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## #1 Re: Help Me ! » integrating trigonometric functions » 2006-04-01 08:02:13

i understand now, i borrowed some books from a local library and now i understand.
thank you George

## #2 Re: Help Me ! » integrating trigonometric functions » 2006-03-30 22:46:15

whoops sorry it's supposed to say..
1-(Cos^2)2x = 1+Cos4x

## #3 Help Me ! » integrating trigonometric functions » 2006-03-30 22:45:00

lkomarci
Replies: 5

i've been doing integrals of trig functions because it's my weakest point.
there are a few basic things that i don't understand.

for eg. i KNOW that:

sin^2x=(1-Cos2x)/2
cos^2x=(1+Cos2x)/2

but please explain this, why is...

1-(Cos^2)2x = 1-Cos4x

and

(Cos^3)2x = (1-(Sin^2)2x)Cos2x

## #4 Re: Help Me ! » bunch of stuff » 2006-03-25 21:24:00

am i really not going to get ANY help?
jeez thanx

## #5 Re: Help Me ! » bunch of stuff » 2006-03-25 04:40:04

thank you for that MathIsFun..
next thing you could help me with is making someone to help me with these problems

## #6 Re: Help Me ! » bunch of stuff » 2006-03-23 23:55:20

i've edited my post now...i hope you can get to the images now.
i used images upload before that and they didn't appear

## #7 Re: Help Me ! » bunch of stuff » 2006-03-23 23:43:50

eermm..guys...i've attached 2 pictures but...it seems they didn't appear in the post. how's so?

## #8 Help Me ! » bunch of stuff » 2006-03-23 23:42:42

lkomarci
Replies: 6

first of all, i have to inform you that this is not any sort of homeworks. i'm a student and i have an exam at the end of next month. analysis isn't really my teritory, neither is math overall. :)

most of the stuff i put on this paper is...harder for me to solve. some of it are implicit derivative which should be easy but i just need your help to clarify some stuff.

i would really appreciate if you could take some time to help me out, at least with some of the exercises.

Image I

Image II

## #9 Re: Introductions » not THAT new » 2006-03-23 23:07:52

well...i guess it's not THAT bad afterall. always some new challenges.

## #10 Re: Help Me ! » finding the crosspoint(?) of two functions » 2006-03-23 07:07:17

i just need to find the point where they meet. without that i won't be able to calculate the size of space these two functions "embrace."

## #11 Help Me ! » finding the crosspoint(?) of two functions » 2006-03-22 23:26:20

lkomarci
Replies: 2

hey guys, i don't know if i've used the right term (crosspoint)..
i need to find the point where these two functions meet.

f1(x)=lnx
f2(x)=x^2/2e

f1(x)=f2(x)
y=?

i'm pretty sure that i have to use "ln" on this but i'm not very good with LNs so please help

## #12 Re: Help Me ! » help...integrals of trigonometric functions » 2006-03-22 23:21:39

fgarb: no problem, it happens
George: thanx man, i can't believe i didn't realize what i was doing wrong..

## #13 Help Me ! » help...integrals of trigonometric functions » 2006-03-19 02:29:57

lkomarci
Replies: 4

hey guys,

i have a problem with integrals of trig functions.

let me explain on an example...

∫ ((Cos^3)x/Sinx^1/2)dx...

i get to the point where i use the substitution method.... Sinx=t--->cosxdx=dt

when i include this into my function i get this.... ∫ ((1-t^2)/t^1/2)dt...

i don't get the next part of the procedure... --->  ∫(t^-1/2)dt - ∫(t^3/2)dt....
this last part i don't understand...

i don't understand how they get the t^3/2...explanation please

btw i know the stuff such as (cos^2)x=1-(sin^2)x and that so that i understand

## #14 Re: Introductions » not THAT new » 2006-02-16 01:36:55

oh yea i'm loving it already

## #15 Introductions » not THAT new » 2006-02-15 21:58:29

lkomarci
Replies: 10

hey all,
i'm not really THAT new on this forum but still haven't introduced myself properly.

so...what the hell am i doing on this forum? (OMG it's a math forum! and WHO(!) likes math?!!)
yea well i really do hate it. but i gotta study some math aswell for my university. btw i study computer science, but they're gonna MURDER me with all this math. it's like i'm studying math and not computer science.

i just had my second math exam for the third frickin time! and almost EVERY TIME they "invent" some field of math that i don't know. btw the exam is about limes functions, integrals, derivatives and financial math. i went through every type of this stuff but like i said, they add some new math.

i'm slowly but surely going nuts because of math. i don't have the will and the power to start learning for my other exams before i finish math. besides, in my case, math probably burns more calories than boxing, so i'm always way too tired to go and grab another book after studying math.

## #16 Help Me ! » mathematica drawing graphs » 2006-02-15 21:49:35

lkomarci
Replies: 2

hey all...i'm really new to mathematica so i don't know how to make graphs of functions..

can someone please do this for me?

f(x)= (1-4x^2) / (4x^2(1+x^2))

thank you

## #17 Re: Help Me ! » Homework problems » 2005-09-13 21:58:47

no no no no WCY.. not correct

watch:

x/(x-1) + 3 = 1/(x-1)    / (x-1)

x + 3(x-1) = 1

x + 3x - 3 = 1

x+ 3x = 1 + 3

4x = 4 /:4

x = 1

## #18 Re: Help Me ! » Homework problems » 2005-09-13 21:55:27

i have to say i'm proud that even i can help out someone:

i got the solution to the 2 task...it actually couldn't be easier.

6/(x-2) - 1/(x+5) = 2/x^2+3x-10

6/(x-2) - 1/(x+5) = 2/(x-2)(x+5)     /(x-2)(x+5)

6(x+5) - (x-2) = 2

6x+30-x+2=2

5x = -30  /:5

x = -6

see?? not so hard...the only thing that you HAVE to notice is x^2+3x-10, which can also be written in form of (x-2)(x+5)

when u multiply the first two variables of each bracket (x,x) you get the X^2, when you sum up the second two variables (-2, 5) you get the 3x, and when u multiply them (-2, 5) you get the -10.
that's the way this thing works. you practice that, you have to be experienced enough to spot this as soon as you look at the exercise.

## #19 Re: Help Me ! » Integrals: substitution method » 2005-09-13 21:20:17

Excellent indeed
and once more...the day is saved by kylekatarn. THANK YOU

## #20 Re: Help Me ! » Integrals: substitution method » 2005-09-12 04:33:28

hey kyle...i got another one for you. i want u to solve this one, because i think that the solution in the book is not correct.  the book was written by some of my teachers, they probably wrote the darn book in one night. so many mistakes.

ok here it goes:

∫ 3dx / 4(2-5x)^4/5 =

the solution given in the book goes like this:

I=-3(2-5x)^1/5  /  4 + C

## #21 Re: Help Me ! » Integrals: substitution method » 2005-09-11 08:03:18

i dunno why i didn't think of writing that root as en exponent. i do that a lot

thanx again. i'll write back if i got some more problems

## #22 Help Me ! » Integrals: substitution method » 2005-09-11 01:20:48

lkomarci
Replies: 13

hi everyone, plz don't laugh at me, i'm a math noob, i'm currently learning integrals and i got a problem. i dont understand how to solve this exercise:

∫ 4th sqrt(3x-5)³ dx    (it's not 4 x, it's 4th root)

1.  now...i usually take this under the brackets and derivate it:
3x - 5=t
3dx=dt /:3
dx=dt/3

the solution should be:   I = 4 X 4th sqrt (3x-5)^7  / 21 + C

my math-english is not that good so.. this 4th sqrt should mean fourth square root

## #23 Help Me ! » derivative » 2005-08-24 18:53:14

lkomarci
Replies: 1

is there someone on msn who is good at derivating? :-D
i would need some help plz, it's not just one exercise but more types of em...if someone is willing to help a bit...add me to msn, it's sport_billy007@hotmail.com

cheers