Sunday, September 12, 2010

Post 3

Okay. So I would like to state a couple of things that Abbey and I learned while doing that partner thing:

1. When, after doing trig sub, you get either SIN^2 or COS^2, that would be when you use the following Reduction Formulas:

cos^2 = (1+cos2x)/2
sin^2 = (1-cos2x)/2

The way I remember that is, sin is negative because it's negative that its not sine. and cosine is cosine, so it's positive.

2. So instead of doing by parts like the textbook says to do when you get S of sec to an odd power, you should just memorize s=S sec^3(x) because it shows up quite often. I think it's something like 1/3secxtanx + 1/3ln(secx + tanx) + C

3. When doing trig sub, you have to be able to see basic trig concepts, so to speak. Say I have S 1/sec^2(x). One of the problems Abbey had was realizing that that was the same as cos^2(x). That's just something you have to watch out for.

4. Also, I'm pretty sure everyone could brush up on some things...ie trig formulas.

5. Memorize your triangles...I'm thinking.

6. In normal trig integration. You have to realize that sometimes...say you have tan and sec in an integral...if you can some how get a du (ie sec^2 or sectan) out of it, thats the way you want to go because by doing so, you just have to do normal substitution integration again. Got it? good.

Okay, for stuff you can comment on...umm...

Could someone explain the process of doing definite integrals for me? I'm kinda sketchy on the part where you plug the original values in?? it's all kind of a blur. Anyways..thanks!!

2 comments:

  1. Me too Mal Pal, I'm looking at an example right now. I get you still do the whole x, dx, square root thing, plug everything in. Now you have the integral. I think you go back to your x and plug in you pounds for theta and solve? Somehow I don't know, but that is what I'm interpreting. Maybe me and you partner should go over this next time..

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  2. idk what you mean by doing definite integrals... so i'll go back to the basics.
    when you are done integrating,
    b
    S x^2
    a

    then f(b) - f(a) is your answer... haha

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