AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

8. Basic Integration Rules In this section we will review various integration strategies. Strategies: I. Separate the integrand into two fractions. + 3 4 II. Rewrite the integrate in an equivalent form. + e (Hint: Add and subtract e in the numerator and then rewrite as the sum of two fractions. OR Multiply the fraction by e e. )

III. Complete the square. (Hint: When you complete the square in the denominator, you may end up with an integral for arc sin, arc tan, or arc sec.) IV. Divide a Rational Function. + (Hint: If the degree of the numerator is greater than or equal to the degree of the denominator, it may be helpful to divide the numerator by the denominator.)

V. Epand the epression. ( + e ) VI. Use Trigonometric Identities. cot (Hint: Use a trig identity to rewrite the integrand. In this case, use a Pythagorean Identity.)

VII. Multiply and Divide by the Conjugate. + sin (Hint: Multiply numerator and denominator by ( sin ) )

8. Integration by Parts Question: How do we integrate epressions such as ln or e or e sin??? We will use a new method, integration by parts, to integrate these complicated looking integrals. The product rule is the basis for deriving the formula for integration by parts. Given u() and v(), the derivative of the product u( ) v( ) is given by: Integrating both sides of the equation gives: d u ( ) v ( ) = u ( ) v '( ) + v ( ) u '( ) d [ u() v() ] = [ u() v'() + v() u '()] u() v() = u() v'() + v() u'() u() v() = u() dv + du v() u() v() = u() dv + v() du Rearrange the equation to get the formula for Integration by Parts: u()dv = u() v() v()du Using a simpler notation we get: u dv = uv v du The new formula epresses the original integral in terms of another integral that should be easier to evaluate. To apply the Integration by Parts formula:. Determine the appropriate substitution for u and dv.. Integrate dv to solve for v, and differentiate u to solve for du. 3. Substitute the values for u, v, du, and dv into the formula. 4. Solve the new integral.

Eample sin Courtesy of www.askamathgeek.blogspot.com Eample e Eample 3 e cos

Eample 4 ln The Tabular Method This method is useful when you must apply integration by parts multiple times. For instance, when integrating any of the forms below, the tabular method is helpful. Eample 5 n sin a or n cosa or n e a sin 4 Let u = and dv = sin 4 The sign applies to Differentiate until you get to 0. Integrate. u and u'. Alternating Sign +/- u and du dv and v The solution is obtained by adding the signed product of the diagonal entries.

8.3 Trigonometric Integrals The following trig identities will prove quite useful in this section. Pythagorean Identity sin + cos = Power Reduction Formulas sin cos = cos + cos = Notice that the Power Reduction Formulas are just another arrangement of the Double Angle Identities from trig: cos = sin and cos = cos Some simple trig integrals that you already have done Use U- sub.. sin 3 cos. tan 4 sec 3. sec 3 sec tan

Now let s make it a little more challenging In this section we will be solving integrals like: sin m cosn. When the power of either sin or cos is odd and positive, keep one factor of sin and convert all other factors to cos, OR keep one factor of cos and convert all other factors to sin. Use sin or cos as your du for u- substitution. Eample: sin 3 cos 4. When the powers of both sine and/or cosine are even (you can't use the first strategy) make repeated use of the power reducing identities above. Eample: cos 4

In this section we will be solving integrals of the form: sec m tan n Strategies:. If the power of secant is even and positive, save a secant squared factor for the du and convert the remaining factors to tangent. Use the identity: tan + = sec Eample: sec 4 tan 3. If the power of tangent is odd and positive, save a secant- tangent factor for the du, and convert the remaining factors to secants. Eample: tan 3 sec

3. If there are no secant factors and the power of tangent is even and positive, convert a tangent- squared factor to a secant- squared factor. Let the secant- squared factor be your du. Eample: tan 4 4. If there are no tangent factors and the power of secant is odd and positive, use Integration by parts. Eample: sec 3 5. If none of the 4 strategies above apply, try converting to sines and cosines! sec Eample: tan

8.5 Partial Fractions The decomposition of a rational function into simpler rational fractions, otherwise known as partial fractions can be helpful in the integration of difficult rational functions.. Divide an improper fraction (degree of numerator is greater than or equal to the degree of the denominator).. Decompose rational functions into partial fractions. (see all cases below) 3. Integrate the resulting terms. ) 4 ) 3 6

3) 5 3

8.7 Indeterminate Forms and L'Hopital's Rule Eamples of some indeterminate forms: 0 0 0,,,, 0,,, 0, 0 In the process of determining a limit we occasionally arrive at one of the indeterminate forms listed above. We can only speculate about the meaning of epressions such as these. Previously, we might have used algebraic techniques or a graphing calculator to solve such a dilemma, but now we will learn to use L'Hopital's rule to determine the value of an indeterminate form when it is the solution to a given limit problem. First, we'll review techniques with which you are already familiar. Algebraic Method Eample : 3 3 0 lim = 0 by direct substitution Rewrite the fraction using algebra and try evaluating the limit again. 3 lim ( )( + ) = lim 3( + ) = 6 ( ) Eample : 5 lim + = by direct substitution 6 7 Rewrite the fraction using algebra and try evaluating the limit again. 5 5 lim lim + 6 7 = 7 6 + = 5 6 Graphing Calculator Method e 0 Eample 3: lim = 0 0 by direct substitution From the graph and the table of values in the neighborhood of = 0, you can approimate the limit. e lim 0 =

L'Hopital's Rule Let f and g be functions that are differentiable on an open interval (a, b) containing c, ecept possibly at c itself. Assume that g'() is not equal to 0 for all in (a, b), ecept possibly at c itself. If the limit f ( ) as approaches c produces the indeterminate form g( ) 0/0, then lim c f () g() = lim c f '() g'() provided the limit on the right eists or is infinite. This result also applies if the limit of f ( ) g( ) as approaches c produces any one of the indeterminate forms,,,. We can also apply L'Hopital's Rule to limits approaching. f lim ( ) f lim '( = ) g( ) g'( ) It is important to note that L'Hopital's rule applies only to limits that yield the indeterminate forms 0 / 0 or ± / ±. If a limit yields one of the other indeterminate forms listed above and can be rewritten to yield the form 0 / 0 or ± / ±, then you can apply L'Hopital's rule to the new form. Note also, there are other forms known to be "determinate" forms such as: + + 0 0 0 The last determinate form is perhaps the most difficult to accept. It is not intuitively obvious. A proof of the last determinate form follows at the end of the notes for the particularly inquisitive Calculus student. Let's use L'Hopital's Rule to determine the limit in Eample 3 from page. Eample 3: (again) e lim 0 =

Eample 4: lim sin 0 = Eample 5: lim cos 0 = Eample 6: lim 0 4 + 4 = Eample 7: sin8 lim = 0 Eample 8: lim e = Eample 9: ln lim Eample 0: lim sin

Eample : lim e Eample : lim ( sin ) HINT: Set the limit equal to y and 0 + take the ln of both sides of the equation. Eample 3: lim( cos ) 0 Eample 4: lim + ln NOTE: When possible, try checking these limits using a graphing calculator. J

8.8 Improper Integrals You will recall that when we evaluate a definite integral, the interval [a,b] is a closed interval. In other words, the limits of integration are usually two constants. In addition, the Fundamental Theorem of Calculus requires that a function be everywhere continuous in order to integrate the function. There are two types of integrals that defy these typical requirements for integration. We call them Improper Integrals. We will first consider integrals that have one or both limits of integration equal to or -. Then we will consider integrals that have a discontinuity at or between the limits of integration. Infinite Limits of Integration Case I: If f is continuous on the interval [a, ), then a f () = lim b b a f () Case II: If f is continuous on the interval (-,b], then b f () = lim b a a f () Case III: If f is continuous on the interval (-, ), then f () = f () + f () where c is any real number. Rewrite as the sum of two limits c = lim c a a c f () + lim b b c f () In the first two cases, the improper integral converges if the limit eists - otherwise, the improper integral diverges and we can not evaluate it. In the third case, the improper integral on the left converges only when both integrals on the right have a limit and are therefore convergent. Eample : Rewrite the integral as a limit: lim b Now, evaluate the integral and then determine the limit. b Eample : Rewrite the integral as a limit: lim b b Compare the graphs of f ( ) = and f ( ) =. The graphs are very similar and yet one integral converges while the other diverges. Can you eplain why?

Eample 3: e value of a, let's say 0 in this case. lim a e a Rewrite the integral as the sum of two limits, choosing a convenient 0 + lim b e b 0 Integrals with a Discontinuity at or within the Limits of Integration Case I: If f is continuous on the interval [a, b) and has an infinite discontinuity at b, then b a f () = lim c b c f (). a Case II: If f is continuous on the interval (a, b] and has an infinite discontinuity at a, then b a f () = lim c a + b f (). c Case III: If f is continuous on the interval [a,b] ecept for some c in (a,b) at which f has an infinite discontinuity, then Rewrite as the sum of limits b f () = f () + f (). a c a = lim d c d a b c f () + lim e c + b e f () In the first two cases, the improper integral converges if the limit eists - otherwise, the improper integral diverges. In the third case, the improper integral on the left converges only if both integrals on the right converge. Eample 4: 4 Rewrite as a limit: lim 4 c 4 c 4

Eample 5: Rewrite as a limit: lim c 0 + 0 c Eample 6: 8 Rewrite as the sum of two limits: 3 lim c 0 c 3 + lim c 0 + 8 c 3

Chapter 8 Review 8. Basic Integration Methods. (tan + sec ). + 4 3 3. 4. + 4 + 5 e + e 5. sec tan 6. 4 7. + cos 8. sin csc

8. Integration by Parts 9. sin e 0. e. sin. e 3. csc 4. ln 5. ln 6. sin

7. ln 8. cos e 8.3 Trigonometric Integrals 9. sin 5 0. sin θ dθ π /4 π /8. cos 3. tan sec

8.5 Method of Partial Fractions 3. 3 ( + 3) 4. + 5 + 6 8.7 L Hopital s Rule 5. lim e 0 6. lim ln 7. lim 0 8e /8 (8 + ) 8. lim 0 + sin

π 9. lim π sin sin 30. lim 0 3 e 3 e 5 3. lim 0 5 3. lim e 5 33. lim 0 + ln( +) 8.8 Improper Integrals 34. 3 35. 0 4 ( 3)

36. e 37. 0 + 38. e 0