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Help spotting trig identities to simplify integration

  1. Aug 13, 2012 #1

    Say I'm working with ∫ sqrt(1-cos(t)) dt

    I end up with a substitution of u = 1-cos(t) and dt = du/sin(t)

    sub back in: ∫ sqrt(u) / sin(t) du

    Still got a t in there ... hrrmmm

    So I go to wolfram alpha for some inspiration and 'show steps':

    After some bashing about I see what they are doing is valid - but have no idea how to go about spotting what they have done in between the substitution and ending up with an integral only in terms of u.

    I need a little sub-'show-working' section - anyone care to show me how the 'ah-hah, that'll work!' moment actually comes about ?

    It's simple ?

    Last edited: Aug 14, 2012
  2. jcsd
  3. Aug 14, 2012 #2

    Simon Bridge

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    <hunted look> erm, simple? ... no.... not usually... it takes lots and lots of practice. The computer is your friend.

    In the above case you were on the right track, you just needed to figure out how to write sin(t) in terms of u.

    I'm guessing:
    [itex]u=1-\cos(t) \Rightarrow t=\arccos(1-u)[/itex]
    [itex]du=\sin(t)dt = \sin(\arccos(1-u)) =\sqrt{1-(1-u)^2}[/itex]
    ... which is where you just have to know the identity: [itex]\sin(\arccos(a))=\sqrt{1-a^2}[/itex]

    ... what usually happens is that you get desperate and pour through a table of trig identities, trying out likely candidates, until you get one that makes all the headaches go away. The trouble with the inverse trig functions is that they only exist for a limited range.

    If you get stuck on trigs, you can always brute-force it by using complex exponentials.
  4. Aug 14, 2012 #3

    Thanks for that - exactly what I needed :tongue2:

    I wonder if there is a resource of examples of most/all of these tricks ?

    Many textbooks and coursebooks show you a couple of them in (unrelated) examples and then leave you on your own.

    I'm at a new university now and it seems there is an expectation this stuff is prior learning. Got to catch up and fast
  5. Aug 14, 2012 #4

    Simon Bridge

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    If you didn't realise about the arccos substitution or didn't want to use it, you may just decide to get rid of those pesky trig functions and see what happens:

    [itex]x = r\cos(t)[/itex] ... definition of cosine
    [itex]dx=r\sin(t)dt = ry/r = y = \sqrt{r^2-x^2}[/itex] ... pythagoras.
    ... note: similarity to trig ID if r=1

    put [itex]r=1[/itex] ... needed for x=cos(t).
    integrand becomes:

    [tex]\frac{dx}{\sqrt{1-x^2}}\frac{1}{\sqrt{1-x}}[/tex]... then see if it helps any.
    eg. factorize (1-x2)

    Several substitutions you may end up with something you can integrate.
    If you keep track of the substitutions you may be able to write it up as one transformation and look super-clever.

    With experience comes foresight.
    Until then, you have to do a lot of messing about.
  6. Aug 14, 2012 #5
    its been a while tis all...

    catching my tail everyday
  7. Aug 14, 2012 #6

    Simon Bridge

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    yep - know the feeling.
    Note. 1st year college is mostly about establishing a baseline of prior knowledge ... if you missed or forgot something, the course gives you a heads-up and you get some important experience. You are only expected to be able to look things up ;)

    You've got it easy ... I had the same thing without the internet.
  8. Aug 14, 2012 #7
    I ended up with t then u, s and p substitutions !

    but I found a much easier way using:


    (the part in parentheses)

    Wolfram-Alpha led me down a longer and more arcane path (in terms of the trig identities used) ... bah !!

    ...guess in the long run I don't mind the exercise :rolleyes:
  9. Aug 15, 2012 #8

    Simon Bridge

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    Builds character.
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