# Tricky integral

## Homework Statement

Solve dW/dx=W(4-2W)^(1/2)

## The Attempt at a Solution

dW/[W(4-2W)^(1/2)]=dx
dW/[W(4-2W)^(1/2)]=x+C
I'm unsure of whether to use a u-sub, partial fractions, or a trig substition to integrate. I looked in the back of the book and the answer has a sech in the answer but how to get there?

## Answers and Replies

First try u = 4 - 2W

I did that and was left with dW/[Wu^(1/2)]. I know du would give me -2. I thought about using du/[(u-4)u^(1/2)], but wasn't sure what to do from there.

HallsofIvy
Science Advisor
Homework Helper
Then you didn't do it right! When you substitute, you cannot leave the "old" variable in the integral, not even in the differential. If u= 4-2W, then 2W= 4- u so W= (4-u)/2 and dW= (-1/2)du.

You need to "replace" or substitute all W's with u's in the integrand. Using u = 4 - 2W, solve for W and substitute that for W in the integrand. You also need du in your integrand, not dW.

So, I'm left with :
-du/[(4-u)u^(1/2)]
Then I'm stuck

Use another substitution that gets rid of the square root.

So, using a=u^(1/2) and a^2=u, then du=2ada and 4-u=4-a^2
-2ada/[(4-a^2)a]=2da/(a^2-4). This leads to a trig answer because I think there is a trig law with a^2+c^2?

You could use a trig substitution; however, partial fractions would be better.

So, I got 1/2ln[(4-2w)^(1/2)-2]-1/2ln[(4-2w)^(1/2)+2]=x+c

After checking the answer on wolframalpha, it looks like you're missing a couple numbers in the ln's.

I went from:
partial fractions A=1/2, B=-1/2
1/2ln(a-2)-1/2ln(a+2)
1/2ln(u^1/2-2)-1/2ln(u^1/2+2)
So, I got 1/2ln[(4-2w)^(1/2)-2]-1/2ln[(4-2w)^(1/2)+2]=x+c
Where would I have gone wrong to need these extra numbers?

Differentiating your result does give the same thing you started with, so your answer is correct.

ok thanks