# Hard integrationWatch

Thread starter 10 years ago
#1
How do you integrate ?

Let ,
So ,

=
=

Now let and
So and

=
=
At the end, I just get
=

This is actually not the full question, the actual question is

Anyone has clue how to do it? Thank you.
0
10 years ago
#2
Nothing obvious. Where does it come from?
0
Thread starter 10 years ago
#3
(Original post by DFranklin)
Nothing obvious. Where does it come from?
It's from 'problem plus' of the book Calculus by James Stewart. They are the challenging exercises in the book.

http://www.amazon.com/Calculus-James...5437323&sr=8-2
0
10 years ago
#4
Well, the integral can't be found in terms of elementary functions.

You could use the beta function, but I don't know if that will give a nice answer...
0
10 years ago
#5
when you intergrate by parts twice, you have to use the same terms for u and (dv/dx) if you don't you basically undo the intergration you done, and go back to the origianl eqation
0
10 years ago
#6
(Original post by Civ-217)
It's from 'problem plus' of the book Calculus by James Stewart. They are the challenging exercises in the book.

http://www.amazon.com/Calculus-James...5437323&sr=8-2
Can you tell me which chapter/question it is? I know someone who has this book and you (and I) will be interested to know how they did it.
0
10 years ago
#7
Maple can't do it (it wants to use Hypergeometrics), I haven't got Axiom installed so I can't try with that unfortunately.
0
10 years ago
#8
If there's a nice answer, my gut tells me it's going to be zero.
0
10 years ago
#9
Perhaps this is of help?

Thus, ...

I probably have made a very obvious mistake somewhere, though.
0
10 years ago
#10
(Original post by DFranklin)
If there's a nice answer, my gut tells me it's going to be zero.
That's what I got from using the beta function.
0
10 years ago
#11
Hey those limits weren't there before >_<. That makes the question so much easier lol...
0
10 years ago
#12
Nope, that formula only gives
b-a = 1-0=1 and
f (a+(b-a)x) = f(0+(1-0)x) = f(x)
hence it just gives you the original integrand

Maybe trying u=x^7?
0
10 years ago
#13
i do think the answer is 0, i mean, when you look at the (1-x^7)^1/3

whenever you put 1 into it, it becomes 0
and if its being multiplied by a power of x, whne you put the 0 in, it also becomes 0.

so my guess is that it equals 0
0
10 years ago
#14
Let . and so the integral becomes
. Integrate by parts with du = (1-y)^(-6/7):
Do a t = 1-y sub and get

But this equals by a similar argument to the first two lines, but subbing
0
10 years ago
#15
I plugged it into my calculator, and it crashed
10 years ago
#16
(Original post by DFranklin)
Let . and so the integral becomes
. Integrate by parts with du = (1-y)^(-6/7):
Do a t = 1-y sub and get

But this equals by a similar argument to the first two lines, but subbing
Nice.
0
10 years ago
#17
Well, it's really just an explicit version of doing it with beta functions... Oh, and I differentiated y^1/3 wrong (now fixed).
0
10 years ago
#18
(Original post by DFranklin)
Well, it's really just an explicit version of doing it with beta functions... Oh, and I've differentiated y^1/3 wrong - need to fix that...
Yeah, I realised that after I posted.

Quite a cool result, I think.
0
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