Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

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

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

Offline

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

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.

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,320

((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**

Offline

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

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.

Offline

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

fgarb: no problem, it happens

George: thanx man, i can't believe i didn't realize what i was doing wrong..

Offline

Pages: **1**