anonymous
  • anonymous
need help in the derivation of integration formula:
Mathematics
chestercat
  • chestercat
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misty1212
  • misty1212
HI!!
misty1212
  • misty1212
i didn't know there was such a formula
anonymous
  • anonymous
\[\int\limits_{.}^{.} \sqrt{x^{2}-a^{2}}dx =\frac{x }{ 2 }\sqrt{x ^{2}-a ^{2}}-\frac{ a ^{2} }{ 2 }\log \left| x+\sqrt{x ^{2}+a^2} \right|+C\]

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anonymous
  • anonymous
solve by taking x=a sect
anonymous
  • anonymous
no problem @misty1212
misty1212
  • misty1212
this one is a pain in the neck i think, but the sub is right
misty1212
  • misty1212
you are going to get \[a^2\int \sec^3(u)-\sec(u)du\]
misty1212
  • misty1212
did you get to that part, or no?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i dont know what to do after that.....
misty1212
  • misty1212
this is the kind of thing you look up in the back of the text because it is boring beyond belief
anonymous
  • anonymous
can u go after that
misty1212
  • misty1212
there is a "reduction formula" for secant
misty1212
  • misty1212
are you allowed to use it?
anonymous
  • anonymous
u can take the snapshot for the things u solved in the notebook instead of typing them. it could be easier, if u can do it.....
anonymous
  • anonymous
what is reduction formula
misty1212
  • misty1212
you mean in general or "what is the reduction formula for \(\int \sec^n(x)dx\)?
anonymous
  • anonymous
then what is that?
misty1212
  • misty1212
here are a bunch of them http://archives.math.utk.edu/visual.calculus/4/recursion.2/
misty1212
  • misty1212
in your case \(n=3\) so you get \[\int sec^3(u)du=\frac{1}{2}\tan(u)\sec(u)+\frac{1}{2}\int \sec(u)du\]
anonymous
  • anonymous
hav u done it by taking integration by parts?
misty1212
  • misty1212
it looks like that right? but it is not it comes from doing some trig business and a u sub
misty1212
  • misty1212
like the same trick they use for \(\int\sin^n(x)dx\) we can work through it if you like although it is not that interesting
misty1212
  • misty1212
oh to answer your question, no i used nothing but the"reduction formula" with \(n=3\) the one i sent the link to
misty1212
  • misty1212
ignoring the annoying \(a^2\) out front, and combining like terms, we should be at \[\frac{a^2}{2}\sec(u)tan(u)-\frac{a^2}{2}\int \sec(u)du\]
anonymous
  • anonymous
i dont know these reduction formulas...it's not in my book....in my book, the derivation was given by the method of integration by parts and in the end it was written that u can take x=sec t to solve the problem...so i did it by doing the substitution and got clutched in the middle...
misty1212
  • misty1212
now how to integrate secant, again it is not interesting, best to memorize it however, you want a good explanation, easier than i can write here, click on this http://math2.org/math/integrals/tableof.htm then go to "proof"
anonymous
  • anonymous
i can prove by the method of integration by parts...so no problem, when i would learn about these reduction formulas then i would go by that method also...
misty1212
  • misty1212
i think you use parts for \(\int \sec^3(x)dx\) but not for \(\int \sec(x)dx\)
misty1212
  • misty1212
it is just something they do to make a formula is all some people like formulas
ganeshie8
  • ganeshie8
if you like partial fractions, \[\sec x =\dfrac{\cos x}{\cos^2x} = \dfrac{\cos x}{1-\sin^2x}\]
misty1212
  • misty1212
oh cool, lots easier than the un-intuitive multiplying top and bottom business
anonymous
  • anonymous
please give a more detail where to start from....@ganeshie8
ganeshie8
  • ganeshie8
same can be extended to \(\sec^3x\) too i think \[\sec^3 x =\dfrac{\cos x}{\cos^4x} = \dfrac{\cos x}{(1-\sin^2x)^2}\]
misty1212
  • misty1212
@ganeshie8 is proving \[\int\sec(x)dx=\ln(\sec(x)+\tan(x))\]
anonymous
  • anonymous
i dont know how to integrate that partial fraction....@ganeshie8
misty1212
  • misty1212
\[\int \frac{\cos(x)}{1-\sin^2(x)}dx\] put \(u=\sin(x)\) and integrate \[-\int \frac{du}{1-u^2}\] using partial fractions
misty1212
  • misty1212
@ganeshie8 that is correct yes? never saw it done this way
anonymous
  • anonymous
oh yes i got that...
ganeshie8
  • ganeshie8
looks good to me !
ganeshie8
  • ganeshie8
for \(\int\sec^3x\,dx\) i would try reduction formula though as partial fractions looks a bit lengty http://www.wolframalpha.com/input/?i=%5Cint+1%2F%281-u%5E2%29%5E2
anonymous
  • anonymous
the problem is that i dont know these reduction formulas and the link that @ganeshie8 has given....i am a beginner in integration i only know the formulas given in my book....i wold rather use integration by parts from the beginning than using all such formulas...that is easiest...
anonymous
  • anonymous
the statement in the middle had initiated me...@ganeshie8
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anonymous
  • anonymous

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