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anonymous

  • one year ago

need help in the derivation of integration formula:

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    i didn't know there was such a formula

  3. anonymous
    • one year ago
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    \[\int\limits_{.}^{.} \sqrt{x^{2}-a^{2}}dx =\frac{x }{ 2 }\sqrt{x ^{2}-a ^{2}}-\frac{ a ^{2} }{ 2 }\log \left| x+\sqrt{x ^{2}+a^2} \right|+C\]

  4. anonymous
    • one year ago
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    solve by taking x=a sect

  5. anonymous
    • one year ago
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    no problem @misty1212

  6. misty1212
    • one year ago
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    this one is a pain in the neck i think, but the sub is right

  7. misty1212
    • one year ago
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    you are going to get \[a^2\int \sec^3(u)-\sec(u)du\]

  8. misty1212
    • one year ago
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    did you get to that part, or no?

  9. anonymous
    • one year ago
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    yes

  10. anonymous
    • one year ago
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    i dont know what to do after that.....

  11. misty1212
    • one year ago
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    this is the kind of thing you look up in the back of the text because it is boring beyond belief

  12. anonymous
    • one year ago
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    can u go after that

  13. misty1212
    • one year ago
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    there is a "reduction formula" for secant

  14. misty1212
    • one year ago
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    are you allowed to use it?

  15. anonymous
    • one year ago
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    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.....

  16. anonymous
    • one year ago
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    what is reduction formula

  17. misty1212
    • one year ago
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    you mean in general or "what is the reduction formula for \(\int \sec^n(x)dx\)?

  18. anonymous
    • one year ago
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    then what is that?

  19. misty1212
    • one year ago
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    here are a bunch of them http://archives.math.utk.edu/visual.calculus/4/recursion.2/

  20. misty1212
    • one year ago
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    in your case \(n=3\) so you get \[\int sec^3(u)du=\frac{1}{2}\tan(u)\sec(u)+\frac{1}{2}\int \sec(u)du\]

  21. anonymous
    • one year ago
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    hav u done it by taking integration by parts?

  22. misty1212
    • one year ago
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    it looks like that right? but it is not it comes from doing some trig business and a u sub

  23. misty1212
    • one year ago
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    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

  24. misty1212
    • one year ago
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    oh to answer your question, no i used nothing but the"reduction formula" with \(n=3\) the one i sent the link to

  25. misty1212
    • one year ago
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    ignoring the annoying \(a^2\) out front, and combining like terms, we should be at \[\frac{a^2}{2}\sec(u)tan(u)-\frac{a^2}{2}\int \sec(u)du\]

  26. anonymous
    • one year ago
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    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...

  27. misty1212
    • one year ago
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    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"

  28. anonymous
    • one year ago
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    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...

  29. misty1212
    • one year ago
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    i think you use parts for \(\int \sec^3(x)dx\) but not for \(\int \sec(x)dx\)

  30. misty1212
    • one year ago
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    it is just something they do to make a formula is all some people like formulas

  31. ganeshie8
    • one year ago
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    if you like partial fractions, \[\sec x =\dfrac{\cos x}{\cos^2x} = \dfrac{\cos x}{1-\sin^2x}\]

  32. misty1212
    • one year ago
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    oh cool, lots easier than the un-intuitive multiplying top and bottom business

  33. anonymous
    • one year ago
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    please give a more detail where to start from....@ganeshie8

  34. ganeshie8
    • one year ago
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    same can be extended to \(\sec^3x\) too i think \[\sec^3 x =\dfrac{\cos x}{\cos^4x} = \dfrac{\cos x}{(1-\sin^2x)^2}\]

  35. misty1212
    • one year ago
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    @ganeshie8 is proving \[\int\sec(x)dx=\ln(\sec(x)+\tan(x))\]

  36. anonymous
    • one year ago
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    i dont know how to integrate that partial fraction....@ganeshie8

  37. misty1212
    • one year ago
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    \[\int \frac{\cos(x)}{1-\sin^2(x)}dx\] put \(u=\sin(x)\) and integrate \[-\int \frac{du}{1-u^2}\] using partial fractions

  38. misty1212
    • one year ago
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    @ganeshie8 that is correct yes? never saw it done this way

  39. anonymous
    • one year ago
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    oh yes i got that...

  40. ganeshie8
    • one year ago
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    looks good to me !

  41. ganeshie8
    • one year ago
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    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

  42. anonymous
    • one year ago
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    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...

  43. anonymous
    • one year ago
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    the statement in the middle had initiated me...@ganeshie8

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  44. anonymous
    • one year ago
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    @ganeshie8

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