Homework Help: Calculus

Posted by O_o Rion on Saturday, December 5, 2009 at 8:31pm.

Hi.
In an integration solution, the integral of (1/(sqrt (8-u squared)) is written as arcsin(u/sqrt 8), but I dont see how they got it. When I did it I got (1/8)*(arcsin(u*sqrt8)). What I did was take sqrt8 common in the denominator to get it in the form sqrt(1-u sq.) and then find the integral (arcsin u) and then since my 'u' was sqrt.8 * u, I divided sqrt.8 from the whole thing. Where did I go wrong?

Calculus - MathMate, Sunday, December 6, 2009 at 9:20am
"take sqrt8 common in the denominator to get it in the form sqrt(1-u sq.)"
Shouldn't it be
1/sqrt(8-u^2)
=(1/sqrt(8))/(1-(u/sqrt(8))^2)

When you do a substitution, it is always advisable to use a different letter. You are likely to confuse yourself if you use the same letter, namely, substitute u by u/sqrt(8).

Try:
w=u/sqrt(8)
then
dw=du/sqrt(8)

and
∫ du/sqrt(8-u^2)
=∫ du/(sqrt(8))/(1-(u/sqrt(8)))
=∫ dw/(1-w^2)
=asin(w)
=asin(u/sqrt(8))
Calculus - Laurie, Thursday, May 24, 2012 at 12:44pm
square root of (1-u squared)

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