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#1 2008-01-21 10:47:33
- MarkusD
- Member
- Registered: 2006-10-08
- Posts: 28
Integration Question (Im very rusty!)
How do I integrate the following:
∫3x/√(1-x²)dx
I tried integration by parts where u=3x, du=3, v=(1-x²)^-½, dv=-½(1-x²)^(-3/2)-2x(1-x²)^-½
it works out ok, but I get stuck with the ∫vdu part which is ∫3(1-x²)^-½
Any help?
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#2 2008-01-21 11:13:55
- mathsyperson
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- Registered: 2005-06-22
- Posts: 4,900
Re: Integration Question (Im very rusty!)
Not entirely sure that this works, by try substituting x = sin u in there.
That √(1-x²) is begging for some kind of trigonometric substitution.
Why did the vector cross the road?
It wanted to be normal.
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#3 2008-01-22 00:59:51
- TheDude
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- Registered: 2007-10-23
- Posts: 361
Re: Integration Question (Im very rusty!)
Set u = 1 - x^2, then you get
It should be straightforward from there.
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