Integral of root(1-cosx)

  • Thread starter luznyr
  • Start date
  • #1
luznyr
28
0

Attachments

  • eqn.gif
    eqn.gif
    263 bytes · Views: 2,642
  • working.gif
    working.gif
    899 bytes · Views: 2,962
Last edited:

Answers and Replies

  • #3
luznyr
28
0
thanks for the link, when i used mathematica 5.2 i got (1/2 - cosx/2)x. is that the same as -2sqrt(1-cosx).cot(x/2) (the answer from the integrator) ?

What i was really after was how you would integrate the original function algebraically.

Edit:

Sorry i put the eqn in mathematica wrong, it does give the same ans as the integrator as expected. my end result was -2sqrt(2).cos(x/2) = -2sqrt(1-cosx).cot(x/2)). thanks for all your help
 
Last edited:
  • #4
christianjb
530
1
thanks for the link, when i used mathematica 5.2 i got (1/2 - cosx/2)x. is that the same as -2sqrt(1-cosx).cot(x/2) (the answer from the integrator) ?

What i was really after was how you would integrate the original function algebraically.

I would do it using Mathematica. Failing that, I would recognize that cos(2x)=1-2sin^2(x)

-> cos(x)=1-2sin^2(x/2)

-> 1-cos(x)=2sin^2(x/2)

-> sqrt(1-cos(x))=sqrt(2) sin(x/2)

etc. etc.
 
  • #5
luznyr
28
0
thanks that should help heaps
 
  • #6
christianjb
530
1
There are probably about 236 ways of expressing the final answer, so don't be discouraged if it doesn't look like any of the above.

You should try to see if they agree though for random values of x.
 
  • #7
luznyr
28
0
thanks that should help heaps
 
  • #8
Gib Z
Homework Helper
3,352
6
Even if it doesn't looking anything similar, find the derivative and if its the original integrand, you are done :) christianjb's way is also good, but the 2 expressions may differ by a constant so watch out for that.
 
  • #9
christianjb
530
1
Even if it doesn't looking anything similar, find the derivative and if its the original integrand, you are done :) christianjb's way is also good, but the 2 expressions may differ by a constant so watch out for that.

But my way is

1) Use Mathematica,
2) if that doesn't work- wait for Gib Z to solve it.
 
  • #10
luznyr
28
0
hahahahahaha
 
  • #11
Schrodinger's Dog
817
6
My answers don't often look like the result, but just to show how fluid an anwer can be here's

Mathcad.

[itex](2-2cos(x))^\frac{1}{2}.sin(x)\frac{2^\frac{1}{2}}{-1+cos(x)}+C[/itex]

and

http://www.calc101.com/webMathematica/integrals.jsp#topdoit

[itex]-2\sqrt{cos(x)+1}+C[/itex]

and Wolfram which agrees with this.

Both answers return -2.482 with x=1.

hehe.
 
Last edited:
  • #12
Gib Z
Homework Helper
3,352
6
I don't know why Mathcad does not simplify it's answers like calc101 does. Even without any trigonometric manipulations, Mathcad's answer can be simplified to [tex]\frac{ -2\sin x}{\sqrt{1-\cos x}}[/tex], and yes in this case both anti derivatives are identical in the sense that when equated, the Constants are equal to 0.

EDIT: Some members of the forum wish to revive one of my old threads, and I'm not complaining, so here it is in case your interested :) https://www.physicsforums.com/showthread.php?t=149706&page=14

The original purpose was for people to post up integrals (usually indefinite) for me to solve. However the renewed purpose is for anyone to post up a particularly difficult problem for anyone to solve. The problems should be able to be worked out with no more than CalcII knowledge please. It would be wondering you you all would participate :)
 
Last edited:

Suggested for: Integral of root(1-cosx)

  • Last Post
2
Replies
54
Views
2K
Replies
7
Views
614
Replies
3
Views
672
  • Last Post
Replies
2
Views
266
Replies
5
Views
315
  • Last Post
Replies
12
Views
559
Replies
16
Views
138
Replies
8
Views
894
Replies
7
Views
316
  • Last Post
Replies
5
Views
252
Top