Tricky integral

  • Thread starter kathrynag
  • Start date
  • #1
598
0

Homework Statement


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



Homework Equations





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

  • #2
867
0
First try u = 4 - 2W
 
  • #3
598
0
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
HallsofIvy
Science Advisor
Homework Helper
41,833
964
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
867
0
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
598
0
So, I'm left with :
-du/[(4-u)u^(1/2)]
Then I'm stuck
 
  • #7
867
0
Use another substitution that gets rid of the square root.
 
  • #8
598
0
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
867
0
You could use a trig substitution; however, partial fractions would be better.
 
  • #10
598
0
So, I got 1/2ln[(4-2w)^(1/2)-2]-1/2ln[(4-2w)^(1/2)+2]=x+c
 
  • #11
867
0
After checking the answer on wolframalpha, it looks like you're missing a couple numbers in the ln's.
 
  • #12
598
0
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
867
0
Differentiating your result does give the same thing you started with, so your answer is correct.
 
  • #14
598
0
ok thanks
 

Related Threads on Tricky integral

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
969
  • Last Post
Replies
1
Views
928
  • Last Post
Replies
2
Views
541
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
998
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
9
Views
1K
Top