Integral of root(1-cosx)

luznyr

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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

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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

Homework Helper
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

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

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.

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Homework Helper
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 :) 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 :)

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