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help_with_math

  • 4 years ago

Integration help! integral of sec^5 x dx

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  1. Walleye
    • 4 years ago
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    /barf im guessing u sub will help you here

  2. across
    • 4 years ago
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    Integration by parts!

  3. across
    • 4 years ago
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    Whoops, I misread.

  4. help_with_math
    • 4 years ago
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    yes integration by parts

  5. Roachie
    • 4 years ago
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    (sec^2x)(sec^2)(sec x)

  6. across
    • 4 years ago
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    I read it as sec^5(x)x for a moment there. Hmm, in this case, you have to use a reduction formula.

  7. Walleye
    • 4 years ago
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    red formula will work if you have it

  8. help_with_math
    • 4 years ago
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    well i havent learned taht so is tehre anoterh way to do it by parts

  9. Roachie
    • 4 years ago
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    sec^2x = 1 + tan ^2

  10. Roachie
    • 4 years ago
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    ( 1 + tan ^2 x ) ( 1 + tan^2 x ) (sec x )

  11. across
    • 4 years ago
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    I had to look it up\[\frac{\sin(x)\sec^{m-1}(x)}{m-1}+\frac{m+2}{m-1}\int\sec^{m-2}(x)dx\]

  12. help_with_math
    • 4 years ago
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    this is wat i have so far: \[\int\limits_{}^{}\sec^3 x (\sec^2 dx)\] \[\int\limits_{}^{}\sec^3 x (1+\tan^2 x dx)\] \[\int\limits_{}^{}(\sec^3 x +\sec^3 x \tan^2 x )dx\]

  13. Roachie
    • 4 years ago
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    derive tan x = sec^2x

  14. Roachie
    • 4 years ago
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    so do a u sub with u = tan x , du = sec^2 x dx

  15. help_with_math
    • 4 years ago
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    \[\int\limits_{}^{}(\sec^3 x) + \int\limits_{}^{}(\sec^3 x \tan^2 x) dx\] can i use u sub after this step

  16. Roachie
    • 4 years ago
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    you will prolly end up with a ( 1 + u ) ^(to some power ) then it a simple integration

  17. help_with_math
    • 4 years ago
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    can u show me the steps after the one i just did

  18. help_with_math
    • 4 years ago
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    or wat to do?

  19. Roachie
    • 4 years ago
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    hang on...

  20. Roachie
    • 4 years ago
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    im a little rusty

  21. help_with_math
    • 4 years ago
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    ughh i hate problems liek this!! XD...requires too much work

  22. Roachie
    • 4 years ago
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    yea, it may have to be broken up into parts I cant get rid of one of the sec^x check out: http://www.wolframalpha.com/input/?i=integrate+sec^5+x

  23. Roachie
    • 4 years ago
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    this IS a reduction formula for powers of sec.

  24. help_with_math
    • 4 years ago
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    i did n idont understand it b.c it uses smthng i didnt leard which is teh reduction formula thing

  25. help_with_math
    • 4 years ago
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    then again do u mind tellin me wat that is

  26. help_with_math
    • 4 years ago
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    or how to use it

  27. Roachie
    • 4 years ago
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    yea, your instructor may not accept a reduction formula answer, it the quick way, but there is a trigonometric solution for this.

  28. help_with_math
    • 4 years ago
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    cool i will prob understand that if u explain

  29. help_with_math
    • 4 years ago
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    the trig solution i mean

  30. Roachie
    • 4 years ago
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    there is an identity that will lead to an integral of u. anywa here is a like with the reduction formula for secant, and others http://www.sosmath.com/calculus/integration/moretrigpower/moretrigpower.html

  31. Roachie
    • 4 years ago
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    as a matter of fact I bet your textbook has a table of integrals for integrals of powers of secant

  32. help_with_math
    • 4 years ago
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    what is the dffnce between sex^n (x) and sec^n (ax)

  33. Roachie
    • 4 years ago
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    a=1

  34. help_with_math
    • 4 years ago
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    i wud use the first formula right? sec^5 (x) = sec ^5-2 (x) sec^2(x) = sec^3 (tan^2 x * sec^2 x -1)

  35. help_with_math
    • 4 years ago
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    where do u go on from there

  36. help_with_math
    • 4 years ago
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    omg i havent done definite integrals!

  37. Roachie
    • 4 years ago
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    no, I mistyped

  38. Roachie
    • 4 years ago
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    ∫sec^5dx=1/4sec^3xtanx+3/4∫sec^3xdx

  39. Roachie
    • 4 years ago
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    no definite integrals here

  40. help_with_math
    • 4 years ago
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    hmm how did u get 1/4 sec ...etc

  41. Roachie
    • 4 years ago
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    then apply the reduction formula ( or integrate ) \[\int\limits \sec ^ 3 x dx\]

  42. help_with_math
    • 4 years ago
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    im confused how u got the previous step thou

  43. Roachie
    • 4 years ago
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    its from the reduction formula, it ends with an integral of ∫sec^3xdx

  44. help_with_math
    • 4 years ago
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    i mean teh 1/4 sec part im confused about

  45. Roachie
    • 4 years ago
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    in the reduction formula it 1/(a(n-1) n=5, the power of secant a = 1

  46. help_with_math
    • 4 years ago
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    ok so now how did u get 3/4?

  47. Roachie
    • 4 years ago
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    from the reduction formula: (n-2)/(n-1) again n = 5, the power of the secant

  48. Roachie
    • 4 years ago
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    you instructor will be either impressed or put off that you are using the integral tables. :)

  49. help_with_math
    • 4 years ago
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    lol...he will b impressed hopefully...i have a test 2mm n im trying to udnerstand probs idk

  50. Roachie
    • 4 years ago
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    yea the trigonometric integrals take practice( obviously )

  51. help_with_math
    • 4 years ago
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    srry this is new to me so after taht step wat do u do..u still have a integral of sec^3 dx to deal with

  52. Roachie
    • 4 years ago
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    right! so integrate the sec^x or use the formula again, but this time with n = 3, see?

  53. Roachie
    • 4 years ago
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    sec^x is short, but it indefinite integral has many parts.

  54. help_with_math
    • 4 years ago
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    so after using n=3 do u get 1/2 sec x dx

  55. Roachie
    • 4 years ago
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    right ( n-2)/(n-1) this time the formula will end with \[\int\limits \sec x dx\]

  56. help_with_math
    • 4 years ago
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    wait but isnt is half secant

  57. help_with_math
    • 4 years ago
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    so how is teh integral jsut sec

  58. Roachie
    • 4 years ago
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    no the formula will end ith ∫secxdx, integrate that and you are done.

  59. Roachie
    • 4 years ago
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    \[\int\limits \sec x dx = \ln ( \sec x +tanx ) + C\]

  60. help_with_math
    • 4 years ago
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    so teh final answer is: 1/4 sec^3 x tanx + 3/4 ...idk teh rest

  61. Roachie
    • 4 years ago
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    1/4 sec^3 x tanx + 3/4∫sec^3xdx then apply the formula again for ∫sec^3xdx that answer will end with ∫sec xdx integrate ∫secxdx, done!

  62. help_with_math
    • 4 years ago
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    so the final asnwer is: 1/4 sec^3 x tanx + 3/8 sec x tanx + tn (secx+tanx) + C

  63. help_with_math
    • 4 years ago
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    SRY TAHTS SUPPOOSED TO be ln not tn

  64. Roachie
    • 4 years ago
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    right, distribute the 3/4, I didnt work it all out but it looks right. Those tables are in you textbook right, in the appendix?

  65. help_with_math
    • 4 years ago
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    let me check now

  66. Roachie
    • 4 years ago
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    "table of integrals"

  67. help_with_math
    • 4 years ago
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    under wat heading shud i looj for te hreduction formulas

  68. help_with_math
    • 4 years ago
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    i see basic forms, forms invlving sqrt of (a^2+u^2, forms involving (a^2 - u^2), forms involvung sqrt of (u^2 - a^2), forms involving a+bu, trig forms, inverse trig forms, exp and log forms, hyperbolic forms

  69. Roachie
    • 4 years ago
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    Tables of integrals, Lists of Integrals usually in the appendix, could be on the inside covers.

  70. help_with_math
    • 4 years ago
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    yeah i see them in the appenxix...jsut under wat heading of integrals will i fimd teh reduction formulas

  71. help_with_math
    • 4 years ago
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    or wher u jsut letting me know taht tehre was a table of integrtals

  72. Roachie
    • 4 years ago
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    that really what we were using, an integration formula, so you can integrate by parts or look up the formula.

  73. Roachie
    • 4 years ago
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    there might be one for ∫sec^n (ax)dx

  74. Roachie
    • 4 years ago
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    might not http://integral-table.com/

  75. help_with_math
    • 4 years ago
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    ths is wat is in my bk...sec^n du = 1/n tan u sec^n-2 + (n-2/n-1) *integral of sec^n-2 u du

  76. help_with_math
    • 4 years ago
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    i guess tahts anoterh form of te one u showed me

  77. Roachie
    • 4 years ago
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    right that one dont account for a coefficient on x. so, yea, you seem to have it. Im out. ( gotta brush up on my integrals ;) )

  78. help_with_math
    • 4 years ago
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    THANK U SOOOOO MUCH!!! PLUS THE WEBSITE LOOKS SUPER HELPFUL =)

  79. Roachie
    • 4 years ago
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    peace

  80. help_with_math
    • 4 years ago
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    ii imma go study sum more

  81. help_with_math
    • 4 years ago
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    gnite

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