INTEGRATION STRATEGIES 1. Introduction It Should Be Clear Now

INTEGRATION STRATEGIES 1. Introduction It should be clear now that it is a whole lot easier to differentiate a function than integrating one. Typically the formula used to determine the derivative of a function is apparent. With integration, the appropriate formula won't necessarily be obvious. So far we have reviewed each of the basic techniques of integration: with substitutions (ch. 5.5, 7.2, 7.3 ), integration by parts (ch. 7.1), and partial fractions (ch 7.4). More realistically, integrals will be given outside of the context of a textbook chapter and the challenge is to identify which formula(s) to use. We will review several integrals at random and suggest strategies for determining which formula to evaluate the integral. These strategies will only be helpful if one has a knowledge of basic integration formulas. Listed in the table below are an expanded list of the formulas for commonly used integrals. Many of these can be derived using basic principles it will be helpful to memorize these. In lieu of formulas 19 and 20 one could use partial fractions and trigonometric substitutions respectively. Figure 1. Table of Integration Formulas Constant of integration have been omitted. 1 2 INTEGRATION STRATEGIES 2. Strategies If an integral does not fit into this basic list of integration formulas, try the following strategy: (1) Simplify the Integrand if Possible: Algebraic manipulation or trigonometric identities can simplify the integrand and allow for integration in its new form. For example, p p p R x(1 + x)dx = R ( x + x)dx; R tan θ R sin θ 2 R R sec2 θ dθ = cos θ cos θdθ = sin θ cos θdθ = frac12 sin 2θdθ; R (sin x + cos x)2dx = R (sin2 x + 2 sin x cos x + cos2 x)dx = R (1 + 2 sin x cos x)dx; (2) Look for an Obvious Substitution: Try substitutions u = g(x) in the integrand for which the differential du = g0(x)dx also appears in the integran. For example, the substitution v = u2 − 1 simplifies the integral significantly: Z x Z 1 dv dx ! : x2 − 1 2 v (3) Classify the Integrand According to Its Form: If Steps 1 and 2 have failed to produce a solution, determine if the integrand fits into these four categories • Trigonometric Functions - If f(x) is a product of powers of sin x and cos x, or, tan x and sec x, or, cot x and csc x, then try treating these as trigonometric integrals. • Rational Functions - If f is a rational function, use partial fractions to write this as a sum of simpler integrals. • Integration by Parts - If f(x) is the product of a power of a polyno- mial, and a transcendental function (e.g., trigonometric, exponential, or logarithmic functions), then try integration by parts choosing u and dv so that du is simpler than u, and dv's anti-derivative is known. • Radicals p- When a radical appears there are two suggested approaches: (a) If ±x2 ± a2 use the table in the trigonometric substitution section. 1 (b) If (ax+b) n appears, use the rationalizing substitution u = (ax+ 1 n 1 n u −b n b) so that x = a . If one is fortunate this will work for g(x) in an integrand. (4) Try Again: Seriously. There are really only two methods of integration: substitution and parts; if the first three steps have not helped you might want to give them another go. • Try Substitution - If all of the obvious substitutions try to think of alternatives, even if they seem like a long shot. • Try Parts: Usually we use integration by parts on products of functions, but it can be successful on single functions as well (i.e., arctanx; arxsinx, and ln x as integrands.). • Manipulate the integrand : Rationalizing the denominator, using trig identities or any other algebraic manipulation will hep to simplify the integral. Like any good puzzle the manipulations required to simplify INTEGRATION STRATEGIES 3 a function may rely on ingenuity. As an example Z dx Z 1 1 + sin x Z 1 + sin x dx = dx = dx 1 − sin x 1 − sin x 1 + sin x 1 − sin2 x Z 1 + sin x Z sin = dx = sec2 x + dx cos2 x cos2 • Relate to Previous Problems: Using our prior knowledge one might be able to use a method for a similar integral on the current integral with some modifications. Alternatively one might be able to express the given integral in terms of known one. For example, Z Z Z Z tan2 x sec xdx = (sec2 x − 1) sec xdx = sec3 xdx − sec xdx the integral on the right hand side has a known solution as a trigonometric integral. • Mix and Match Methods We have seen where an integral requires several applications of substitutions and integration by parts. It will hap- pen that two or three methods are required to integrate an expression. 3. Examples In the following examples we will show how this four-step strategy leads to an equivalent but simpler integral whose solution is known. R tan3 x Example 3.1. Simplify cos3 x dx Proof. We rewrite the integral in the simpler form Z tan3 x Z dx = tan3 x sec3 xdx cos3 x This integral has a known solution as an trigonometric integral. As an alternative, we could simplify this as well Z tan3 x Z sin3 x Z (1 − cos2 x) dx = dx = sin xdx cos3 x cos6 x cos6 x Z 1 − u2 Z = (−du) = (u−4 − u−6)du u6 p Example 3.2. Rewrite R e xdx. p Proof. We have a radical here, so let us try u = x, then x = u3, dx = 2du, and Z p Z e xdx = 2 ueudu This can be solved using integration by parts. R x5+1 Example 3.3. What formula can one use for x3−3x2−10x dx? Proof. Steps 1 and 2 will not help here. However, the integrand is a rational function and so we can use partial fractions to simplify this. R dx Example 3.4. Simplify x(ln x)3 4 INTEGRATION STRATEGIES dx Proof. We need only apply step 2 here, trying u = ln x the differential is du = x which also appears in the integrand. R q 1−x Example 3.5. Rewrite 1+x dx q 1−x Proof. The rationalizing substitution u = 1+x would work after a lot of algebra. Instead, rationalize the numerator, then Z r1 − x Z 1 − x dx = p dx 1 + x 1 − x2 Z 1 Z x = p dx − p dx 1 − x2 1 − x2 p = arcsin x + 1 − x2 + C 4. Are All Continuous Functions Integrable? We have seen many examples where an integral can be evaluated. One should wonder if all integrals involving a continuous function can be evaluated with this 2 strategy. For example can we write down a closed form expression for R ex dx? It turns out that we cannot, or at least not in terms of the functions we are familiar with. Up until now we have only seen elementary functions, like polynomials, rational functions, power functions (xa), exponential functions (ax), logarithmic functions, trigonometric and inverse trigonometric functions, hyperbolic and inverse hyperbolic functions, and all functions that can be porduced by these using addi- tion, subtraction, multiplication, division, and composition. As an example r x2 − 1 f(x) = − xesin 2x x3 + 2x − 1 is an elementary function. If f is an elementary function, its derivative f 0 will be an elementary function but it is possible that R f(x)dx will not be an elementary function. For example 2 the function ex is continuous, and so its integral exists. Denoting the function F by x Z 2 F (x) = et dt: 0 We know from the Fundamental Theorem of Calculus that 2 F 0(x) = ex 2 and so f(x) = ex has an anti-derivative F , but it may be proven that F is not an elementary function. We can never write down a closed form for this function in terms of the functions we know, however we can express this as an infinite series. Similarly the following functions x R e ; R sin(x2)dx; R cos(ex)dx px R 3 R 1 R sin x x + 1dx; ln x dx; x dx The truth is that most elementary functions do not have elementary anti-derivatives..

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