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Two qn about integration (Understanding)

  1. Sep 21, 2008 #1
    Hi everyone. I got two questions I want to ask.

    1) I had just learning all the basic integration techniques (e.g. by parts, trigo etc) and just reading on my own about trigonometric substitution. I found this when I was having some difficulty integrating some basic trigo f(x) and was searching for some generic method when I stumbled upon this topic. I started reading but really couldnt understand it. From this passage, especially this part:

    Trigonometric functions The six trigonometric functions of x may be expressed
    in terms of cos x and sin x, so that the basic trigonometric polynomial integral takes
    the form
    sinm x cosn xdx. We can also allow m or n to be negative.
    Case m odd We put u = cos x and du = -sin xdx and use sin2 x = 1 - u2 on the
    remaining even powers of sin x, to get a rational function of u.
    Case n odd We put u = sin x and du = cos xdx and use cos2 x = 1 - u2 on the
    remaining even powers of cos x, to get a rational function of u.

    From :http://www.math.jhu.edu/~jmb/note/methint.pdf

    I just dont know the rationale for this. How can n be odd and then there is the "remaining even powers of cos x"? Sorry for being a newbie and all, but I really hope someone can explain the application. (or a simpler way of seeing it)

    2)Also, on another different case,

    in my lecture notes, integral of 1/(a 2-x2) is 1/2a ln abs ((a+x)/(a-x)) but integral of 1/ (x2 -a2) is 1/2a ln abs ((x-a)/(x+a)). I thought it was just a matter of removing the minus sign out and putting it as a power of the integrated function which results in the numerator and denominator switching but this is clearly not the case here.
  2. jcsd
  3. Sep 21, 2008 #2
    1. When n is odd, you take ONE of the sine or cosine and together with dx it forms du. The rest is even powers of sine/cosine.

    2. The two logarithms are indeed related by negative since their arguments are reciprocal:

    abs ((a+x)/(a-x)) = 1 / abs ((x-a)/(x+a)) ( note that abs(x-a) = abs(a-x)) and there is a general formula

    ln (something) = - ln (1/something)
  4. Sep 21, 2008 #3


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    Hi qazxsw11111! :smile:

    Suppose you have sin7xcosnx dx.

    Using u = cos x and du = -sin xdx gives (1 - u2)3un du …

    the left-hand part only has even powers of u (because one of the sinx's was absorbed into the du). :wink:
    erm … one answer is minus the other … log(p/q) = -log(q/p), and |x-a| = |a-x| … no problemo! :biggrin:
  5. Sep 22, 2008 #4
    Ok.....So let me repeat again what I think I understand. We always try to make one even power and one odd (since if it is both even we can use double angle formula)? Then apply the above formula? But one thing I dont understand is that how does the sinx gets absorbed into the dx?

    From what I understand, the extra sine will be converted into du/dx dx which will 'times' together to become du. But since du/dx=-sinx, wouldnt there be a minus sign or is there a terrible misunderstanding on my part?

    As for the second question, excellent responses! I just realized how blur I am lolx.

    Thanks guys for your help!
  6. Sep 22, 2008 #5


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    Yes … if one is odd, we can use this u substitution … if both are even, we use the double angle formula until every term has one odd. :smile:
    oops! yes, I forgot the minus! :redface:
  7. Sep 22, 2008 #6
    Thanks for all your replies. I read on further and understood the trigo subs. But it mentions "If n is negative, the substitution u = tan x, du = sec2 xdx can be useful." What is the purpose of this? When is this applicable since I convert all to sin and cos? And what should I do if there is "For integrals of the form sinmxsin nxdx" (from the text)

    At the double angled part I stumbled again. For example, if I have sin6x cos 8, then even I use double angle, it would result in a product of the two double angle formula with powers. Do I need to use substitution in this case? Since expanding them out would probably not do any good as it will result in sometiming like cos n2x. What is the easiest take? Thanks.
    Last edited: Sep 22, 2008
  8. Sep 22, 2008 #7


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    For example, if you have n = -(m+2), then it's tanmx sec2x, which is d/dx of tanm+1x/(m+1) …

    you can always use the previous method … but sometimes using sec and tan is obviously quicker!! :smile:
    2sinmxsin nx = cos(m-n)x - cos(m+n)x …

    you must memorise all your trigonometric identities ! :wink:
    That becomes (1 - cos2x)3(1 + cos2x)4/128 …

    the odd powers of cos2x are then easy, and for the even powers, you have to use the double angle formula again. :smile:
  9. Sep 22, 2008 #8


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  10. Sep 22, 2008 #9
    Yes. I did until that step. Then I couldnt integrate it further as they are products. For products I only know the formula for f(x)^n f'(x) and by parts.
    Last edited: Sep 22, 2008
  11. Sep 22, 2008 #10


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    cos72x is easy (7 is odd).

    cos62x = (1 + cos4x)3/8 …

    which gives you a cos4x and a cos34x, which are easy, and a (1 + cos8x). :smile:
  12. Sep 22, 2008 #11
    Im sorry, but how do you get these? Did you expand them out or something?:confused:

  13. Sep 22, 2008 #12


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    cos2x = (1 + cos2x)/2

    sin2x = (1 - cos2x)/2

    learn them! :smile:
  14. Sep 22, 2008 #13
    Yeah, I know that. But what has it got to do with (1 - cos2x)3(1 + cos2x)4/128 to become cos72x and cos 62x
  15. Sep 22, 2008 #14


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    'cos …

    'cos (1 - cos2x)3(1 + cos2x)4 has all the powers of cos2x up to the seventh,

    and 'cos even powers of cos2x (or sin2x) can be written in terms of cos4x, even powers of cos4x can be written in terms of cos8x, etc. :smile:
  16. Sep 22, 2008 #15
    You mean I need to use binomial to expand them out? Thats pretty tedious.
  17. Sep 23, 2008 #16


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    Yes, but you only need the x term … in the limit as x –> ∞, you can ignore x2 terms and higher. :smile:
  18. Sep 23, 2008 #17
    Im sorry. I dont get you. I dont see any x2 terms when I expand them out. I can only see the expansion as a string of numbers with cos n 2x (with quite a number of terms by the way).
  19. Sep 23, 2008 #18


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    oops! I replied to the wrong thread …

    someone else was asking about the binomial theorem also …

    please ignore last post! :redface:

    sensible response …
    Yes, I agree! :biggrin:

    Sometimes you just have to slog away … :smile:
  20. Sep 23, 2008 #19
    Arg.... Thats wickedly tiring!! Is there any other method?
  21. Sep 23, 2008 #20


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    It is not a matter of "making" one power odd. If either is odd, then you can factor out a single sine or cosine, leaving an even power so that you can use either sin2n(x)= (sin2(x))n= (1- cos2)n or cos2n(x)= (cos2(x))n= (1- sin2(x))n. In the first case, since you already have the "sin(x)dx" from the one you factored out, you can let u= cos(x) and have (1- u2)du, with, of course, additional powers of u from powers of cos(x).

    But you cannot make a power of sine or cosine odd. Either it is or it isn't to start with. If you have an integral with only even powers of sine and cosine, say [itex]\int sin^2(x)cos^2(x)dx[/itex], you cannot use that method. In that case, you can use the "double angle formulas": sin2(x)= (1/2)(1- cos(2x)) and cos2(x)= (1/2)(1+ cos(2x)).

  22. Sep 23, 2008 #21
    But look at the question above. If it has a high power such as 6 or 8, expanding them out is absolutely crazy. Any althernative methods?
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