Unleashing the Beast: Solving the Challenging Integral from Hell"

[lubuntu]

Integral from Hell!

Homework Statement

$$\int<br /> \frac{\sqrt{ln(9-x}}{\sqrt{ln(9-x} + \sqrt{ln(x+3)}}<br /> dx$$

Homework Equations

woe = me

The Attempt at a Solution

Tried all the algerbraic manipulations I could think of, can't figure out a substitution either.

 

[letmeknow]

Have you tried multipying by the conjugate? Just a guess ( not a hint guess, but a real guess)

 

[lubuntu]

Tried didn't get me anywhere seemingly

 

[gabbagabbahey]

Is this supposed to be an indefinite integral, or are you given limits? Also, are you sure it is exactly as written above?

 

[lubuntu]

limits from 2 to 4, figured i'd just get help getting the anti derivative

 

[gabbagabbahey]

I don't think there is a general antiderivative; but I suspect that the limits allow you to use some trick (like in the 1/(1+tan^b (x) ) thread)...hmmmm, let me see...

 

[lubuntu]

Holy crap this is linear! anyone care to help me figure out why?

 

[gabbagabbahey]

Really?!...Doesn't look exactly linear to me...It looks close to a straight line, but not really.

 

[lubuntu]

I could have been mistaken

 

[gabbagabbahey]

I'd start by dividing the numerator and denominator by $\sqrt{\ln(9-x)}$...

 

[lubuntu]

nice one! how didn't i see that!

 

[gabbagabbahey]

nice one! how didn't i see that!

Lol , actually, I'm still trying to see why the area under $$\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$ is the same as the area under $$\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}$$

Is there some proof I'm not remembering?

 

[lubuntu]

not that i know of, I'm kind of stuck at that point now I don't see any substitutions that would make things simpler from here.

 

[lubuntu]

$$\frac{\frac{\ln(3+x)}{\ln(9-x)}^b}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$

 

[lubuntu]

I don't see that going anywhere, anyone got ideas?

 

[gabbagabbahey]

$$\frac{\frac{\ln(3+x)}{\ln(9-x)}^b}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$

Sure, $$\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}= \frac{\frac{\ln(3+x)}{\ln(9-x)}^b}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$

so $$\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}+\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}=1$$

But that doesn't explain why the area under $$\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$ is equal to the area under $$\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}$$ over that interval (although I know from mathematica that it is)

 

[gabbagabbahey]

Hmmm...what do you get when you flip the function $$\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$ over the line $x=3$?

You may need to look in your linear algebra text, under the transformations/reflections section for this one)

 

[lubuntu]

I think I'm giving up! This was one of 4 challenge problems my professor gave... I think I got 2 of the others I don't know on this one haha.

 

[gabbagabbahey]

Say you have any function y=f(x) on some interval [a,b], and you reflect the graph of that function over the mid-way line of the interval (x=(a+b)/2)...would the area under the function over that interval change under such a reflection?

http://img4.imageshack.us/img4/4063/92703524.jpg
http://g.imageshack.us/img4/92703524.jpg/1/

http://img510.imageshack.us/img510/9892/88557644.jpg
http://g.imageshack.us/img510/88557644.jpg/1/

 

[lubuntu]

Of course not, nice job finding that out, not sure if it is going to help me compute the integral.

 

[gabbagabbahey]

Of course not, nice job finding that out, not sure if it is going to help me compute the integral.

It should!

Hmmm...what do you get when you flip the function $$\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$ over the line $x=3$?

You may need to look in your linear algebra text, under the transformations/reflections section for this one)

Well...what does the function become?

 

[lubuntu]

I'm not familar with linear algebra :(, haven't take that course yet!

 

[gabbagabbahey]

Do you know how to flip a graph over the line x=0?

 

[lubuntu]

Sure wouldn't you just solve it for -x?

 

[lubuntu]

Ok I see your point about the relfection now after doing it out, but that doesn't make the integral any easier?

 

[gabbagabbahey]

Sure wouldn't you just solve it for -x?

What do you mean by 'solve it'?

A refection over the line x=0 (y-axis) corresponds to the transformation $x\to-x$ and $y\to y$; that means that every point (x,y) becomes (-x,y). Therefor, if you reflect f(x) over the line x=0 it becomes f(-x).

Follow?

Now, to reflect a function over the line x=c, you make the transformation $(x-c)\to-(x-c)$ which implies $f(x)\to f(2c-x)$

Follow?

If so, apply that to the integrand...

 

[gabbagabbahey]

Ok I see your point about the relfection now after doing it out, but that doesn't make the integral any easier?

Why not, what do you get when you reflect the integrand over x=3?

 

[lubuntu]

Right... that's what I just did, so you flip the numerator and denomontator, how does that get us closer to the antiderviative thou?

 

[gabbagabbahey]

Right... that's what I just did, so you flip the numerator and denomontator, how does that get us closer to the antiderviative thou?

Well, if $$\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}$$ is what you get by flipping $$\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$ over the line x=3, what can you say about the area under each of those on the interval [2,4]?

 

[lubuntu]

The are equal