# Thread: Finding the integral of an inverse function

1. ## Finding the integral of an inverse function

Hi,

I have a function $f:x \to y$ for which I would like to find the integral. The function "f" is:

$y=k e^{\frac{\arccos{\left(\frac{2x}{N}-1\right) z}}{n \pi}}$

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, $f^{-1}:y \to x$, that is to say, the integral of $x=\frac{N}{2}\left(\cos(\frac{n \pi \ln \frac{y}{k}}{z})+1\right)$. Might it be possible to derive the integral of $f$ from the integral of $f^{-1}$?

Thanks

2. Try to use this:

If f and g are continuous & inverse functions of each other, then :

$\int_a^b f(x) \, dx = b \cdot f(b) - a \cdot f(a) - \int_{f(a)}^{f(b)} g(x) \, dx$

3. Not that it makes a huge difference, but is the z inside the argument of the inverse cosine function?

4. I would let u equal the argument of the exponential function. See what you get from doing that.

5. Originally Posted by BayernMunich
Try to use this:

If f and g are continuous & inverse functions of each other, then :

$\int_a^b f(x) \, dx = b \cdot f(b) - a \cdot f(a) - \int_{f(a)}^{f(b)} g(x) \, dx$
This exactly the sort of thing I was looking for. Where did you get this? Thanks a lot.

6. Originally Posted by Ackbeet
I would let u equal the argument of the exponential function. See what you get from doing that.
Yes z is outside of the arccosine argument.

As far as I can tell, letting u equal the argument of the exp function gives me:

$k \int - \frac{n \pi N\sqrt{1-\left(\frac{2x}{N}-1 \right)^2}}{2z} \ e^u \ du$

It's possible I've made a mistake, but it looks like this doesn't help me too much.

7. Originally Posted by rainer
Originally Posted by BayernMunich
Try to use this:

If f and g are continuous & inverse functions of each other, then :

$\int_a^b f(x) \, dx = b \cdot f(b) - a \cdot f(a) - \int_{f(a)}^{f(b)} g(x) \, dx$
This exactly the sort of thing I was looking for. Where did you get this? Thanks a lot.
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

$(b - a) \cdot f(b)$

and

$a \cdot [f(a) - f(b)]$

(adding one and subtracting the other from the integral of g.)

[Edit: I was mistaken. See here.]

Anyway, for the indefinate integral of f...

Originally Posted by rainer
It's possible I've made a mistake, but it looks like this doesn't help me too much.
It's fine. But express it fully in terms of u, and you'll have

$\displaystyle{- \frac{k N n \pi}{2 z} \int e^u sin(\frac{n \pi}z u)\ du$

Also, I'll try and edit in a pic later, but most of it will be like here which is for the similar

$\displaystyle{\int e^x sin(2x)\ \rmmath{d}x$

Also, are you aware of this site ? (Click on 'show steps')

However...

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

8. 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.

9. Originally Posted by rainer
This exactly the sort of thing I was looking for. Where did you get this? Thanks a lot.
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