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?
"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))
square root of (1-u squared)
To find the integral of (1/(√(8-u^2)), you can use a trigonometric substitution.
Let's start by rewriting the integral:
∫(1/(√(8-u^2))) du
Now, we can make the substitution u = √8 sinθ. This substitution will help us simplify the integral.
Differentiating both sides with respect to u, we get:
du = √8 cosθ dθ
Now we can substitute u and du into the integral:
∫(1/(√(8-u^2))) du = ∫(1/(√(8 - (√8 sinθ)^2))) (√8 cosθ) dθ
= ∫(1/(√(8 - 8sin^2θ))) (√8 cosθ) dθ
= ∫(1/(√(8(1 - sin^2θ)))) (√8 cosθ) dθ
= ∫(1/(√(8cos^2θ))) (√8 cosθ) dθ
= ∫(1/(2cosθ)) dθ
= ∫(secθ) dθ
Now, we need to solve this integral. Recall that the derivative of arcsec(x) is 1/(x√(x^2 - 1)).
So, we have:
∫(secθ) dθ = arcsec(secθ) + C
= arcsec(u/√8) + C
Therefore, the correct integral is arcsin(u/√8) + C, not (1/8) * arcsin(u√8) + C as you calculated.
You went wrong when you divided √8 from the entire expression. Instead, you need to use the trigonometric substitution and follow the steps mentioned above to simplify and evaluate the integral correctly.