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Calculus Section 2: Techniques of Integration. S1: Motivation S2

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Calculus Section 2: Techniques of Integration. S1: Motivation S2 MATH 1231 S2 2010: Calculus For use in Dr Chris Tisdell’s lectures Section 2: Techniques of integration. Created and compiled by Chris Tisdell S1: Motivation S2: What you should already know S3: Integrals of trig functions S4: Reduction formulae S5: Trig & hyperbolic substitutions S6: Partial fractions S7: Rationalising substitutions S8: Weierstrass substitutions S9: Appendix (more on what you should already know) Images from“Thomas’ calculus”by Thomas, Wier, Hass & Giordano, 2008, Pearson Education, Inc. 1 1. Motivation. Why study integration? The theory of integration is a cornerstone of calculus. In- tegration finds a useful place in many disciplines, such as: engineering; physics; biology; economics and beyond. Where are we going? We will develop a number of important integration techniques that will be useful in the study of applied problems. Throughout our discussions we will see HOW integration nat- urally arises in the analysis of applied problems and in mod- elling. This is critical to motivate the ideas and also to build the intuition. 2 2. What you should already know. You studied integration at school and in MATH 1131. You should be comfortable with: • using a table of integrals • integrating by inspection • integration by parts • integration by simple substitution. Just in case you’ve forgotten these techniques, I have included some examples to refresh your memory and to test your skills in the Appendix. You will find this course much easier if you can integrate with ease. Your hard work will pay–off later in the session!! 3 4 5 3. Integrals of trig functions. You should be familiar with: The above integrals are derived by using the “double angle formulas”: cos 2x = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x and for future reference sin 2x = 2 sin x cos x. 6 Integrals of products of sin and/or cos. We first discuss how such integrals naturally arise. Applications matter! So–called “Fourier series” play an important role in the study of heat flow. A finite Fourier series of a function f is given by the sum N X f(x) := an sin nx (1) n=1 = a1 sin x + a2 sin 2x + ··· aN sin Nx where the an are some numbers that are to be determined. That is, we aim to write f(x) as a sum of sin functions. 7 We now see how integration of trig functions is central to the method. We claim that the mth coefficient am is given by 1 Z π am = f(x) sin mx dx. (2) π −π If we multiply both sides of (1) by sin mx and then integrate over [−π, π] we obtain N Z Z π X f(x) sin mx dx = an sin nx sin mx dx −π n=1 N X Z π = an sin nx sin mx dx . n=1 −π Thus we have obtained an integral that involves a product of sin functions. How do we evaulate such an integral? 8 Integrals of products of sin and/or cos These are done using the so–called product to sum formulae: sin A cos B = [sin(A − B) + sin(A + B)]/2 sin A sin B = [cos(A − B) − cos(A + B)]/2 cos A cos B = [cos(A − B) + cos(A + B)]/2 Ex: R sin 4x cos 2x dx = Independent learning ex: Now return to, and prove, (2). 9 Integrals of the form R cosm x sinn x dx If m or n are odd then our aim is to transform the integral into one of the types: Z 1 sink x cos x dx = sink+1 x + C k + 1 Z 1 cosk x sin x dx = − cosk+1 x + C. k + 1 At the heart of the approach is to factor out a sin x or cos x from the odd power term and then use the identity cos2 x + sin2 x = 1 to transform the integral into one which can be directly eva- lutated. Q: How can you verify the above integrals? 10 Ex. Evaluate Z I := cos4 x sin3 x dx. 11 Ex: Evaluate Z I := cos5 x sin5 x dx 12 If both m and n are even, then the integral is slightly harder. We will see shortly a recursive method for dealing with such an integral, but for smaller values of m and n one can use the identities 1 + cos 2x cos2 x = 2 1 − cos 2x sin2 x = . 2 Ex. Evaluate I := R cos4 x sin2 x dx. See that Z I = (cos2 x)2 sin2 x dx Z 1 + cos 2x2 1 − cos 2x = dx 2 2 1 Z = 1 + cos 2x − cos2 2x − cos3 2x dx 8 Z = 1 1 + cos 2x − cos2 2x − (1 − sin2 2x) cos 2x dx 8 Z = 1 1 − cos2 2x + sin2 2x cos 2x dx 8 h i = 1 x − 1 sin 4x + 1 sin3 2x + C. 8 2 8 6 13 Applications matter! An important problem in applied math, engineering and physics is to calculate the total mass of an given object. Consider a thin plate occupying the unit quarter–disk in the first quadrant. If the density δ(x, y) of the plate at any point (x, y) is given by δ(x, y) = x2y4 then the total mass of the plate is given by √ Z 1 "Z 1−x2 # M = x2y4 dy dx. 0 0 To make this integral easier, we switch to polar co–ordinates: x = r cos θ; y = r sin θ; dy dx = r dr dθ with a substitution giving Z π/2 Z 1 M = (r2 cos2 θ)(r4 sin4 θ) rdrdθ 0 0 Z 1 Z π/2 = r7 dr (cos2 θ)(sin4 θ) dθ. 0 0 14 4. Reduction formulae Repeated integration by parts can be a long process! We can shorten the process by applying so–called “reduction formu- lae”. Ex: The reduction formula Z 1 Z tann x dx = tann−1 x − tann−2 x dx (∗) n − 1 can be used in an iterative fashion to calcuate Z tan5 x dx R n If In := tan x dx then (*) may be compactly written as 1 n−1 In := tan x − In−2, n ≥ 2. n − 1 15 R π/2 n Ex. Let In := 0 sin x dx. Use the reduction formula n − 1 In := I 2, n ≥ 2 n n− to calculate I7. 16 Ex: Construct a reduction formula for Z n x In := x e dx. 17 Ex: Find the reduction formula for Z π/2 n In := sin x dx. 0 The reduction formula for the integral of cosn x is similar. 18 Ex: Construct a reduction formulae from Z π/4 n In := sec x dx. 0 A similar method is used to obtain the reduction formula for R π/4 n the integral In := 0 tan x dx. 19 Ex. [Q1, Class Test 1, 2002] Let Z π/4 n In := tan θ sec θ dθ. 0 Show that 1 √ In := 2 − (n − 1)I 2 , for n ≥ 2. n n− Note that d sec θ = sec θ tan θ. dθ 20 A Two parameter recurrence: Consider Z π/2 m n Im,n = cos x sin x dx 0 Z π/2 = [cosm−1 x][sinn x cos x] dx 0 and use integration by parts. Choose u = cosm−1 x and v0 = sinn x cos x. Thus, 1 u0 = −(m−1) sin x cosm−2 x, v = sinn+1 x. n + 1 Integration by parts then leads to m − 1 Im,n = I 2 , m ≥ 2. m + n m− ,n In a similar way, one could also obtain the recurrence n − 1 Im,n = I 2. n ≥ 2. m + n m,n− In applying the above formulae we must reach one of I1,1 = 1 π 2,I1,0 = I0,1 = 1 or I0,0 = 2. You are not expected to memorise this formula. 21 5. Trig & Hyperbolic Substitutions. You will have already seen integration by simple substitution. With integrals involving square roots of quadratics, the idea is to make a suitable trigonometric or hyperbolic substitution that greatly simplifies the integral. 22 Integrals involving square roots of quadratics can be worked out using the following trigonometric or hyperbolic substitu- tions. q a2 − x2 try x = a sin θ q a2 + x2 try x = a tan θ or x = a sinh θ q x2 − a2 try x = a sec θ or x = a cosh θ 23 Ex: Evaluate Z 4 q I = x2 16 − x2 dx. 0 24 Ex: Evaluate Z dx I := . (a2 + x2)3/2 25 Ex: Evaluate Z dx I := q x2 x2 − 1 via the substitution x = sec θ (This last integral can also be done using a cosh θ substi- tution but it is not easy.) 26 Applications matter! An important problem in applied math, engineering and physics is to calculate the total mass of an given object. Consider a thin wire of constant density δ (mass per unit length) that lies in the XY –plane along the curve y = f(x) := x2/2, a ≤ x ≤ b. It can be shown that the total mass M of the wire is Z b q M = δ 1 + [f 0(x)]2 dx a Z b q = δ 1 + x2 dx. a q See how an integral of 1 + x2 naturally arises? 27 6. Partial Fractions. f(x) A rational function is a function of the form where f g(x) and g are polynomials. Ex: Which of the following are rational functions? 2x + 1 (a) 3x2 − 4 cos x (b) x2 + 1 x2 (c) x − 1 ex (d) .
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