# Tricky integral

1. Nov 6, 2009

### kathrynag

1. The problem statement, all variables and given/known data
Solve dW/dx=W(4-2W)^(1/2)

2. Relevant equations

3. 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?

2. Nov 6, 2009

### Bohrok

First try u = 4 - 2W

3. Nov 6, 2009

### kathrynag

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.

4. Nov 6, 2009

### HallsofIvy

Staff Emeritus
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.

5. Nov 6, 2009

### Bohrok

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.

6. Nov 6, 2009

### kathrynag

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

7. Nov 6, 2009

### Bohrok

Use another substitution that gets rid of the square root.

8. Nov 6, 2009

### kathrynag

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?

9. Nov 6, 2009

### Bohrok

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

10. Nov 6, 2009

### kathrynag

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

11. Nov 6, 2009

### Bohrok

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

12. Nov 6, 2009

### kathrynag

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?

13. Nov 6, 2009