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## INTEGRATION TECHNIQUES:

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**INTEGRATION… WHAT DO WE KNOW SO FAR?**• In terms of getting integrals, we know how to do basic anti-differentiation with power functions, trig functions and exponential functions. • We started some techniques using “u-substitution” to solve integrals. More or less, “a reverse chain rule.” • We also did area, volume, and other physical applications using the definite integral. • This chapter is devoted to the indefinite integral.**1) U-SUBSTITUTIONREVISITED**• Just to freshen your minds, let’s do a u-substitution problem. There is a little catch to it… • For this problem, it is preferable to pick u = x+2, since du can never be in the denominator. • Use algebra to give a name for x+1. Since x=u-2, then x+1 would be u -2+1 or u-1. • After putting everything in, do the integration. • Note that in the final answer, the constant 2 does not need to be there, since C is more general.**PRODUCT RULE FLASHBACK**• Remember back in Chapter 2, we mentioned the product rule? Looked like the following… In regular form…. Or differential form….**2) INTEGRATION BY PARTS**• If you play around with the differentials, you will get the following. • If you integrate both sides, you get what is known as the INTEGRATION BY PARTS FORMULA**EXAMPLE#1**• Integrate x*exdx. • STEP 1: • Pick your “u” and “dv.” • TRICK: L.I.P.E.T. This determines what u should be initially. Logarithm, Inverse, Power, Exponential, and Trig functions. • In this problem, a power function and an exponential function are present. Since power (P) comes before exponential (E), u should equal x. • From the u, find du. From the dv, find v. • Therefore u=x, then du=dx! Don’t forget that. • Simply plug this into the “parts formula.”**EXAMPLE#1**• I = integral to be solved. • Plug in u, v, du, and dv. • If the integral on the right looks easy to compute, then simply integrate it. • Don’t forget the +C!**EXAMPLE#2**• Integrate the function…**EXAMPLE#2**• Step 1: Find the u and dv: • An exponential and a trig function is present. • E (exponential) comes before T (trig). • Use the exponential function for u.**EXAMPLE#2**• Step#2: Plug u, du, v and dv into the formula • Simplify… • If the last integral is easy to compute, compute it. • However, it is not easy. In fact, we are in more mess than we started in. Looks like we have to use the integration by parts rule again.**EXAMPLE#2**• Use integration by parts again. Since there is an exponential function, call that “u.” After finding u, du, v, and dv, plug those values in the “Parts equation” and see what happens… • We have even more of a mess……but wait! The integral of exsin(x) is supposed to be equal to I (the integral we wanted to solve for in the first place!!). • So we can replace the integral of exsin(x) with I and add it to both sides. • You will see that it works out. • Always add the constant **RULE OF THUMB WITH INTEGRATION BY PARTS PROBLEMS**• Always pick the right “u.” If the problem is getting really difficult, maybe you picked the wrong “u.” Just like in u-substitution. You had to pick the right “u” to work with the problem. • If the integral on the right does not look easy to compute, then do integration by parts for that integral only. • If you see the resulting integral looks like the integral you are asked to solve for in the first place, then simply combine the two like integrals and use algebra to solve for the integral.**3) TRIGONOMETRIC INTEGRALS**• You will always get the situation of funny combination of integrals. Trig functions are as such that you can translate from one function to a function with just sines and cosines. For example, you can always write tan, sec, csc, cot in terms of either sine or cosine. • Remember the following identities from PRE-CALCULUS!!!**PROCEDURE**• Well… I’m afraid to say it, but there is really no procedure or real template in attacking these problems except proper planning. • This takes a great deal of practice**EXAMPLE# 1**• Given: • Best thing to do is to break the cosine function down to a 2nd degree multiplied by a 1st degree cosine. • Since cos2x=1-sin2x, you can replace it. • Use u-substitution to solve the integral.**EXAMPLE#2**• Given • Break the 4th degree sine to two 2nd degre sines. • Use the sin2x theorem. • Expand the binomial squared. • For the cos22x, use the cos2x theorem. • Simplify • Don’t forget the +C!**EXAMPLE #3**• Whenever you see a tan, sec, csc, or cot, always convert them to sines and cosines. This way, you can cancel or combine whenever necessary. • In this case, tan2x=sin2x/cos2x. We are also lucky that the cos2x cancels. • Using the sin2x theorem, we can simply integrate.**NOTE ABOUT TRIGONOMETRIC INTEGRALS**• There is no real rule for such integrals. But always remember: • 1) If there is a mix of sines and cosines, break them up until they resemble an easier form • 2) Use any trig theorem that would be relevant to make a problem simpler. • 3) Convert everything to sines or cosines.**4) TRIGONOMETRIC SUBSTITUTION**• Remember when we took derivatives of inverse trigonometric functions, we commonly dealt with sums or differences of squares. • Similarly, integrating sums of differences of square will lead us to the inverse trig functions. • However we need a stepping stone to integrate such functions.**GENERAL PROCEDURE**Given that a is a constant, and u is a function, then follow the IF YOU HAVE THIS.. THEN USE THIS FORMULA**THETA?**• Since we are working with a substitution, theta would be the variable to use subsitution… • Doesn’t make sense? Let’s do an example problem…**EXAMPLE #1**• Initially a very bad looking problem… • Focus on the denominator, inside the radical, you have 9-x2. In effect, that is a2-u2, a being 3 and u being x. If a2-u2 is used, then according to the table in the last slide, we would use u=a sinq. • You already found a name for x. You need to give a name for “dx.” Differentiate x with respect to q. Solve for dx.**EXAMPLE #1**• Here is the original problem… • With the substitutions of x and dx, here is the original problem… • Simplify a little… • Pull out constants when needed.**EXAMPLE #1**• The denominator is actually cos2x, according to the trig theorem. (Memorize them!) • Simplify • Integrate • We have our answer in terms of q!!! We need it in terms of x!**EXAMPLE #1**• Don’t forget what we said earlier. That x=3 sin q. We need to know what q is in order to find out the solution in terms of x.**EXAMPLE #1**• Simply replace all the q expressions with x expressions. • Simplify • Add constant!! • Sighs!! We’re finished!!!**That was a LOT of work!!!**• Here are trig substitution steps… • 1) Find the correct equality statement using the table. • 2) Make the proper substitutions. Remember to have a substitute for x as well as dx. • 3) Integrate in terms of q. • 4) Convert all q terms to x terms.**5) RATIONAL FUNCTIONS**• Of course, there will always be functions in the form of a ratio of two functions. • Two integrate most rational functions, the method of partial fractions come into play. • <<Break from Calculus… entering Algebra Territory>>**PARTIAL FRACTIONS**• This means you take a fraction and break it down into a sum of many fractions. • This way, we can add up the integrals of simpler easier fractions.**EXAMPLE**• Given • The denominator could be factored to (x+5)(x-2). This way we could have new denominators for the two new fractions. • Add these new fractions and distribute. Make sure you bring all x terms together, as well as bringing all the constants together.**EXAMPLE**• The coefficient of x on the right side is 1. In order to keep the equality true, the coefficient of x on the left side should also equal 1. • A+B=1 • Same thing with the constant. If the equality holds true, then -2A+5B must equal -9. • To solve for A and B, you use methods from algebra. <System of linear equations>. • If you multiply A+B=1 by 2, you will see that B=-1. Therefore A=2.**EXAMPLE**• Since A=2 and B=-1, we can simply plug them in. • And integrate!!! • And the final answer!!**ANOTHER EXAMPLE**• Given • Note: If the numerator has a higher degree than the denominator, then do long polynomial division. • If you actually do the long division, you will get x-1-1/(x+1). This is very easy to integrate.**POINTERS OF PARTIAL FRACTIONS**• 1) Check if the top degree is bigger than the bottom. If so, perform long division • 2) If the denominator is factorable, then assume that the denominators of the new fractions will be those factors.**6) QUADRATIC DENOMINATOR PROBLEMS**• This is really no different than trigonometric substitution. • Strictly rational functions with a quadratic denominator that cannot be reduced. • To make the denominator easier to work with, you must complete the square**COMPLETING THE SQUARE**• Given problem: • Look at the denominator. Take the coefficient of x and divide it by 2. • Take this result and square it. • 4/2 =22=4 • This result would form a perfect square when added to x2+4x. • The perfect square would be (x+2)2. • However, we have a 5. If you add and subtract 4, combining 5 and -4 will yield 1. • You have a form of u2+a2! Time for trig substitution!**EXAMPLE**• Since we have u2+a2, we must use the fact of u=a*tan(q). • u=x+2 while a=1 • Substitute the values in appropriate spots.**POINTERS**• 1) Make the denominator into one of the three forms that allows trig substitution by the use of completing the square. • 2) Follow rules of trig substitution.**7) IMPROPER INTEGRALS**• This is not an integral evaluating technique. • An improper integral is basically an integral that has infinity as its limits or has a discontinuity within its limits.**IMPROPER INTEGRALS**• Examples of improper integrals:**IMPROPER INTEGRALS**• With limits of infinity, just use a letter to replace the infinity and treat as a limit. • And integrate as if nothing ever happened • Don’t forget to use the limit. • Amazing! As we start from 0 to infinity, we get closer to 1 square unit of area! We say that it converges to 1.**IMPROPER INTEGRALS**• Since we have a discontinuity in this function at x=-2. To take this into account, we must split the integral into two parts. In addition, we cannot go exactly -2, but we have to get there pretty darn close. Therefore, we must use the one-sided limits from Chapter 1 to represent this. Two integrals: one from -3 to a little before -2, and a little after -2 to 2. • Other than that, simply integrate • Notice how we got an answer that don’t exist!! D.N.E (does not exist)! This means that this integral diverges. Also if an integral goes to infinity, it diverges.**POINTERS OF IMPROPER INTEGRALS**• Remember to identify all the points of discontinuity. Remember to use limits before and after the points of discontinuity. • If you have infinity as your limit, remember to use infinity as your limit. • Other than that, use ALL of the previous techniques of integration mentioned.**FUNCTIONS WITHOUT AN ANTIDERIVATIVE**• Besides three more chapters, this is the last of the single variable calculus. That is to say: y=f(x) in 2-dimensional x,y graph. • Before moving on, I must admit… even though all continuous functions have derivatives, not all continuous functions have simple integrals in terms of elementary functions. • Elementary functions are adding, subtracting, multiplication, division, power, rooting, exponential, logarithmic, trigonometric, all of their inverses as well as combinations or composition functions. Basically, all the functions you ever used were composed of elementary functions. • Some functions do not have elementary antiderivatives. For example… the classic (sin x)/x problem. • No matter what method you tried. Neither by u-substitution, integration by parts, trig substitution, partial fractions, or even guess and check will get you an antiderivative. • From my experience from differential equations class last year, the integral of sin(x)/x is Si(x) also known as the sine integral! • YOU DON’T NEED TO KNOW THAT!!!!**OUTTA THIS WORLD FUNCTIONS!!!!!!!**• You will be dealing with functions like erf(x), Si(x), Ci(x), Shi(x), Chi(x), FresnelS(x), and FresnelC(x). Take their derivatives and you’ll get regular “sane” functions. AAHHH!!! HARI BOL!!!!**SUMMARY**• Actually, for once, looking at the length and material of this chapter. I am quite amazed to say that I have no words to summarize this chapter. There has been so many methods of integration. Namely u-substitution, integration by parts, how to deal with trig integrals, trig substitution, partial fractions, quadratic denominators, and improper integrals. • All I can say is that review this material over again!!! • Like I said previously, there is no set way to do these problems. There are more than one way of doing it. • You have to know what to do when which problem arrives at you.