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## Topic review (newest first)

lkomarci
2006-03-23 22:21:39

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

fgarb
2006-03-20 17:46:25

Oops, I guess I read the post a bit too fast. You already had included all the information I asked for, lkomarci, sorry about that.

George,Y
2006-03-20 13:49:11

((1-t^2)/t^1/2) = 1/t^1/2 - t^2/t^1/2 =t^-1/2 -t^3/2

t^a/t^b=t^a-b and 1/t^a=t^-a

fgarb
2006-03-20 07:34:00

Instead of looking at the answer, try just writing everything in terms of t using the t = sinx substitution. Tell us what you get when you do that and if your answer comes out wrong, write down the steps you used. I'm guessing you'll figure it out for yourself if you go through this exercise.

lkomarci
2006-03-20 01:29:57

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