to be a little more resourceful. gain, we complete the square, giving (7.50) x 3 x 4x 5 x 3 x If only that x 3 were x, we could use Proposition 7.c, with u x. Well, since x 3 x 5, there is no problem: (7.5) x 3 x 4x 5 x x 5 x ln x 5arctan x C x Example 7.7 x 6x 4? First, we complete the square in the denominator: x 6x 4 x 3 5. Now, write the numerator in terms of x 3 : x x 3 7. This gives the expansion: (7.5) x x 6x 4 7 x 6x 4 x 3 x 6x 4

9 7.3 Partial Fractions 5 so, using Proposition 7.: (7.53) x x 6x 4 7 x 3 5 x 3 x 3 5 (7.54) 7 5 arctan x 3 5 ln x 3 5 C x Example 7.8? x x Here we have to expect each of the terms in Proposition 7. to appear, so we try an expression of the form (7.55) x x x x B x Cx x Clearing the denominators on the right, we are led to the equation (7.56) x x Bx Cx Setting x 0 gives. But we have no more roots to substitute to find B and C, so instead we equate coefficients. The coefficient of x on the left is 0, and on the right is C, so C 0; since, we learn that C. Comparing coefficients of x we learn that B. Thus 7.55 becomes (7.57) and our integral is (7.58) x x x x x x x x x x ln x arctanx ln x C x Example 7.9? x x 4x 5 The denominator is x x, so we expect a partial fractions expansion of the form (7.59) x x x 4x 5 x Clearing of denominators, we obtain the equation B x C x x (7.60) x x Bx C x x For x 0, we obtain 5, so 5. Comparing coefficients of x we obtain C, so C 5. Comparing coefficients of x we obtain 0 4 B C, so B 5, so B 5 and 7.59 becomes (7.6) x x x 4x 5 5 x 5 x x 5 x

10 Chapter 7 Techniques of Integration 6 which we can integrate to (7.6) x x x 4x 5 5 ln x 5 arctan x 0 ln x 4x 5 C Multiple Roots If the denominator has a multiple root, that is there is a factor x r raised to a power, then we have to allow for the possibility of terms in the partial fraction of the form x r raised to the same power. But then the numerator can be (as we have seen above in the case of quadratic factors) a polynomial of degree as much as one less than the power. This is best explained through a few examples. x Example 7.0? x 3 x We have to allow for the possibility of a term of the form x Bx C x 3, or, what is the same, an expansion of the form (7.63) x x 3 x x B x C x 3 D x Clearing of denominators, we obtain (7.64) x x x Bx x C x Dx 3 Substituting x 0 we obtain C, so C. Substituting x, we obtain D. To find and B we have to compare coefficients of powers of x. Equating coefficients of x 3, we have 0 D, so. Equating coefficients of x, we have B, so B. Thus the expansion 7.63 is (7.65) x x 3 x x x x 3 x which we can integrate term by term: (7.66) x x 3 x ln x x x ln x C 7.4. Trigonometric Methods Now, although the above techniques are all that one needs to know in order to use a Table of Integrals, there is one form which appears so often, that it is worthwhile seeing how the integration formulae are found. Expressions involving the square root of a quadratic function occur quite frequently in practice. How do we integrate x or x? When the expressions involve a square root of a quadratic, we can convert to trigonometric functions using the substitutions suggested by figure 7..

11 7.4 Trigonometric Methods 7 Figure 7. PSfrag replacements x x x u u x () (B) Example 7. To find x, we use the substitution of figure 7.: x sinu cosudu x cosu. Then (7.67) x cos udu Now, we use the half-angle formula: cos u cosu : (7.68) x cosu du u sinu C 4 Now, to return to the original variable x, we have to use the double angle formula: sinu sinucosu x x, and we finally have the answer: (7.69) x arcsin x x x C 4 Example 7. To find x, we use the substitution of figure 7.B: x tanu sec udu x secu. Then (7.70) x sec 3 udu This is still a hard integral, but we can discover it by an integration by parts (see Practice Problem set 4, problem 6) to be (7.7) sec 3 du secutanu ln secu tanu C Now, we return to figure 7.B to write this in terms of x: tanu x secu x We finally obtain (7.7) x x x ln x x C Example 7.3 x x?

12 Chapter 7 Techniques of Integration 8 Don t be misled: always try simple substitution first; in this case the substitution u x du x leads to the formula (7.73) x x u du 3 x 3 C Example 7.4 x x? Here simple substitution fails, and we use the substitution of figure 7.: x sinu cosudu x cosu. Then (7.74) x x sin ucos udu This integration now follows from use of double- and half-angle formulae: (7.75) sin ucos udu 4 sin u du cos 4u du 8 sin 4u u C 8 4 Now, sin 4u sin u cos u 4sinucosu sin u 4x x x Finally (7.76) x x arcsinx 8 x x x C For the remainder of this course, we shall assume that you have a table of integrals available, and know how to use it. There are several handbooks, and every Calculus text has a table of integrals on the inside back cover. There are a few tables on the web: /table of integrals.htm /calculus/calculus table integrals content.htm jrp/ma00/website/int /node5.html

### Techniques of Integration

8 Techniques of Integration Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Sometimes this is a simple problem, since it

### 5 Indefinite integral

5 Indefinite integral The most of the mathematical operations have inverse operations: the inverse operation of addition is subtraction, the inverse operation of multiplication is division, the inverse

### Techniques of Integration

0 Techniques of Integration ½¼º½ ÈÓÛ Ö Ó Ò Ò Ó Ò Functions consisting of products of the sine and cosine can be integrated by using substitution and trigonometric identities. These can sometimes be tedious,

### The Method of Partial Fractions Math 121 Calculus II Spring 2015

Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method

### 1.7. Partial Fractions. 1.7.1. Rational Functions and Partial Fractions. A rational function is a quotient of two polynomials: R(x) = P (x) Q(x).

.7. PRTIL FRCTIONS 3.7. Partial Fractions.7.. Rational Functions and Partial Fractions. rational function is a quotient of two polynomials: R(x) = P (x) Q(x). Here we discuss how to integrate rational

### Integrals of Rational Functions

Integrals of Rational Functions Scott R. Fulton Overview A rational function has the form where p and q are polynomials. For example, r(x) = p(x) q(x) f(x) = x2 3 x 4 + 3, g(t) = t6 + 4t 2 3, 7t 5 + 3t

### Section 4.4. Using the Fundamental Theorem. Difference Equations to Differential Equations

Difference Equations to Differential Equations Section 4.4 Using the Fundamental Theorem As we saw in Section 4.3, using the Fundamental Theorem of Integral Calculus reduces the problem of evaluating a

### Analytic Trigonometry

Name Chapter 5 Analytic Trigonometry Section 5.1 Using Fundamental Identities Objective: In this lesson you learned how to use fundamental trigonometric identities to evaluate trigonometric functions and

### 2 Integrating Both Sides

2 Integrating Both Sides So far, the only general method we have for solving differential equations involves equations of the form y = f(x), where f(x) is any function of x. The solution to such an equation

### Chapter 11. Techniques of Integration

Chapter Techniques of Integration Chapter 6 introduced the integral. There it was defined numerically, as the limit of approximating Riemann sums. Evaluating integrals by applying this basic definition

### 1 Lecture: Integration of rational functions by decomposition

Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.

### Trigonometry Review with the Unit Circle: All the trig. you ll ever need to know in Calculus

Trigonometry Review with the Unit Circle: All the trig. you ll ever need to know in Calculus Objectives: This is your review of trigonometry: angles, six trig. functions, identities and formulas, graphs:

### 6 Further differentiation and integration techniques

56 6 Further differentiation and integration techniques Here are three more rules for differentiation and two more integration techniques. 6.1 The product rule for differentiation Textbook: Section 2.7

### Notes and questions to aid A-level Mathematics revision

Notes and questions to aid A-level Mathematics revision Robert Bowles University College London October 4, 5 Introduction Introduction There are some students who find the first year s study at UCL and

### 2 2 [ 1, 1]. Thus there exists a function sin 1 or arcsin. sin x arcsin x(= sin 1 x) arcsin(x) = y sin(y) = x. arcsin(sin x) = x if π 2 x π 2

7.4 Inverse Trigonometric Functions From looking at the graphs of the trig functions, we see that they fail the horizontal line test spectacularly. However, if you restrict their domain, you can find an

### PRACTICE FINAL. Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 10cm.

PRACTICE FINAL Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 1cm. Solution. Let x be the distance between the center of the circle

### EXAM. Practice Questions for Exam #2. Math 3350, Spring April 3, 2004 ANSWERS

EXAM Practice Questions for Exam #2 Math 3350, Spring 2004 April 3, 2004 ANSWERS i Problem 1. Find the general solution. A. D 3 (D 2)(D 3) 2 y = 0. The characteristic polynomial is λ 3 (λ 2)(λ 3) 2. Thus,

### Algebra. Exponents. Absolute Value. Simplify each of the following as much as possible. 2x y x + y y. xxx 3. x x x xx x. 1. Evaluate 5 and 123

Algebra Eponents Simplify each of the following as much as possible. 1 4 9 4 y + y y. 1 5. 1 5 4. y + y 4 5 6 5. + 1 4 9 10 1 7 9 0 Absolute Value Evaluate 5 and 1. Eliminate the absolute value bars from

### Verifying Trigonometric Identities. Introduction. is true for all real numbers x. So, it is an identity. Verifying Trigonometric Identities

333202_0502.qxd 382 2/5/05 Chapter 5 5.2 9:0 AM Page 382 Analytic Trigonometry Verifying Trigonometric Identities What you should learn Verify trigonometric identities. Why you should learn it You can

### Review Solutions MAT V1102. 1. (a) If u = 4 x, then du = dx. Hence, substitution implies 1. dx = du = 2 u + C = 2 4 x + C.

Review Solutions MAT V. (a) If u 4 x, then du dx. Hence, substitution implies dx du u + C 4 x + C. 4 x u (b) If u e t + e t, then du (e t e t )dt. Thus, by substitution, we have e t e t dt e t + e t u

### 6.6 The Inverse Trigonometric Functions. Outline

6.6 The Inverse Trigonometric Functions Tom Lewis Fall Semester 2015 Outline The inverse sine function The inverse cosine function The inverse tangent function The other inverse trig functions Miscellaneous

### Trigonometric Identities and Conditional Equations C

Trigonometric Identities and Conditional Equations C TRIGONOMETRIC functions are widely used in solving real-world problems and in the development of mathematics. Whatever their use, it is often of value

### y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx

Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE Evaluate cos 3 x dx.

### MATH SOLUTIONS TO PRACTICE FINAL EXAM. (x 2)(x + 2) (x 2)(x 3) = x + 2. x 2 x 2 5x + 6 = = 4.

MATH 55 SOLUTIONS TO PRACTICE FINAL EXAM x 2 4.Compute x 2 x 2 5x + 6. When x 2, So x 2 4 x 2 5x + 6 = (x 2)(x + 2) (x 2)(x 3) = x + 2 x 3. x 2 4 x 2 x 2 5x + 6 = 2 + 2 2 3 = 4. x 2 9 2. Compute x + sin

### 9.1 Trigonometric Identities

9.1 Trigonometric Identities r y x θ x -y -θ r sin (θ) = y and sin (-θ) = -y r r so, sin (-θ) = - sin (θ) and cos (θ) = x and cos (-θ) = x r r so, cos (-θ) = cos (θ) And, Tan (-θ) = sin (-θ) = - sin (θ)

### The graph of. horizontal line between-1 and 1 that the sine function is not 1-1 and therefore does not have an inverse.

Inverse Trigonometric Functions The graph of If we look at the graph of we can see that if you draw a horizontal line between-1 and 1 that the sine function is not 1-1 and therefore does not have an inverse.

### Differentiation and Integration

This material is a supplement to Appendix G of Stewart. You should read the appendix, except the last section on complex exponentials, before this material. Differentiation and Integration Suppose we have

### pp. 4 8: Examples 1 6 Quick Check 1 6 Exercises 1, 2, 20, 42, 43, 64

Semester 1 Text: Chapter 1: Tools of Algebra Lesson 1-1: Properties of Real Numbers Day 1 Part 1: Graphing and Ordering Real Numbers Part 1: Graphing and Ordering Real Numbers Lesson 1-2: Algebraic Expressions

### y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx y 1 u 2 du u 1 3u 3 C

Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE Evaluate cos 3 x dx.

### Partial Fractions. Combining fractions over a common denominator is a familiar operation from algebra:

Partial Fractions Combining fractions over a common denominator is a familiar operation from algebra: From the standpoint of integration, the left side of Equation 1 would be much easier to work with than

### Taylor Polynomials and Taylor Series Math 126

Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will

### Find all of the real numbers x that satisfy the algebraic equation:

Appendix C: Factoring Algebraic Expressions Factoring algebraic equations is the reverse of expanding algebraic expressions discussed in Appendix B. Factoring algebraic equations can be a great help when

### Partial Fractions. p(x) q(x)

Partial Fractions Introduction to Partial Fractions Given a rational function of the form p(x) q(x) where the degree of p(x) is less than the degree of q(x), the method of partial fractions seeks to break

### Solutions to Exercises, Section 5.1

Instructor s Solutions Manual, Section 5.1 Exercise 1 Solutions to Exercises, Section 5.1 1. Find all numbers t such that ( 1 3,t) is a point on the unit circle. For ( 1 3,t)to be a point on the unit circle

### MORE TRIG IDENTITIES MEMORIZE!

MORE TRIG IDENTITIES MEMORIZE! SUM IDENTITIES Memorize: sin( u + v)= sinu cosv + cosu sinv Think: Sum of the mixed-up products (Multiplication and addition are commutative, but start with the sinu cosv

### Section 6-3 Double-Angle and Half-Angle Identities

6-3 Double-Angle and Half-Angle Identities 47 Section 6-3 Double-Angle and Half-Angle Identities Double-Angle Identities Half-Angle Identities This section develops another important set of identities

### M3 PRECALCULUS PACKET 1 FOR UNIT 5 SECTIONS 5.1 TO = to see another form of this identity.

M3 PRECALCULUS PACKET FOR UNIT 5 SECTIONS 5. TO 5.3 5. USING FUNDAMENTAL IDENTITIES 5. Part : Pythagorean Identities. Recall the Pythagorean Identity sin θ cos θ + =. a. Subtract cos θ from both sides

### GRE Prep: Precalculus

GRE Prep: Precalculus Franklin H.J. Kenter 1 Introduction These are the notes for the Precalculus section for the GRE Prep session held at UCSD in August 2011. These notes are in no way intended to teach

### Integration Involving Trigonometric Functions and Trigonometric Substitution

Integration Involving Trigonometric Functions and Trigonometric Substitution Dr. Philippe B. Laval Kennesaw State University September 7, 005 Abstract This handout describes techniques of integration involving

### Sample Problems. Practice Problems

Lecture Notes Partial Fractions page Sample Problems Compute each of the following integrals.. x dx. x + x (x + ) (x ) (x ) dx 8. x x dx... x (x + ) (x + ) dx x + x x dx x + x x + 6x x dx + x 6. 7. x (x

### Partial Fractions. (x 1)(x 2 + 1)

Partial Fractions Adding rational functions involves finding a common denominator, rewriting each fraction so that it has that denominator, then adding. For example, 3x x 1 3x(x 1) (x + 1)(x 1) + 1(x +

### Solutions to Homework 10

Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x

### The Derivative. Philippe B. Laval Kennesaw State University

The Derivative Philippe B. Laval Kennesaw State University Abstract This handout is a summary of the material students should know regarding the definition and computation of the derivative 1 Definition

### 29 Wyner PreCalculus Fall 2016

9 Wyner PreCalculus Fall 016 CHAPTER THREE: TRIGONOMETRIC EQUATIONS Review November 8 Test November 17 Trigonometric equations can be solved graphically or algebraically. Solving algebraically involves

### Senior Secondary Australian Curriculum

Senior Secondary Australian Curriculum Mathematical Methods Glossary Unit 1 Functions and graphs Asymptote A line is an asymptote to a curve if the distance between the line and the curve approaches zero

### Blue Pelican Calculus First Semester

Blue Pelican Calculus First Semester Teacher Version 1.01 Copyright 2011-2013 by Charles E. Cook; Refugio, Tx Edited by Jacob Cobb (All rights reserved) Calculus AP Syllabus (First Semester) Unit 1: Function

### Trigonometric Identities

Trigonometric Identities Dr. Philippe B. Laval Kennesaw STate University April 0, 005 Abstract This handout dpresents some of the most useful trigonometric identities. It also explains how to derive new

### 1 TRIGONOMETRY. 1.0 Introduction. 1.1 Sum and product formulae. Objectives

TRIGONOMETRY Chapter Trigonometry Objectives After studying this chapter you should be able to handle with confidence a wide range of trigonometric identities; be able to express linear combinations of

### x 2 + y 2 = 1 y 1 = x 2 + 2x y = x 2 + 2x + 1

Implicit Functions Defining Implicit Functions Up until now in this course, we have only talked about functions, which assign to every real number x in their domain exactly one real number f(x). The graphs

### First Order Non-Linear Equations

First Order Non-Linear Equations We will briefly consider non-linear equations. In general, these may be much more difficult to solve than linear equations, but in some cases we will still be able to solve

### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Assume that the cities lie on the same north-south line and that the radius of the earth

### Mathematical Procedures

CHAPTER 6 Mathematical Procedures 168 CHAPTER 6 Mathematical Procedures The multidisciplinary approach to medicine has incorporated a wide variety of mathematical procedures from the fields of physics,

### Limit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x)

SEC. 4.1 TAYLOR SERIES AND CALCULATION OF FUNCTIONS 187 Taylor Series 4.1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. For example, the derivative f f (x + h) f

### South Carolina College- and Career-Ready (SCCCR) Pre-Calculus

South Carolina College- and Career-Ready (SCCCR) Pre-Calculus Key Concepts Arithmetic with Polynomials and Rational Expressions PC.AAPR.2 PC.AAPR.3 PC.AAPR.4 PC.AAPR.5 PC.AAPR.6 PC.AAPR.7 Standards Know

### Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: f (x) =

Vertical Asymptotes Definition of Vertical Asymptote The line x = a is called a vertical asymptote of f (x) if at least one of the following is true: lim f (x) = x a lim f (x) = lim x a lim f (x) = x a

### Week 13 Trigonometric Form of Complex Numbers

Week Trigonometric Form of Complex Numbers Overview In this week of the course, which is the last week if you are not going to take calculus, we will look at how Trigonometry can sometimes help in working

### Second Order Linear Differential Equations

CHAPTER 2 Second Order Linear Differential Equations 2.. Homogeneous Equations A differential equation is a relation involving variables x y y y. A solution is a function f x such that the substitution

5.3 SOLVING TRIGONOMETRIC EQUATIONS Copyright Cengage Learning. All rights reserved. What You Should Learn Use standard algebraic techniques to solve trigonometric equations. Solve trigonometric equations

### Inverse Trigonometric Functions - Trigonometric Equations

Inverse Trigonometric Functions - Trigonometric Equations Dr. Philippe B. Laval Kennesaw STate University April 0, 005 Abstract This handout defines the inverse of the sine, cosine and tangent functions.

### Approximating functions by Taylor Polynomials.

Chapter 4 Approximating functions by Taylor Polynomials. 4.1 Linear Approximations We have already seen how to approximate a function using its tangent line. This was the key idea in Euler s method. If

### Calculus 1: Sample Questions, Final Exam, Solutions

Calculus : Sample Questions, Final Exam, Solutions. Short answer. Put your answer in the blank. NO PARTIAL CREDIT! (a) (b) (c) (d) (e) e 3 e Evaluate dx. Your answer should be in the x form of an integer.

### Algebra 2/ Trigonometry Extended Scope and Sequence (revised )

Algebra 2/ Trigonometry Extended Scope and Sequence (revised 2012 2013) Unit 1: Operations with Radicals and Complex Numbers 9 days 1. Operations with radicals (p.88, 94, 98, 101) a. Simplifying radicals

### 106 Chapter 5 Curve Sketching. If f(x) has a local extremum at x = a and. THEOREM 5.1.1 Fermat s Theorem f is differentiable at a, then f (a) = 0.

5 Curve Sketching Whether we are interested in a function as a purely mathematical object or in connection with some application to the real world, it is often useful to know what the graph of the function

PRE-CALCULUS GRADE 12 [C] Communication Trigonometry General Outcome: Develop trigonometric reasoning. A1. Demonstrate an understanding of angles in standard position, expressed in degrees and radians.

### Section 6-4 Product Sum and Sum Product Identities

480 6 TRIGONOMETRIC IDENTITIES AND CONDITIONAL EQUATIONS Section 6-4 Product Sum and Sum Product Identities Product Sum Identities Sum Product Identities Our work with identities is concluded by developing

### Partial Inverse Trigonometric Functions

Partial nverse Trigonometric Functions Although the three basic trigonometric functions sin, cos, and tan do not have inverse functions, they do have what can be called partial inverses. These partial

### Section 7.4 Trigonometric Identities

9 Section 7.4 Trigonometric Identities Definition Two functions f and g are said to be identically equal if f(x) g(x) for all x in the domain of f and g. We say that f(x) g(x) is an identity. Determine

### Section 8.1: The Inverse Sine, Cosine, and Tangent Functions

Section 8.1: The Inverse Sine, Cosine, and Tangent Functions The function y = sin x doesn t pass the horizontal line test, so it doesn t have an inverse for every real number. But if we restrict the function

### INVERSE TRIGONOMETRIC FUNCTIONS. Colin Cox

INVERSE TRIGONOMETRIC FUNCTIONS Colin Cox WHAT IS AN INVERSE TRIG FUNCTION? Used to solve for the angle when you know two sides of a right triangle. For example if a ramp is resting against a trailer,

### Unit 8 Inverse Trig & Polar Form of Complex Nums.

HARTFIELD PRECALCULUS UNIT 8 NOTES PAGE 1 Unit 8 Inverse Trig & Polar Form of Complex Nums. This is a SCIENTIFIC OR GRAPHING CALCULATORS ALLOWED unit. () Inverse Functions (3) Invertibility of Trigonometric

### Chapter 11 - Curve Sketching. Lecture 17. MATH10070 - Introduction to Calculus. maths.ucd.ie/modules/math10070. Kevin Hutchinson.

Lecture 17 MATH10070 - Introduction to Calculus maths.ucd.ie/modules/math10070 Kevin Hutchinson 28th October 2010 Z Chain Rule (I): If y = f (u) and u = g(x) dy dx = dy du du dx Z Chain rule (II): d dx

### Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

### 39 Verifying Trigonometric Identities

39 Verifying Trigonometric Identities In this section, you will learn how to use trigonometric identities to simplify trigonometric expressions. Equations such as (x 2)(x + 2) x 2 4 or x2 x x + are referred

### Lecture 5 : Continuous Functions Definition 1 We say the function f is continuous at a number a if

Lecture 5 : Continuous Functions Definition We say the function f is continuous at a number a if f(x) = f(a). (i.e. we can make the value of f(x) as close as we like to f(a) by taking x sufficiently close

### correct-choice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were:

Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that

### Calculating Areas Section 6.1

A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics Calculating Areas Section 6.1 Dr. John Ehrke Department of Mathematics Fall 2012 Measuring Area By Slicing We first defined

### SUM AND DIFFERENCE FORMULAS

SUM AND DIFFERENCE FORMULAS Introduction We have several identities that we are concentrating on in this section: o Difference Identities for Cosine o Sum Identities for Cosine o Cofunction Identities

### Thnkwell s Homeschool Precalculus Course Lesson Plan: 36 weeks

Thnkwell s Homeschool Precalculus Course Lesson Plan: 36 weeks Welcome to Thinkwell s Homeschool Precalculus! We re thrilled that you ve decided to make us part of your homeschool curriculum. This lesson

### is identically equal to x 2 +3x +2

Partial fractions 3.6 Introduction It is often helpful to break down a complicated algebraic fraction into a sum of simpler fractions. 4x+7 For example it can be shown that has the same value as 1 + 3

### 3.3. Solving Polynomial Equations. Introduction. Prerequisites. Learning Outcomes

Solving Polynomial Equations 3.3 Introduction Linear and quadratic equations, dealt within Sections 3.1 and 3.2, are members of a class of equations, called polynomial equations. These have the general

### 4.4 CURVE SKETCHING. there is no horizontal asymptote. Since sx 1 l 0 as x l 1 and f x is always positive, we have

SECTION 4.4 CURE SKETCHING 4.4 CURE SKETCHING EXAMPLE A Sketch the graph of f x. sx A. Domain x x 0 x x, B. The x- and y-intercepts are both 0. C. Symmetry: None D. Since x l sx there is no horizontal

### Euler s Formula Math 220

Euler s Formula Math 0 last change: Sept 3, 05 Complex numbers A complex number is an expression of the form x+iy where x and y are real numbers and i is the imaginary square root of. For example, + 3i

### D f = (2, ) (x + 1)(x 3) (b) g(x) = x 1 solution: We need the thing inside the root to be greater than or equal to 0. So we set up a sign table.

. Find the domains of the following functions: (a) f(x) = ln(x ) We need x > 0, or x >. Thus D f = (, ) (x + )(x 3) (b) g(x) = x We need the thing inside the root to be greater than or equal to 0. So we

### UNIT 1: ANALYTICAL METHODS FOR ENGINEERS

UNIT : ANALYTICAL METHODS FOR ENGINEERS Unit code: A/60/40 QCF Level: 4 Credit value: 5 OUTCOME 3 - CALCULUS TUTORIAL DIFFERENTIATION 3 Be able to analyse and model engineering situations and solve problems

### CALCULUS 2. 0 Repetition. tutorials 2015/ Find limits of the following sequences or prove that they are divergent.

CALCULUS tutorials 5/6 Repetition. Find limits of the following sequences or prove that they are divergent. a n = n( ) n, a n = n 3 7 n 5 n +, a n = ( n n 4n + 7 ), a n = n3 5n + 3 4n 7 3n, 3 ( ) 3n 6n

### FURTHER TRIGONOMETRIC IDENTITIES AND EQUATIONS

Mathematics Revision Guides Further Trigonometric Identities and Equations Page of 7 M.K. HOME TUITION Mathematics Revision Guides Level: AS / A Level AQA : C4 Edexcel: C3 OCR: C3 OCR MEI: C4 FURTHER TRIGONOMETRIC

### Higher Education Math Placement

Higher Education Math Placement Placement Assessment Problem Types 1. Whole Numbers, Fractions, and Decimals 1.1 Operations with Whole Numbers Addition with carry Subtraction with borrowing Multiplication

### 10 Polar Coordinates, Parametric Equations

Polar Coordinates, Parametric Equations ½¼º½ ÈÓÐ Ö ÓÓÖ Ò Ø Coordinate systems are tools that let us use algebraic methods to understand geometry While the rectangular (also called Cartesian) coordinates

### Inverse Circular Function and Trigonometric Equation

Inverse Circular Function and Trigonometric Equation 1 2 Caution The 1 in f 1 is not an exponent. 3 Inverse Sine Function 4 Inverse Cosine Function 5 Inverse Tangent Function 6 Domain and Range of Inverse

### Trigonometric Identities and Equations

Chapter 4 Trigonometric Identities and Equations Trigonometric identities describe equalities between related trigonometric expressions while trigonometric equations ask us to determine the specific values

### Solutions to modified 2 nd Midterm

Math 125 Solutions to moifie 2 n Miterm 1. For each of the functions f(x) given below, fin f (x)). (a) 4 points f(x) = x 5 + 5x 4 + 4x 2 + 9 Solution: f (x) = 5x 4 + 20x 3 + 8x (b) 4 points f(x) = x 8

### MATHEMATICS (CLASSES XI XII)

MATHEMATICS (CLASSES XI XII) General Guidelines (i) All concepts/identities must be illustrated by situational examples. (ii) The language of word problems must be clear, simple and unambiguous. (iii)

### Partial differentiation

Chapter 1 Partial differentiation Example 1.1 What is the maximal domain of the real function g defined b g(x) = x 2 + 3x + 2? : The ke point is that the square root onl gives a real result if the argument

### Semester 2, Unit 4: Activity 21

Resources: SpringBoard- PreCalculus Online Resources: PreCalculus Springboard Text Unit 4 Vocabulary: Identity Pythagorean Identity Trigonometric Identity Cofunction Identity Sum and Difference Identities

### Focusing properties of spherical and parabolic mirrors

Physics 5B Winter 2009 Focusing properties of spherical and parabolic mirrors 1 General considerations Consider a curved mirror surface that is constructed as follows Start with a curve, denoted by y()

### Chapter 1 Quadratic Equations in One Unknown (I)

Tin Ka Ping Secondary School 015-016 F. Mathematics Compulsory Part Teaching Syllabus Chapter 1 Quadratic in One Unknown (I) 1 1.1 Real Number System A Integers B nal Numbers C Irrational Numbers D Real

### Unit 6 Introduction to Trigonometry Inverse Trig Functions (Unit 6.6)

Unit 6 Introduction to Trigonometry Inverse Trig Functions (Unit 6.6) William (Bill) Finch Mathematics Department Denton High School Lesson Goals When you have completed this lesson you will: Understand