Integration Techniques: A Comprehensive Review, Study notes of Analytical Geometry and Calculus

A review of various integration techniques, including integration by parts, trigonometric integrals, trigonometric substitution, and partial fractions. It covers formulas, strategies, and examples for each method, as well as guidelines for choosing the appropriate technique for different types of integrals.

Typology: Study notes

Pre 2010

Uploaded on 03/10/2009

koofers-user-obe-2
koofers-user-obe-2 🇺🇸

10 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
7.1 - 7.5 Review - Integration Techniques
My reviews and review sheets are not meant to be your only form of studying. It is vital to your success
on the exams that you carefully go through and understand ALL the homework problems, worksheets
and lecture material. Hopefully this review sheet will remind you of some of the key ideas of these
sections.
1. Integration by Parts
Integration by Parts Formula: Rudv =uv Rvdu and Rb
audv =uv|b
aRb
avdu
Understand how to perform integration by parts and how to choose your u. Remember u
should be ‘easy’ to differentiate and dv should be ‘easy’ to antidifferentiate. To help you
decide we discussed LIPET (Logs, Inverse trig, Powers of x, Exponentials, Trig); let uthe
part of the integrand that occurs earlier in LIPET.
There is no one rule that tells you when to use which methods. But here are some instances
when integration by parts tends to work:
Products - specifically powers of xtimes sin(x), cos(x), or ex. It also works with ex
times sin(x) or cos(x) (but you have to pay attention to when the process repeats).
Logs - integrals involving logs (take u= ln(x))
Inverse Trig - integrals involving inverse trigonometric functions (take u= ‘inv. trig’)
2. Trigonometric Integrals
There is one general theme: Rewrite the integral using identities and use u-substitution (if
needed).
The identities that are most useful are:
cos2(x)+sin
2(x) = 1 , which we may use as cos2(x)=1sin2(x) or sin2(x)=1cos2(x)
Dividing these by cos2(x) gives more useful identities:
1 + tan2(x) = sec2(x) , which we may use as 1 = sec2(x)tan2(x) or tan2(x) = sec2(x)1
We will also need the half angle identities:
sin2=1
21
2cos(2x) , cos2(x)= 1
2+1
2cos(2x) , and sin(x) cos(x)= 1
2sin(2x)
Page 484 and 486 discuss strategies for when to use which identities and why. For me, it
is most useful to attempt to set up a u-substitution, this gives three possibilities for sin(x)
and cos(x) and three possibilities for sec(x) andtan(x):
Integrals with sin(x) and cos(x)
(a) Using u= sin(x) and du = cos(x)dx, look for an odd power of cos(x) and pull out one
factor. Then use the identites to turn the rest of the problem into sin(x)’s.
(b) Using u= cos(x) and du =sin(x)dx, look for an odd power of sin(x) and pull out
one factor. Then use the identites to turn the rest of the problem into cos(x)’s.
(c) If both powers are even, the above two methods don’t work. Use the half angle identities
to simplify the problem.
Integrals with sec(x) and tan(x)
(a) Using u= tan(x) and du = sec2(x)dx, look for an even power of sec(x) and pull out
two factors. Then use the identites to turn the rest of the problem into tan(x)’s.
(b) Using u= sec(x) and du = sec(x) tan(x)dx, look for an odd power of tan(x) and pull
out one factor of tan(x) and one factor of sec(x). Then use the identites to turn the
rest of the problem into sec(x)’s.
1
pf3

Partial preview of the text

Download Integration Techniques: A Comprehensive Review and more Study notes Analytical Geometry and Calculus in PDF only on Docsity!

7.1 - 7.5 Review - Integration Techniques

My reviews and review sheets are not meant to be your only form of studying. It is vital to your success on the exams that you carefully go through and understand ALL the homework problems, worksheets and lecture material. Hopefully this review sheet will remind you of some of the key ideas of these sections.

  1. Integration by Parts
    • Integration by Parts Formula:

udv = uv −

vdu and

∫ (^) b a udv^ =^ uv|

b a −^

∫ (^) b a vdu

  • Understand how to perform integration by parts and how to choose your u. Remember u should be ‘easy’ to differentiate and dv should be ‘easy’ to antidifferentiate. To help you decide we discussed LIPET (Logs, Inverse trig, Powers of x, Exponentials, Trig); let u the part of the integrand that occurs earlier in LIPET.
  • There is no one rule that tells you when to use which methods. But here are some instances when integration by parts tends to work: - Products - specifically powers of x times sin(x), cos(x), or ex. It also works with ex times sin(x) or cos(x) (but you have to pay attention to when the process repeats). - Logs - integrals involving logs (take u = ln(x)) - Inverse Trig - integrals involving inverse trigonometric functions (take u = ‘inv. trig’)
  1. Trigonometric Integrals
  • There is one general theme: Rewrite the integral using identities and use u-substitution (if needed).
  • The identities that are most useful are:

cos^2 (x) + sin^2 (x) = 1 , which we may use as cos^2 (x) = 1 − sin^2 (x) or sin^2 (x) = 1 − cos^2 (x)

Dividing these by cos^2 (x) gives more useful identities:

1 + tan^2 (x) = sec^2 (x) , which we may use as 1 = sec^2 (x) − tan^2 (x) or tan^2 (x) = sec^2 (x) − 1

We will also need the half angle identities:

sin^2 =

cos(2x) , cos^2 (x) =

cos(2x) , and sin(x) cos(x) =

sin(2x)

  • Page 484 and 486 discuss strategies for when to use which identities and why. For me, it is most useful to attempt to set up a u-substitution, this gives three possibilities for sin(x) and cos(x) and three possibilities for sec(x) andtan(x):
  • Integrals with sin(x) and cos(x) (a) Using u = sin(x) and du = cos(x)dx, look for an odd power of cos(x) and pull out one factor. Then use the identites to turn the rest of the problem into sin(x)’s. (b) Using u = cos(x) and du = − sin(x)dx, look for an odd power of sin(x) and pull out one factor. Then use the identites to turn the rest of the problem into cos(x)’s. (c) If both powers are even, the above two methods don’t work. Use the half angle identities to simplify the problem.
  • Integrals with sec(x) and tan(x) (a) Using u = tan(x) and du = sec^2 (x)dx, look for an even power of sec(x) and pull out two factors. Then use the identites to turn the rest of the problem into tan(x)’s. (b) Using u = sec(x) and du = sec(x) tan(x)dx, look for an odd power of tan(x) and pull out one factor of tan(x) and one factor of sec(x). Then use the identites to turn the rest of the problem into sec(x)’s.

(c) If neither of the above cases work, you may need to use identites or integration by parts. Also it may be useful to have the following integrals: ∫ tan(x) dx = ln | sec(x)| + C and

sec(x) dx = ln | sec(x) + tan(x)| + C

  1. Trigometric Substitution
    • This is a different type of substitution. We let x = a sin(θ), x = a tan(θ), or x = a sec(θ). These methods often work well for the specific integrals that have

a^2 − x^2 ,

√ a^2 +^ x^2 , or x^2 − a^2 somewhere in the problem.

  • The basic idea: Replace x with a trigonometric function, do the substitution, use a trig identity and hope for the best. (the result is a trig. integral which we studied in 7.2). These steps will ensure that you get rid of the square root.
  • Page 490 of the text has a table reminding you of which trig. function to use along with the corresponding identity.
  • At the end of the problem you will have a solution in terms of θ. To get back to x, you need to know the ‘triangle trick’ as is discussed in the text (and we will discuss it in class).
  1. Partial Fractions
  • The basic idea: Break up rational functions into sums of factors and integrate each sepa- rately. This is successfully because each individual part can be integrated easily. If we have a quadratic that factors, then we immediately factor it from the beginning. For example, 1 x^2 − 6 x + 5

(x − 1)(x − 5)

Then we will have a method for breaking this up and integration each part separately by using the first set of integrals below. If we have a quadratic that doesn’t factor, then we will need to complete the square and use what we know about the second integrals below. For example, 1 x^2 + x + 1

x^2 + x + 1/4 + 1/4 + 1

(x + 1/2)^2 + 5/ 4

Then we will use u-substitution along with second and third integrals below.

  • Thus, we are trying to simplify into integrals in an “easier form”. We will use all the following 2 facts: (a) ∫ 1 x − b dx = ln |x − b| + C ∫ 1 (x − b)^2

dx = −

x − b

+ C

(x − b)^3

dx = −

x − b

2

  • C

And so on. (b) ∫ 1 u^2 + a^2

du =

a

tan−^1

( (^) u a

+ C

(c) ∫ x x^2 + a^2

dx =

ln |x^2 + a^2 | + C