Asked by O_o Rion
Hi.
In an integration solution, the integral of (1/(sqrt (8-u squared)) is written as arcsin(u/sqrt 8), but I don't 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?
In an integration solution, the integral of (1/(sqrt (8-u squared)) is written as arcsin(u/sqrt 8), but I don't 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?
Answers
Answered by
MathMate
"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<b>/sqrt(8)</b>)^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))
Shouldn't it be
1/sqrt(8-u^2)
=(1/sqrt(8))/(1-(u<b>/sqrt(8)</b>)^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))
Answered by
Laurie
square root of (1-u squared)
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