Unleashing the Beast: Solving the Challenging Integral from Hell"

[gabbagabbahey]

Serious how did you even thing to start down that line of thinking?

It looked similar in form to your $$\int_0^{\pi/2} \frac{1}{1+\tan\theta^p}dx$$ integral...so, once I realized that $$\left[\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}+ \frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}\right]=1$$ , it was just a matter of justifying that the two individual integrals were equal. So, I graphed both integrands and noticed that one was a reflection of the other.

 

[gabbagabbahey]

Evidently, a general theorem can be inferred from this:

$$\int_a^b \frac{1}{1+\left(\frac{f(x)}{f(a+b-x)}\right)^c}dx=\frac{1}{2} (b-a)$$

For all real valued functions $f$. (might require one-to-oneness and integrability)

I wonder if this theorem has a name...

 

[snipez90]

Prove the identity

$$\int_a^b \ f(x)}dx=\int_a^b \ f(b+a-x)}dx$$

using substitution and then try the original integral again.

 

[dirk_mec1]

Use x = 6-t then 2*I=2.

 

[epenguin]

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

It cannot be as written above since the brackets are not closed. They are not (9 - x²) by any chance? :shy: