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Cool integral

by DH
Tags: integral
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DH
#1
Apr29-05, 01:44 AM
P: 22
Determine the integral:
[tex]y = \int_{0}^{1} 1/(u^4+1)du [/tex]
and
[tex]y = \int_{0}^{1} 1/(u^5+1)du [/tex]
and
[tex]y = \int_{0}^{1} 1/(u^6+1)du [/tex]
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Data
#2
Apr29-05, 02:08 AM
P: 998
Just factor the denominator (to quadratics) and use partial fractions.
DH
#3
Apr29-05, 02:14 AM
P: 22
Quote Quote by Data
Just factor the denominator (to quadratics) and use partial fractions.

no no, can't do like that!

Data
#4
Apr29-05, 02:31 AM
P: 998
Cool integral

don't see why not~
DH
#5
Apr29-05, 02:41 AM
P: 22
Quote Quote by Data
don't see why not~
show me the solution!
Data
#6
Apr29-05, 03:01 AM
P: 998
[tex]\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)} [/tex]

[tex]= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right) [/tex]

[tex]= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)[/tex]

and from there it's just minor substitutions to finish.
DH
#7
Apr29-05, 03:04 AM
P: 22
Hix, you did the wrong thing form row 1 -> row 2!!!!
DH
#8
Apr29-05, 03:09 AM
P: 22
Quote Quote by Data
[tex]\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)} [/tex]

[tex]= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right) [/tex]

[tex]= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)[/tex]

and from there it's just minor substitutions to finish.
how about this:
[tex]y = \int_{0}^{1} 1/(u^5+1)du [/tex]
Data
#9
Apr29-05, 03:11 AM
P: 998
Same advice. You can do it yourself this time
DH
#10
Apr29-05, 03:15 AM
P: 22
Quote Quote by Data
Same advice. You can do it yourself this time
but one more time, can you show me the solution! I really want to know the way you solve it to compare with my own method.... Please give me the solution in detail!
Data
#11
Apr29-05, 03:18 AM
P: 998
You can do it almost exactly the same way I did the last one - factor the denominator (in this case, it factors to a product of two irreducible quadractics and the linear factor (x+1)), then use partial fractions to separate it into a sum of functions that you know how to integrate.

Why don't you post an example of a solution using your method? I'm interested now!
DH
#12
Apr29-05, 08:38 AM
P: 22
Quote Quote by Data
[tex]\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)} [/tex]

[tex]= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right) [/tex]

[tex]= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)[/tex]

and from there it's just minor substitutions to finish.
You made a mistake here:
[tex]\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)} [/tex]

[tex]= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right) [/tex]
Data
#13
Apr29-05, 12:04 PM
P: 998
indeed I did, the second line should be

[tex]\int_0^1 \frac{du}{2}\left( \frac{\frac{1}{\sqrt{2}}u + 1}{u^2+\sqrt{2}u+1} - \frac{\frac{1}{\sqrt{2}}u - 1}{u^2-\sqrt{2}u+1}\right),[/tex]

but the simplification after that is still similar, just with more terms.


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