I'm trying to do the following:
int (3x^3 + 4x)/(x^2+1)^2 dx
I let u = x^2+1 and I eventually get:
int 3(u-1)+4/u^2 du/2 When I further break this down, I get:
1/2 int 3u^-1 -u^-2 du am I on the right track? When I integrate this, I'm thinking that I have to be doing something...
I'm trying to integrate the following: int (x^3 + 4x)/sqrt(x^2-4) dx
I let u= x^2-4
so du/2 = xdx
granted, the above also gives me x^2=u+4 so, this gives me:
1/2 int u+8 u^-1/2 du
but the professor has:
1/2 int [u^1/2+8 u^-1/2] du
Did I miss the first u^1/2 somewhere...
I'm trying to figure out the following:
An=1x3x5.... (2n-1)/(2n)^n and I'm trying to determine if it converges or diverges and if it converges, what the limit is. The answer is 1/2n x 3/2n x 5/2n.... (2n-1)/2n and it converges, but I don't understand what they did or how they got to the...
quasar987 & HallsofIvy,
It's late where I am, but I wanted to thank you both. I am sure I am confusing several words in my posts and I'm not ashamed to admit it. That's at least how I learn! Thanks for the clarification. I have been looking at this stuff for quite a bit of time this...
If I take the limit on the sum... I get 1/1 = 1
If the limit does NOT = 0 then sigma f(x) diverges...I'm not quite sure I follow this... Does this mean that in order for the equation to converge, the sum (sigma) must be = to 0?
i'm not quiet sure how to attack this problem:
If I start plugging in #'s for n, then I get:
by this method, I see that it's going to 1/2, but I need another way to 'see' that. Any suggestions?
AKG - After we tried the problem over and over, when we got what we were expecting, you are exactly right! It was funny, when my team started this, we were apprehensive at best, but now we look at them and laugh 'cause they really are easy when you understand the problem!
VietDao29 - THANKS! Your explanation is very clear and was exactly the insight my team needed. We ended up doing just what you suggested before reading your post. The good thing is that we feel confident that we did the project correctly and with your input, it validated our work!
int x^3 cosx dx for this issue
Thanks for the reply...
essientially, we need to set this up and are having problems. We have an example of int x^4ex + dx. We get (Ax^4 + Bx^3 + Cx^2 + Ex +F) e^x + k. But we are stuck for finding the starting equation for int x^3 cosx.
I'm trying to integtrate x^3 sin x without using integration by parts. I have set up the equation to either:
int x^3 cosx dx = (Ax^3 + Bx^2+Cx+D)cos x + K
int x^3 coxs dx = (Ax^3+Bx)sinx + (Cx^2+D)cosx +K
but I'm having trouble... any help would be appreciated!
I do get ur point: tan x = opp/adj :smile:
As for point 2: limx->0 sin(x)/x = 1 vs. sin(x)/x = 1 ARE 2 different statements. Good point and I need to keep this in mind since they say very different things.
I'll search now in google for 'Taylor Series'. IK Wikipedia has also been a...
I understand the first part since (opp/hyp)/(adj/hyp) thus opp/adj = tan. That part makes sense. I don't get sin(x)/x = 1. Is this just something I should accept? I understand sin^2+cos^2=1. Would that be part of it? As for the last part, that makes sense provided they both approach the...
Thank you! I'm not quite sure I understand, but I get tan = sin/cos since tye hypenuses cancel but how can I prove tan(x)/x has the same limit as sin(x)/x? I need to read and understand some of the finer points, or I won't be able to answer some fundamental questions. Any suggestions as to...
ahhh cos(0) = 1 and sin(0) = 0. I should have verified that b4... Given this, then 2/(1+cos2(0)) = 2/2 = 1. Knowing this, there are 2 things that I should have worked through: secx = 1/cosx and that cos(0) = 1.
Thanks very much for your help!
I've stared at this problem for about 5 minutes and I'm not sure how to start it. I see the function and its derivative, but I'm missing something I suspect is simple. Any suggestions as to how to start?
oh wait, re-reading the problem and I think what I'm interpreting as a negative is the dot in the i in sin.... I'm going to work the problem as cos^6x sin^3x dx and I know how to work this... sorry about that!
VietDao, I had a typo: it should have read:
lim x->1 ln x^2/(x^2<b>-</b>1) Sorry for the error. I followed what you had and that makes sense so I think I'm getting it albeit slowly. Using the revised problem, I would get 0/0?
1 more example to make sure I'm understanding this...If I have:
lim x->1 ln x^2/(x^2+1) this can be rewitten as:
lim x->1 (2 ln x)/(x^2+1) hence using the rule, I get:
2 (1/x)/2x which resolves to:
2/2x^2 and since x->1 this equals 1
Am I getting this right?