Derivative and Integral Rules These are on the inside of the back cover of your text.

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2 Derivative and Integral Rules These are on the inside of the back cover of your text. General Derivative Rule General Integral Rule d dx u(x) r = r u(x) r - 1 u(x) u(x)r u(x) dx = u(x) r1 r1 + C r U -1 d dx e u(x) = e u(x) u(x) eu(x) u(x) dx = e u(x) + C d dx b u(x) = ln(b) b u(x) u(x) bu(x) u (x) dx bu(x) ln(b) C d dx log b (u(x)) = u(x) ln(b) u(x) u (x) u(x) dx ln(b) log bu (x) C d dx ln(u(x)) = u(x) u(x) u(x) u(x) dx = ln(u(x)) + C u(x) u(x) dx = ln(u(x)) + C

3 General Derivative Rule General Integral Rule d dx sin(u(x)) = cos(u(x)) u(x) cos(u(x)) u(x) dx = sin(u(x)) + C d dx cos(u(x)) = - sin(u(x)) u(x) sin(u(x)) u(x) dx = - cos(u(x)) + C d dx tan(u(x)) = sec 2 (u(x)) u(x) sec2 (u(x)) u(x) dx = tan(u(x)) + C d dx cot(u(x)) = - csc 2 (u(x)) u(x) csc2 (u(x)) u(x) dx = - cot(u(x)) + C d dx sec(u(x)) = sec(u(x)) tan(u(x)) u(x) sec(u(x)) tan(u(x)) u(x) dx = sec(u(x)) + C d csc(u(x)) dx = - csc(u(x))cot(u(x)) u(x) csc(u(x)) cot(u(x)) u(x) dx = - csc(u(x)) + C

4 Which of the following can you integrate using the Chain Rule and introducing a multiplicative constant? g(x) dx 1 k k ( g(x) dx The object is to express the integral in a form that fits one of the general rules: g(x) dx 1 k u(x) ( f(u(x)) dx

5 a e 5x dx b 2sin(3t) dt c r 1 r 3 dr

6 d 1 t 1/2 dt e t 2 1 t 3 3t5 dt f cos(4s) ds

7 Which can you integrate? What Rule do you use? (a) e 3t dt (b) (c) 2 sin(3t) dt 6 t 3 dt (d) 5 t dt (e) (f) 6 t 2 t dt cos(sin(t)) dt

8 6x 2 x 3 dx (t3) t 2 6t 1 dt 6 cos(t) sin(t) dt

9 Algebraic Manipulation: Trig identities tan2 (t) dt sin3 (t) dt

10 Chain Rule Substitution a sin(t) e 3cos(t) dt b 5e 3t sin(e 3t ) dt c 4 ln(6) ln(5) t 3 dt d e f 5t 2/3 dt 6t3 (t 2 t) 5 dt cos(t) cos(sin(t)) dt

11 Introduction. Evaluating an integral that is not in the form of a basic general integral formula. First approach. Is it possible to manipulate the integrand to obtain an equivalent integral that is of a form that can be directly evaluated. We consider three types of integrals. 1. I = f(x) dx When f(x) has a distinguished sub-term a substitution method may work. If the sub-term u(x) and u(x) is present, when a. u = a + b x r substitute u b. u = e h(x) substitute u c. u = ±a 2 ± x 2 Use trig-substitutions.

12 2. For I = f(x)( g(x) dx When g(x) U f(x) the integral can be transformed by the method of Integration of Parts (IP), based on the product rule (u(v) = u(v + u(v Solving for one term on the right gives so, upon integrating, u(v = 1u(v dx = 1 dx - 1 dx The FTC gives 1(u(v)dx =. So, the IP formula gives the integral of u(v as the term u(v minus the integral of u(v. IP:

13 3. When the integral involves a fraction. The integral I = f(x) g(x) dx can be evaluated directly only if f(x) = or, if g(x) = u r (x) and f(x) = Otherwise, one must reduce the fraction to obtain one of these cases to evaluate I. The basic principle is to write a fraction as the sum of two simpler (partial) fractions: 5/6 = 1/2 + 1/3

14 6.1 Substitutions. Theory. Applying the general integration rules involves making a simple substitution. To integrate an integral I = f(x) dx using one of the General Integration Rules you must recognize a term u(x) and a function F (x) so that the integral has the form F (u) u (x) dx = The integral form of the Chain Rule.

15 If the choice of the function F is not obvious the integral may be evaluated by first manipulating the integrand into a different form. If there is a distinguished sub-term u(x) one can try a substitution u = u(x). To transform the integral with respect to x to an integral with respect to u we need to express f(x) and dx in terms of u and du. If u = u(x) that can be solved for x = g(u). Then and dx = I = f(g(u)) g(u) du The function u is chosen so that f(g(u)) g(u) becomes a nice/simple function of u.

16 This method works only if the resulting u-integral can be easily evaluated using known integral formulas. Then, the antiderivative must be transformed back to a function of x by setting u = g(x).

17 6.1 Substitutions. Examples. Linear substitution. u = ax + b Evaluate I = 1 x( 2 - x) 1/3 dx

18 Exponential Substitution. u = e h(x). Evaluate I = 1 (e 2x - 3) 2 e 2x dx

19 Trigonometric Substitutions for a 2 + x 2, a 2 - x 2, or x 2 - a 2. Integrals involving quadratic functions can often be evaluated by using simple "trigonometric" substitutions. These substitutions "work" is because of the Pythagorean Identities: sin 2 (θ) + cos 2 (θ) = 1 or tan 2 (θ) + 1 = sec 2 (θ) sec 2 (θ) - 1 = tan 2 (θ)

20 There are three basic trig-substitutions, depending on the form of the integrand. SINE SUBSTITUTION: For a 2 - x 2 use the substitution: x = a sin(θ) and dx = Then, a 2 - x 2 = = =. Then x a x a θ = sin -1 ( ) or θ = Arcsin( )

21 TANGENT SUBSTITUTION: For a 2 + x 2 The substitution is: x = a tan(θ) and dx = a sec 2 (θ) dθ, Then, a 2 + x 2 = = =. x in this case θ = tan -1 ( ) a or θ = Arctan( ) x a

22 SECANT SUBSTITUTION: For x 2 - a 2 The substitution: x = a sec(θ) and dx = a sec(θ)tan(θ) dθ Then, x 2 - a 2 = = =. x a x a θ = sec -1 ( ) or θ = Arcsec( )

23 The three trig-substitutions result in integrals with respect to θ. After integrating the antiderivtive must be converted back from the θ-variable to the x-variable. This can be done two ways: i) or ii) using right triangles having angle θ and side lengths x and constant a: a 2 + x 2 a x x x x 2 - a 2 a a 2 - x 2 a x = a tan(0) x = a sin(0) x = a sec(0)

24 Example. Evaluate 1 19x 2 dx

25 Evaluate I = 4x1 (x4) 2 dx

26 6.2 Integration by Parts (IP) Theory. When an integral involves a product of terms but is not of the form F (u(x)) u (x) dx the integral usually can not be evaluated by inspection. An integral of the form I = f(x)( g(x) dx May be evaluated using the method of integration by parts.

27 Integration by Parts, or IP for short, consists of identify the factors f(x) and g(x) as a function u(x) and a derivative v(x). When this is done the integral is evaluated as or I = 1u(v dx = 1(u(v)dx - 1u(v dx I = 1u(v dx = u(v - 1u(v dx What does the IP formula do? When will it work? When will it not work?

28 When you apply the IP formula you must make a choice. Which term is u(x) and which is v(x)? I suggest you make a templet to help organize your choice and work. u = v = u = v = _ Then set I = 1u(v dx = u(v - 1u(v dx and evaluate the integral 1u(v dx.

29 How do you choose u and v? Requirement 1. You must be able to integrate v. Requirement 2. The derivative of u should not be more complex. Requirement 3. You should be able to evaluate 1u(v dx.

30 What do you do if you can not evaluate the resulting integral 1u(v dx? Option 1. Consider the reverse choice of u and v. Option 2. Press forward. Try it again!! Careful: repeated use of the IP method could take you back to the original problem. If this happens reverse your choice for u and v

31 6.2 Integration by Parts (IP) Examples. Evaluate I = 1 3x e -x dx u = v = u = v = _

32 Evaluate I = 1 3x ln(x) dx u = v = u = v = _

33 Evaluate I = 1 sin 2 (2x) dx u = v = u = v = _

34 Evaluate I = 1 sin(2x) e 3x dx u = v = u = v = _

35 Evaluate I = 1 3x 2 e -x dx u = v = u = v = _

36 6.3 Partial Fractions. Theory Only two general integral formulas involve rational expressions, i.e., fractions. The obvious formula is A less obvious formula is

37 To evaluate I = f(x) g(x) dx we must manipulate it to obtain integrals of one of the two formulae types. Four manipulations are frequently useful. 1. Cancellation: If f(x) and g(x) have common factors, cancel them to reduce the complexity.

38 2. Reduction of the fraction: Division: M N A R N when M = A(N + R

39 3. Separate the numerator: A B C A C B C

40 4. Partial Fractions: Write the fraction f/g as the sum of two fractions: This is possible only if the denominator g(x) is the product of two terms, say Q 1 and Q 2. Then, you can write f g f Q 1 (Q 2 P 1 Q 1 P 2 Q 2

41 If g(x) = Q 1 (x)(q 2 (x) how do you express the fraction f/g as a partial fraction? You use the factors Q 1 and Q 2 as the denominators and find the numerators P 1 and P 2 algebraically using the fact that f Q 1 (Q 2 P 1 Q 1 P 2 Q 2 Q 1 (Q 2 if and only if f(x) = for all values of x.

42 A rational function that is a fraction f/g of two polynomials is said to be in reduced form if degree( f ) < degree( g ) If a rational expression is not in reduced form, dividing the denominator into the numerator will give a polynomial plus a fraction that is in reduced form: f(x) g(x) P(x) R(x) g(x) where f(x) = P(x)(g(x) + R(x) and degree( R(x) ) < degree( g(x) )

43 Equating Polynomials Two polynomials are equal if, and only if, their coefficients of each power of x are identical. N i0 a i x i N = ; a i = b i for all i. i0 b i x i

44 What is the form of P(x) if Q(x) is a polynomial? Assuming that f/g is of reduced form, and Q(x) is a simple factor of g(x) then the partial expansion of f/g will have a factor P/Q where P(x) is a polynomial of degree less than the degree of Q(x).

45 E.g., if a factor is Q(x) = 2x then the partial expansion will contain a term of the form AxB 2x 2 1 If a factor is linear, say Q(x) = x + 5 then the partial expansion will contain a term of the form A x5

46 Repeated Factors If the denominator has a factor raised to a power this factor is called a repeated factor. The power is called the factor s multiplicity. When expanding a fraction with a repeated fraction you must include a term with each power of the repeated factor up to its multiplicity.

47 For example, if a factor of g(x) is Q(x) = (x - 1) 3 i.e., the factor (x - 1) with multiplicity 3, then the partial expansion will contain the terms A x1 B (x1) C (x1) 3

48 6.3 Partial Fractions. Examples. Ask these questions: 1. Is cancellation possible? 2. Is it of reduced form? 3. What are the factors of the denominator? 4. What is the form of the partial fraction expansion? Example. Evaluate I = 2x 2 x (2x1)(x 2 4) dx

49 Example Evaluate I = 3x 2 2x5 x 3 dx

50 Example Evaluate I = x 3 3x 2 2x5 dx

51 Evaluate I = (2x1)x 3 (2x1)(x 2 4) dx

52 Evaluate I = 4x1 (x4) 2 dx