Try to use this:
If f and g are continuous & inverse functions of each other, then :
Hi,
I have a function for which I would like to find the integral. The function "f" is:
This is proving rather diffucult for my level of maths skill. Any pointers would be much welcomed.
I have found the integral of the inverse function, , that is to say, the integral of . Might it be possible to derive the integral of from the integral of ?
Thanks
Probably from comparing rectangles in a sketch of f(x) for 0 < a < b, but also (unless I'm mistaken) from assuming f increases. For decreasing f, you'll need to use the rectangles
and
(adding one and subtracting the other from the integral of g.)
[Edit: I was mistaken. See here.]
Anyway, for the indefinate integral of f...
It's fine. But express it fully in terms of u, and you'll have
Also, I'll try and edit in a pic later, but most of it will be like here which is for the similar
Also, are you aware of this site ? (Click on 'show steps')
However...
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
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Oops, my bad. I think letting u be the argument of the arccos function would be more productive. Then you can let v = arccos(u) (draw a triangle to keep track of this substitution), and I think things will be a bit easier to keep track of. The expressions you get at that point are ones you can recognize as trig functions. At that point, you integrate by parts twice to get the same integral back again.
This integral is definitely doable by Calc II methods. It's just that you have to apply more than one method.
Personally I do not know how to prove it.
I posted a thread for its proof.. here is it:
http://www.mathhelpforum.com/math-he...g-formula.html