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appleduardo

  • one year ago

what is the integral of the following function? --> sec^4 (2x+1)dx I hope somebody can help me out.

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  1. appleduardo
    • one year ago
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    \[\int\limits_{}^{}\sec^4(2x+1)dx\]

  2. appleduardo
    • one year ago
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    so far ive done this--> \[\int\limits_{}^{}[1+\tan^2(2x)] [\sec^2(2x+1)]\]

  3. appleduardo
    • one year ago
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    so then I get:\[\int\limits_{}^{}\sec^2(2x+1) +\int\limits_{}^{}\tan^2(2x+1)\sec^2(2x+1)\]

  4. appleduardo
    • one year ago
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    what is the next step? could somebody help me?

  5. mathsmind
    • one year ago
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    Use the reduction formula for m=4 \[\int\limits \sec^m(x)dx=\frac{\sin(x)\sec^m-1(x)}{m-1}+\frac{m-2}{m-1}\int\limits \sec^{m-2}(x)dx\]

  6. mathsmind
    • one year ago
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    now whatever answer u get multiply by 1/2 because \[u =2x+1 \Longrightarrow du=2dx \longrightarrow dx=\frac{1}{2}du\]

  7. mathsmind
    • one year ago
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    \[\frac{1}{2}\int\limits\limits \sec^m(u)du=\frac{\sin(u)\sec^{m-1}(u)}{m-1}+\frac{m-2}{m-1}\int\limits\limits \sec^{m-2}(u)du\]

  8. mathsmind
    • one year ago
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    now let m=4

  9. mathsmind
    • one year ago
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    \[\frac{1}{2}\int\limits\limits\limits\limits \sec^4(u)du=\frac{\sin(u)\sec^{4-1}(u)}{4-1}+\frac{4-2}{4-1}\int\limits\limits\limits\limits \sec^{4-2}(u)du\]

  10. mathsmind
    • one year ago
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    \[=\frac{1}{2}[\frac{\sin(u)\sec^{3}(u)}{3}+\frac{2}{3}\int\limits\limits\limits\limits \sec^{2}(u)du]\]

  11. mathsmind
    • one year ago
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    \[=\frac{1}{6} \tan(u)\sec^2(x)+\frac{1}{3}\int\limits\limits \sec^2(u)du\]

  12. mathsmind
    • one year ago
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    \[=\frac{1}{6} \tan(u)\sec^2(u)+\frac{1}{3}\tan^2(u)+c\]

  13. mathsmind
    • one year ago
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    u =2x+1

  14. mathsmind
    • one year ago
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    \[=\frac{1}{6} \tan(2x+1)\sec^2(2x+1)+\frac{1}{3}\tan^2(2x+1)+c\]

  15. mathsmind
    • one year ago
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    now using the reduction formula is useful because it reduces the amount of repetition of integration by parts, and simplifies the solution with less probability of errors...

  16. mathsmind
    • one year ago
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    An alternative way to solve this problem

  17. mathsmind
    • one year ago
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    there are many trig identities that can help in solving this problem one way is:

  18. mathsmind
    • one year ago
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    \[\int\limits \sec^2(u)\sec^2(u)du=\frac{1}{2}\int\limits \sec^2(u)(1+\tan^2(u))du\]

  19. mathsmind
    • one year ago
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    expand the function and distribute the integral

  20. mathsmind
    • one year ago
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    \[=\frac{1}{2}\int\limits \sec^2(u)du + \frac{1}{2} \int\limits \sec^2(u)\tan^2(u)du\]

  21. mathsmind
    • one year ago
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    \[=\frac{1}{2}\tan(u) +\frac{1}{6} \tan^3(u)+c\]

  22. mathsmind
    • one year ago
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    \[=\frac{1}{2}\tan(2x+1) +\frac{1}{6} \tan^3(2x+1)+c\]

  23. mathsmind
    • one year ago
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    if u think that the answers are different in the two methods, then that is not the case they are both the same u may use the trig identities to justify both solutions...

  24. appleduardo
    • one year ago
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    thank you very much mathsmind! i was kind of stressed, :/ but i have one queation:how did u get tan^3 ??in the final result.

  25. mathsmind
    • one year ago
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    ur welcome! i was expecting ur question i did that on purpose...

  26. mathsmind
    • one year ago
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    now there are two ways to think about this...

  27. mathsmind
    • one year ago
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    when u see a function with its derivative then u use the normal 1/(n+1tan^(n+1) like we do with 1/(n+1)x^(n+1)

  28. mathsmind
    • one year ago
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    the other way to visualize this concept is by assuming that tan(u) =v

  29. mathsmind
    • one year ago
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    dv = sec^2(u)du -> du=dv/sec^(u)

  30. mathsmind
    • one year ago
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    \[\int\limits \tan^2(u)\sec^2(u)du=\int\limits v^2\sec^2(u)\frac{dv}{\sec^2(u)}\]

  31. mathsmind
    • one year ago
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    now can u see the full picture...

  32. mathsmind
    • one year ago
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    (v^3)/3+c

  33. mathsmind
    • one year ago
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    but v=tan(u)

  34. mathsmind
    • one year ago
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    =(1/3)tan^3(u)+c, where u = 2x+1

  35. mathsmind
    • one year ago
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    hope this clarifies the point...

  36. mathsmind
    • one year ago
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    are u there?

  37. appleduardo
    • one year ago
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    yeep i am still here! i am trying to understand what you just typed! :P. uhmm i have another question, uhmm my teacher once told me that we can only use the reduction formula when "n" isnt a pair number, and in this case n=4, so uhmm is it right to solve it with that formula anyways?

  38. mathsmind
    • one year ago
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    Your teacher must have gone into a detailed answer, what happens is u need to consider if the function is an even one or an odd one, so because sec is an even function then u can use an even power...

  39. mathsmind
    • one year ago
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    So u can correct the information for ur teacher, if u think this is what she/he meant...

  40. mathsmind
    • one year ago
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    but anyways i mean if u r not familiar with the even and odd functions then use the 2nd method...

  41. mathsmind
    • one year ago
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    by the way this problem can be solved in many ways...

  42. appleduardo
    • one year ago
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    uhmm yeah i think i have to understand the second method first. i'll have to read it once again

  43. mathsmind
    • one year ago
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    just write it down on a piece of paper, you'll see how easy it is...

  44. appleduardo
    • one year ago
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    omG i just did and i understood it all! :D thank you very much mathsmind!, now id like to understand the reduction method, so uhmm what does "even" and "odd" fuctions mean ..?

  45. mathsmind
    • one year ago
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    if f(x) = (f-x) then that function is said to be even, e.g. x^2, x^4, x^6...etc or you may say x^n, where n = even number, also cos(x) is an even function, but u have to be careful u always need to perform a test and not make a mistake because for example (x+4)^2 is not an even function because \[f(x) \neq f(-x)\]

  46. mathsmind
    • one year ago
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    An odd function on the other hand is defined as -f(x)=f(-x), such as x^3, x^5, x^7, and sin(x) is also an odd function but x^3+1 is not an odd function...

  47. mathsmind
    • one year ago
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    one of the easiest to test if the function is odd or even is visualizing the graph, if the graph is symmetrical about the y-axis then its an even function, but if the graph is symmetrical about y=x then its an odd function, in other words symmetrical about the origin, however there are cases where the functions are neither odd or even...

  48. appleduardo
    • one year ago
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    :o cool!! :D but when i want to make use of the reduction formula.. is the same formula i have to use for an "odd" and an "even" function?

  49. mathsmind
    • one year ago
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    nope they are different and depends, it could be confusing if u don't have experience, i would say try solving the elementary way...

  50. appleduardo
    • one year ago
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    wow there's a lot of things to consider then and thank you so much for all your support. :P i have another question. When you solved this integral by the elementary way, you said tan^3 was the result of 1/(n+1)tan^(n+1) ... but why you did that? i mean, in order to get ...what??

  51. mathsmind
    • one year ago
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    that was the integration

  52. mathsmind
    • one year ago
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    the integral of tan^2(u)sec^(u)du

  53. mathsmind
    • one year ago
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    now when u differentiate tan u get sec

  54. mathsmind
    • one year ago
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    let me show u

  55. mathsmind
    • one year ago
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    \[\frac{1}{2} \int\limits\limits \sec^2(u)\tan^2(u)du\]

  56. mathsmind
    • one year ago
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    let \[v = \tan(u) \therefore dv = \sec^2(u)du, so \space du=\frac{dv}{\sec^2(u)}\]

  57. mathsmind
    • one year ago
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    \[= \frac{1}{2} \int\limits\limits \sec^2(u)v^2\frac{dv}{\sec^2(u)}\]

  58. mathsmind
    • one year ago
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    can u see how sec^2(u) is cancelled so u are left with

  59. mathsmind
    • one year ago
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    \[\frac{1}{2} \int\limits v^2dv\]

  60. mathsmind
    • one year ago
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    and u know the rest, or do u want me to finish it to the end...

  61. appleduardo
    • one year ago
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    yeep its alright! :D thank you! but if tan was originally tan^2 and then you took just tan to be your "u" function, it doesnt change the final result??

  62. mathsmind
    • one year ago
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    yes tan(u)=v therefore tan^2(u)=v^2

  63. mathsmind
    • one year ago
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    is that what u meant? or i misunderstood?

  64. appleduardo
    • one year ago
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    yeep! :D thanks, so if tan^2 were tan^3 then "V" would equal to v^3 in the final result?

  65. mathsmind
    • one year ago
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    well in the case of the question above v^4/4+c

  66. mathsmind
    • one year ago
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    tan^4(u)/4 +c

  67. mathsmind
    • one year ago
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    v^99

  68. mathsmind
    • one year ago
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    tan^100(u)/100+c

  69. mathsmind
    • one year ago
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    if u did not have the sec^2(u), then you can't do that...

  70. mathsmind
    • one year ago
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    that's what i meant of the function multiplied by its derivative

  71. mathsmind
    • one year ago
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    if i tell u to integrate sin(x)cos(x), what answer would i get?

  72. appleduardo
    • one year ago
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    i think i'd use -->:\[\int\limits_{}^{}sen udu\] and since u=sen, and du=cos, then :\[-\cos(x)+c\]

  73. appleduardo
    • one year ago
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    is that correct??

  74. mathsmind
    • one year ago
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    u mean -(1/2)cos^2(x)+c

  75. mathsmind
    • one year ago
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    this is the whole point always when u integrate check if the function has its derivative...

  76. mathsmind
    • one year ago
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    what is the integral of sin(x)

  77. appleduardo
    • one year ago
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    its: -cos (u) +c :D

  78. mathsmind
    • one year ago
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    and what result did u get for sin(x)cos(x) again?

  79. appleduardo
    • one year ago
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    uhmm i thought the result was −cos(x)+c , but if i make use of \[\frac{ 1 }{( n+1)(\cos ^{n+1})} =\frac{ 1 }{ 2\cos^2 }\] , then i'd get:\[-\frac{ 1 }{ 2 }\cos^2(x)+c\]

  80. mathsmind
    • one year ago
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    no never do that

  81. mathsmind
    • one year ago
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    a rcipical of cos never gives u a minus sign be careful never make such a mistake !

  82. mathsmind
    • one year ago
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    \[\frac{1}{2}\frac{1}{\cos^2(x)}=\frac{1}{2}(\cos(x))^{-2}\]

  83. appleduardo
    • one year ago
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    :o oohh, so is that the final result? +c

  84. mathsmind
    • one year ago
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    no that is the result of what you thought it was the result which is not the result in this case

  85. mathsmind
    • one year ago
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    \[\int\limits \sin(x)\cos(x)dx = -\frac{1}{2}\cos^2(x)+c\]

  86. appleduardo
    • one year ago
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    but how did u get the 1/2 and cos^2 ?

  87. mathsmind
    • one year ago
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    use the u sub and check for urself

  88. appleduardo
    • one year ago
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    yeep thanks! i'll try it :D :D

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