Integral of root(1-cosx)

[luznyr]

1. The problem statement, all variables and given/known data


http://www.physicsforums.com/attachment.php?attachmentid=10407&stc=1&d=1183955658
eqn.gif

2. Relevant equations



3. The attempt at a solution

http://www.physicsforums.com/attachment.php?attachmentid=10408&stc=1&d=1183955658
working.gif

any suggestions would be much appreciated.

[christianjb]

The integrator handles it with ease.

http://integrals.wolfram.com/index.jsp

[luznyr]

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

[christianjb]

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.

[luznyr]

thanks that should help heaps

[christianjb]

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.

[luznyr]

thanks that should help heaps

[Gib Z]

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.

[christianjb]

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.

[luznyr]

hahahahahaha

[Schrodinger's Dog]

My answers don't often look like the result, but just to show how fluid an anwer can be here's

Mathcad.

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

and

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

-2\sqrt{cos(x)+1}+C

and Wolfram which agrees with this.

Both answers return -2.482 with x=1.

hehe.

[Gib Z]

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 \frac{ -2\sin x}{\sqrt{1-\cos x}}, 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 :) http://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 :)