With This Integral. Not Home Work Problem

[scottshannon]

Homework Statement

Actually I have the problem solved but I do not understand how the numerator becomes equal to the denominator in the 3rd step.

Homework Equations

I have been looking for an identity for arctan but can't find one that seems to match.

The Attempt at a Solution

My attempt has been to derive a trig identity for arctan (A/B). I found this: arctan[(A+B)/(1−AB)]=arctanA+arctanB

 

[vela]

It's easier to work the other way. Use trig identities to simplify ##\tan(\arctan 3x + \arctan 2x + \arctan x)## and show it's equal to ##\frac{6x(1-x^2)}{1-11x^2}##.

 

[scottshannon]

Hmmm...ok...how do you do that?

 

[haruspex]

Hmmm...ok...how do you do that?

Do you know a formula for expanding tan(A+B)?

 

[scottshannon]

yes I do...I have practiced deriving it...

 

[scottshannon]

here we are dealing with arctan though

 

[haruspex]

yes I do...I have practiced deriving it...

So apply it to tan(arctan3x+arctan2x+arctanx)

 

[haruspex]

here we are dealing with arctan though

Just think of arctan(whatever) as angle A.

 

[scottshannon]

Thank you...I can expand tan (A+B) to tan (A+B+C).
What I don't understand is how did the young man who solved the problem know how to do it the way he did?

 

[haruspex]

Thank you...I can expand tan (A+B) to tan (A+B+C).
What I don't understand is how did the young man who solved the problem know how to do it the way he did?

Please clarify: are you now saying that you can see the two are equal, but you don't understand how anyone guessed that they might be?
If the solution was created by the person who set the problem, it probably went the other way about. They started with the sum of the three arctans, in numerator and denominator, then applied the tan(A+B) expansion twice to the numerator only to disguise the equivalence.