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July 22

Integration by substitution

Using the substitution $u=x+{\frac {1}{x}}$ , evaluate $\int {\frac {x^{2}-1}{x^{4}+3x^{2}+1}}dx$ ?115.178.29.142 (talk) 03:09, 22 July 2010 (UTC)[reply]

So... did you try that? Rckrone (talk) 04:43, 22 July 2010 (UTC)[reply]

It's question 16 here: http://www.tqa.tas.gov.au/4DCGI/_WWW_doc/007087/RND01/MSP5C_report_04.pdf It's the only question in the last 6 years of past exams that I can't do, I've spent heaps of time trying to solve it to no avail, just tell me how to do it please.––115.178.29.142 (talk) 05:24, 22 July 2010 (UTC)[reply]

Probably the first step with any u-substitution is to convert the differential. Differentiating both sides of u = x + 1/x gives you du = (1 - 1/x²)dx, the right side of which should stick out as being very close to the numerator of your integrand. In particular

\int {\frac {(x^{2}-1)dx}{x^{4}+3x^{2}+1}}=\int {\frac {x^{2}du}{x^{4}+3x^{2}+1}}=\int {\frac {du}{x^{2}+3+1/x^{2}}}.

See if you can take it from there.

Also just a note for the future, if you post a question without explanation like you did initially people aren't likely to answer it since the policy (at the top of the page) is that we shouldn't answer homework questions without some evidence that you're tried and gotten stuck and there's no way to tell if your question is homework or not. Rckrone (talk) 05:50, 22 July 2010 (UTC)[reply]

Same guy, different IP (because I'm at home now). Anyway

u^{2}=(x+{\frac {1}{x}})^{2}=x^{2}+2+1/x^{2}

so

\int {\frac {du}{x^{2}+3+1/x^{2}}}=\int {\frac {du}{u^{2}+1}}=\arctan {u}+C=\arctan {(x+{\frac {1}{x}})}+C

thanks for your help ––220.253.172.214 (talk) 06:26, 22 July 2010 (UTC)[reply]

Do we have a Wikipedia article that describes the variety of uses of that particular substitution in algebra and in calculus? Michael Hardy (talk) 19:35, 22 July 2010 (UTC)[reply]

...without that particular substitution I might have tried writing the denominator thus:

x^{4}+3x^{2}+1=\left(x^{2}+{\frac {3+{\sqrt {5}}}{2}}\right)\left(x^{2}+{\frac {3-{\sqrt {5}}}{2}}\right)=\left(x^{2}+\left({\frac {1+{\sqrt {5}}}{2}}\right)^{2}\right)\left(x^{2}+\left({\frac {1-{\sqrt {5}}}{2}}\right)^{2}\right),

and then on to partial fractions and you can see we're headed for arctangents. Then apply the formula for a sum of two arctangents followed by more algebra. Very pedestrian. So you can see that the substitution used above saves some work! Michael Hardy (talk) 19:44, 22 July 2010 (UTC)[reply]

.... I worked through the details, and all the square roots of 5 cancel out of course, and I get

\arctan \left({\frac {-1}{\left(x+{\frac {1}{x}}\right)}}\right)+{\text{constant}},

and this is where one should remember that

\arctan {\frac {-1}{w}}={\frac {-\pi }{2}}+\arctan w.

So the substitution certainly bypasses some complications. Michael Hardy (talk) 20:28, 22 July 2010 (UTC)[reply]

Some details seem uncertain above. Let's do it carefully:

u=x+{\frac {1}{x}}

du=\left(1-{\frac {1}{x^{2}}}\right)\,dx={\frac {x^{2}-1}{x^{2}}}\,dx

\int {\frac {x^{2}-1}{x^{4}+3x^{2}+1}}\,dx=\int {\frac {1}{x^{2}+3+x^{-2}}}\cdot {\frac {x^{2}-1}{x^{2}}}\,dx=\int {\frac {1}{x^{2}+3+x^{-2}}}\,du

This takes us to the point just before getting u² + 1, etc. Michael Hardy (talk) 20:00, 22 July 2010 (UTC)[reply]

Dynamic Programming

Hey am strugling to find out,a good introduction and some terminologies which is used in the topic,dynamic programming of subjet operation research.........If u don mind plz give me answer for the same which i have mentioned above.........I would like to get your reply as fast as possible.

thanking you

yours faith fully jacob jose vazhappily (sd/-) —Preceding unsigned comment added by Jcubjoz (talk • contribs) 10:34, 22 July 2010 (UTC)[reply]

It sounds like you just want our article: dynamic programming. You can reply here if that's not sufficient. --Tardis (talk) 18:41, 22 July 2010 (UTC)[reply]

Barrow on Archimedes

The page Isaac Barrow states that Barrow "lectures for 1667 were published in the same year, and suggest the analysis by which Archimedes was led to his chief results." Does anyone have any additional information on this? Are the 1667 lectures online? Actually, I see that McT gives a slightly different chronology. Tkuvho (talk) 15:11, 22 July 2010 (UTC)[reply]