Integration involving Substitution

[hungryhippo]

1. The problem statement, all variables and given/known data
Find the arc length of the function f(x) = x(sqrt(x/2-x) from 0 to x

Integral forms involving a+bu
--> integral [sqrt(a+bu)]/u^2du

2. Relevant equations
Arc Length = integral sqrt(1+(f'(x))^2)dx

3. The attempt at a solution
First, I took the derrivate
f'(x) = sqrt(x/2-x) + x/[(2-x)^2(sqrt(x/2-x))]
= [x(2-x)+x]/[(2-x)^2(sqrt(x/2-x))]
=[x(3-x)]/[(2-x)^2(sqrt(x/2-x)]

Then, the square of f'(x)
= [x^2(3-x)^2]/[x(2-x)^3]
=[x(3-x)^2]/[(2-x)^3]

Then I added 1 to the square of the derivative, which simplified to
=[8-3x]/[2-x]^3

Then, the square root of the above
=sqrt([8-3x]/[2-x]^3)

I substituted 2-x = t
=sqrt[(3t+2)/(t^3)] <--multiplied top and bottom by sqrt[t]
=sqrt(3t^2+2t)/t^2

I'm stuck here..I'm allowed to use formulas from a table of integrals (i.e. the one specified above), but I can't seem to get it into that form ..

[Dick]

Congratulations on getting that far. That's one nasty integral. I don't have a good integral table so I put sqrt(3t^2+2t)/t^2 into Wolfram Alpha and said "Show steps". It says, i) complete the square in the numerator, ii) do a sec trig substitution and then iii) use the universal tan(x/2) Weierstrass substitution and finally iv) get something you can solve by partial fractions. The result is less than pleasant, but it looks simpler than the steps involved in getting there. There may be a shortcut. But who cares? Life is short. Can you use WA or a computer algebra system instead of an integral table? A system like maxima also seems to handle it ok. And it's free.

[hungryhippo]

Hey, I don't think we're suppose to use the Wolfram alpha application. I asked my prof, and he said there was an "easy" way to simplify, although I have failed to find one yet.

[Dick]

Hey, I don't think we're suppose to use the Wolfram alpha application. I asked my prof, and he said there was an "easy" way to simplify, although I have failed to find one yet.

Ok, so maybe there is an easier way. I guess I'm not seeing one right now either.