Guess what lovely integral just cropped up in my math homework!

A cookie goes to the solver.

8 comments:

harrison
said...

I worked it out to the point where you have to do actually hard stuff. Combine and reduce to a single big expression, then use partial fractions to split into two integrals. One requires polynomial long division, the other requires a trig substitution. Now quite whining and do your homework. Alternatively, plug it into Mathematica.

I plugged that into Mathematica just out of curiosity and I found that the indefinite integral is 'Horrible" looking. I doubt I have the patience to solve that integral manually.

I wonder what the original question was ...

@Paul: I think you meant taking the integral from 1-a to 1, but I doubt the function is symmetric about x = 1 and so that cannot be done.

@Vihang: You are correct. As you can probably tell by the timestamp, it was getting late and I didn't think my comment the whole way through. ^_^ For the record, though, many of the functions and regions Snively and I have been examining in 18.02 have been symmetric, which is why I jumped to that conclusion.

So when I reduced that integral I figured out that if you had one extra x somewhere then the whole thing collapses into a simple trig sub. Are you sure you wrote the problem up right? It's undoable in its current state. You've gotta be missing one more factor of +/- x somewhere.

## 8 comments:

I worked it out to the point where you have to do actually hard stuff. Combine and reduce to a single big expression, then use partial fractions to split into two integrals. One requires polynomial long division, the other requires a trig substitution. Now quite whining and do your homework. Alternatively, plug it into Mathematica.

Snively, what number are you doing? #3, in cylindrical coordinates, perhaps?

Regardless of which one you're doing, I think you can simplify the bounds of integration by taking it from 1-a to 0 and just doubling your answer.

I plugged that into Mathematica just out of curiosity and I found that the indefinite integral is 'Horrible" looking.

I doubt I have the patience to solve that integral manually.

I wonder what the original question was ...

@Paul: I think you meant taking the integral from 1-a to 1, but I doubt the function is symmetric about x = 1 and so that cannot be done.

when I'll study integrals, I'll help you !

@Vihang: You are correct. As you can probably tell by the timestamp, it was getting late and I didn't think my comment the whole way through. ^_^ For the record, though, many of the functions and regions Snively and I have been examining in 18.02 have been symmetric, which is why I jumped to that conclusion.

So when I reduced that integral I figured out that if you had one extra x somewhere then the whole thing collapses into a simple trig sub. Are you sure you wrote the problem up right? It's undoable in its current state. You've gotta be missing one more factor of +/- x somewhere.

Wow. It's like a foreign language. Math-ese or something.

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