CHAPTER EIGHT: INTEGRATION TECHNIQUES:

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CHAPTER EIGHT INTEGRATION TECHNIQUESAS IT IS
Hare Krsna Hare Krsna Krsna Krsna Hare Hare Hare
Rama Hare Rama Rama Rama Hare Hare Jaya Sri Sri
Radha Vijnanasevara (Lord Krsna, the King of Math
and Science) KRSNA CALCULUS PRESENTS

• Released by Krsna Dhenu
• September 28, 2002
• Edited October 7, 2003

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WELCOME BACK!!!

• Hare Krsna everyone!
• Please take a moment to sigh before moving onto
  the next slide.
• This chapter is rated one of the most difficult
  chapters. This contains a great amount of
  material.
• It is strongly suggested that you do NOT start
  this chapter without the strong backgrounds of
  derivative, integral, function behavior, and
  algebra.

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WHAT IS THIS CHAPTER ABOUT?

• Notice when we did derivatives, we only spent one
  whole chapter on how to compute the derivative
  (Chapter 2)
• We spent time using Chapter 3 to do real-world
  examples using derivatives.
• We know how to differentiate ANY function.
• Integrals, however, are more difficult, since
  there are many rules involved. There is no one
  method of computing integrals.
• This chapter is completely devoted to different
  methods of integration. It is important that you
  take this chapter nice and slowly so you can
  build it in.
• Also, you must develop critical thinking in this
  chapter. Without it, this chapter will become
  impossible.

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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.

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1) U-SUBSTITUTIONREVISITED

• Just to freshen your minds, lets do a
  u-substitution problem. There is a little catch
  to it
• For this problem, it is preferable to pick u
  x2, since du can never be in the denominator.
• Use algebra to give a name for x1. Since xu-2,
  then x1 would be u -21 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.

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PRODUCT RULE FLASHBACK

• Remember back in Chapter 2, we mentioned the
  product rule? Looked like the following

In regular form. Or differential form.
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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

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EXAMPLE1

• Integrate xexdx.
• 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 ux, then dudx! Dont forget that.
• Simply plug this into the parts formula.

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EXAMPLE1

• 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.
• Dont forget the C!

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EXAMPLE2

• Integrate the function

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EXAMPLE2

• 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.

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EXAMPLE2

• Step2 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.

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EXAMPLE2

• 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 messbut 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 ?

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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.

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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!!!

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IMPORTANT IDENTITIES
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PROCEDURE

• Well Im 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

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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 cos2x1-sin2x, you can replace it.
• Use u-substitution to solve the integral.

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EXAMPLE2

• 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
• Dont forget the C!

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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, tan2xsin2x/cos2x. We are also
  lucky that the cos2x cancels.
• Using the sin2x theorem, we can simply integrate.

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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.

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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.

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GENERAL PROCEDURE
Given that a is a constant, and u is a function,
then follow the
IF YOU HAVE THIS..
THEN USE THIS FORMULA
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THETA?

• Since we are working with a substitution, theta
  would be the variable to use subsitution
• Doesnt make sense? Lets do an example problem

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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 ua
  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.

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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.

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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!

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EXAMPLE 1

• Dont forget what we said earlier. That x3 sin
  q. We need to know what q is in order to find out
  the solution in terms of x.

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EXAMPLE 1

• Simply replace all the q expressions with x
  expressions.
• Simplify
• Add constant!!
• Sighs!! Were finished!!!

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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.

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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.
• ltltBreak from Calculus entering Algebra
  Territorygtgt

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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.

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EXAMPLE

• Given
• The denominator could be factored to (x5)(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.

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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.
• AB1
• Same thing with the constant. If the equality
  holds true, then -2A5B must equal -9.
• To solve for A and B, you use methods from
  algebra. ltSystem of linear equationsgt.
• If you multiply AB1 by 2, you will see that
  B-1. Therefore A2.

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EXAMPLE

• Since A2 and B-1, we can simply plug them in.
• And integrate!!!
• And the final answer!!

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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/(x1). This is very easy to integrate.

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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.

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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

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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.
• 4/2 224
• This result would form a perfect square when
  added to x24x.
• The perfect square would be (x2)2.
• However, we have a 5. If you add and subtract 4,
  combining 5 and -4 will yield 1.
• You have a form of u2a2! Time for trig
  substitution!

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EXAMPLE

• Since we have u2a2, we must use the fact of
  uatan(q).
• ux2 while a1
• Substitute the values in appropriate spots.

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EXAMPLE (work)
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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.

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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.

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IMPROPER INTEGRALS

• Examples of improper integrals

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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 ?
• Dont 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.

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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 dont exist!!
  D.N.E (does not exist)! This means that this
  integral diverges. Also if an integral goes to
  infinity, it diverges.

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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.

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FUNCTIONS WITHOUT AN ANTIDERIVATIVE

• Besides three more chapters, this is the last of
  the single variable calculus. That is to say
  yf(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 DONT NEED TO KNOW THAT!!!!

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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 youll
  get regular sane functions. ? AAHHH!!! HARI
  BOL!!!!

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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.

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CREDITS

• Dr. A. Moslow
• Dr. W. Menasco
• Mr. G. Chomiak
• Calculus and Early Transcendental Functions 5th
  Ed.
• Finney Calculus
• Single-Variable Calculus (SUNY Buffalo)
• Princeton Review AP Calculus AB

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NEED HELP?

• Need help??
• E-mail kksongs_1_at_hotmail.com
• Please read help statement

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END OF CHAPTER EIGHT

• jaya sri krsna caitanya prabhu nityananda
• sri advaita gadadhara sri vasadi gaura bhakta
  vrnda
• hare krsna hare krsna krsna krsna hare hare
• hare rama hare rama rama rama hare hare

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END OF CHAPTER EIGHT