Download Math 175, Sp16 Notes and Learning Goals Lesson 3

Math 175, Sp16
Notes and Learning Goals
Lesson 3-2
Integral Recognition
In order to be ready for Exam 1 you must be able to quickly and accurately recognize which
antiderivative or integration technique is appropriate to any given problem. There are five situations
that you must recognize:
1. Elementary Antiderivatives.
2. Integration by Substitution.
3. Integration by Parts.
4. Partial Fractions.
5. Trigonometric Substitution.
You must also know how to execute each method. Today’s homework will develop this skill by
asking you to select a method and also execute the first (and most critical) step of the process that
you have selected.
NOTE: Many problems have more than one right answer. It is entirely possible that an integral
could fall in more than one of the above categories. You get full credit for picking any one of the
correct answers.
WARNING: All of the method choices are penalty scored in WebAssign. Here’s how you avoid
WebAssign penalties:
• Decide on a method. But DO NOT click anything.
• Get out paper and execute the first few steps of your method.
• If the problem transforms to something simpler, you are probably right.
• If it gets worse, you probably need a different method.
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The most important recognition skill by far is this:
“You must immediately recognize whether a problem is Elementary or Not.”
Elementary Antiderivatives
• Anything in this list is elementary
Z
1 r+1
xr dx =
x
+ C; r 6= −1
r+1
Z
1
dx = ln |x| + C
x
Z
ex dx = ex + C
Z
sin(x) dx = − cos(x) + C
Z
cos(x) dx = sin(x) + C
Z
(ax + b)r dx =
1
1
·
(ax + b)r+1 + C
a r+1
1
1
dx = ln |ax + b| + C
ax + b
a
Z
1
eax+b dx = eax+b + C
a
Z
1
sin(ax + b) dx = − cos(ax + b) + C
a
Z
1
cos(ax + b) dx = sin(ax + b) + C
a
Z
• These are also “elementary”, but you have to do a little algebra first.
Z
Z √
√
x dx
ax + b dx
Z
Z
1
1
dx
dx
(ax + b)n
xn
• Any constant multiple in front of an elementary antiderivative is still elementary.
• Adding or subtracting several elementary antiderivatives is still elementary.
Example:
Z
(3 + 4x2 − 5x−3 ) dx
Example:
Z 2
dx
15 cos(12x) − √
4x + 1
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Here are the relevant characteristics of the non-elementary types.
Integration by Substitution
• Recognizing integration by substitution is the same as identifying an appropriate choice of u.
• Click here for some suggestions on how to choose u.
• Once you think of u, write it down and execute the substitution process.
• If you end up with an elementary antiderivative then you were right.
• If not, start over with a different technique or a different u.
Integration by Parts
• Recognizing integration by parts is mostly a matter of experience. Work a lot of examples.
• Parts is often indicated by a product, but not all products need parts.
Z
Example 1:
x cos x dx requires parts.
Z
Example 2:
x cos(x2 ) dx requires substitution with u = x2 .
Z
√
Example 3:
x2 1 − x2 dx requires trig substitution with u = sin x.
Z √
Example 4:
x 1 − x2 dx is easiest with u = 1 − x2 , but x = sin u will also work.
• Once you decide that a problem needs integration by parts, write down u and dv.
• Execute the process. If the problem gets simpler (or no worse), you are on the right track.
• Sometimes integration by parts has to be repeated.
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Partial Fractions
• These are pretty easy to recognize. Partial fractions is appropriate when these conditions are
met.
– The integrand is an algebraic fraction.
– Numerator and denominator are both polynomials.1
– The denominator is factorable into linear factors or irreducible quadratics.
– There is more than one factor.
• Once you decide that partial fractions is appropriate, immediately write down the form of the
partial fractions decomposition.
Trig Substitution
• This it triggered by the presence or algebraic structures that are too much for other techniques.
The most common are
1. (a2 − x2 ), where a is a constant. It often, but not always, occurs inside a square root.
Use x = a sin u.
2. (a2 + x2 ), where a is a constant. It often, but not always, occurs in a denominator.
Use x = a tan u.
• Trig substitution is longer and more difficult than other techniques. It is often worth spending
a little time looking for a quicker substitution. See Example 4 above.
• Once you decide that trig substitution is appropriate, execute the substitution and transform
the problem using appropriate trig identities.
√
For x = a sin u you need sin2 u + cos2 u = 1 and cos u = 1 − x2 .
√
For x = a tan u you need tan2 u + 1 = sec2 u and sec u = 1 + x2 .
• Trig substitution transforms an algebraic problem into a trig problem. It’s not much use
unless you are also good at trig integrals.
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The degree of the numerator must be less than the degree of the denominator. This technicality has been kept
out of your homework. You don’t have to worry about it.
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