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#1 2006-03-19 02:29:57

lkomarci
Member
Registered: 2005-08-24
Posts: 23

help...integrals of trigonometric functions

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

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#2 2006-03-19 08:34:00

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: help...integrals of trigonometric functions

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.

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#3 2006-03-19 14:49:11

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

Re: help...integrals of trigonometric functions

((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


X'(y-Xβ)=0

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#4 2006-03-19 18:46:25

fgarb
Member
Registered: 2006-03-03
Posts: 89

Re: help...integrals of trigonometric functions

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.

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#5 2006-03-22 23:21:39

lkomarci
Member
Registered: 2005-08-24
Posts: 23

Re: help...integrals of trigonometric functions

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

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