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roselin Group Title

definite integral from -pi/2 to 0 (2+tan1/2)sec^2 t/2 dt

  • one year ago
  • one year ago

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  1. RadEn Group Title
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    there tan1/2 t right ?

    • one year ago
  2. roselin Group Title
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    sorry, it is actually tan t/2

    • one year ago
  3. RadEn Group Title
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    ok, use by integral substitution u = tan 1/2 t du = .... ?

    • one year ago
  4. RadEn Group Title
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    opss.. u = 2+tan1/2 t, i meant :)

    • one year ago
  5. roselin Group Title
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    where did you take the half from?

    • one year ago
  6. RadEn Group Title
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    u = 2 + tan1/2 t du = 0 + 1/2 sec^2 1/2 t dt right ?

    • one year ago
  7. roselin Group Title
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    definite integral from -pi/2 to 0 (2+tan t/2)sec^2 t/2 dt

    • one year ago
  8. Kainui Group Title
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    When deciding what to make your substitution with, you should consider what in the integral has its derivative in there too. This sort of is the process of reversing the chain rule. In this case you should remember that the derivative of tangent is secant squared.

    • one year ago
  9. RadEn Group Title
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    agreed ^

    • one year ago
  10. roselin Group Title
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    so where did u get that 1/2 from? I did not get it, sorry

    • one year ago
  11. Kainui Group Title
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    \[\frac{ t }{ 2 }=\frac{ 1 }{ 2 }t\]

    • one year ago
  12. Kainui Group Title
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    In the same way 4/2 is equal to 4*(1/2)

    • one year ago
  13. roselin Group Title
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    oh okay,

    • one year ago
  14. RadEn Group Title
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    by using the chain rule : if y=tanAx then y'=A*sec^2 Ax

    • one year ago
  15. RadEn Group Title
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    with A is a constant

    • one year ago
  16. roselin Group Title
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    okay

    • one year ago
  17. RadEn Group Title
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    so, the integration will be int 2u du right ?

    • one year ago
  18. roselin Group Title
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    ye

    • one year ago
  19. RadEn Group Title
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    ok, what's the result it ...?

    • one year ago
  20. RadEn Group Title
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    int 2u du = ... ?

    • one year ago
  21. roselin Group Title
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    sec^2 i/2 t^2/2

    • one year ago
  22. RadEn Group Title
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    int (2u du) = u^2 right ?

    • one year ago
  23. roselin Group Title
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    I am confused,

    • one year ago
  24. roselin Group Title
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    i got it,

    • one year ago
  25. roselin Group Title
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    |dw:1355891028471:dw|

    • one year ago
  26. roselin Group Title
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    yes, and then?

    • one year ago
  27. Kainui Group Title
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    Assuming this is right, you need to remember that the problem you were given to solve had no "u" in it at all, so you must plug in "u" back to the equation to get it in terms of "x".

    • one year ago
  28. roselin Group Title
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    so is that ( tan 1/2 t)^2 ?

    • one year ago
  29. RadEn Group Title
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    (2 + tan 1/2 t)^2 now, plug and calculate for t = 0 and t =-pi/2

    • one year ago
  30. roselin Group Title
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    tan of 0= 0 and tan of -pi/4 = -1

    • one year ago
  31. tanjung Group Title
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    (2 + tan 1/2 (0))^2 - (2 + tan 1/2 (-pi/2))^2 = (2+tan(0))^2 - (2+tan(-pi/4)^2 = (2+0)^2 - (2 - tan(pi/4))^2 = 4 - (2 - 1)^2 = 4 - 1 = 3 if i dont make mistake :)

    • one year ago
  32. roselin Group Title
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    and where do i put te integral sign ?

    • one year ago
  33. tanjung Group Title
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    just remember that int (f(x)) dx [a,b] = [F(x)] [a,b] = F(b)-F(a) after u integrate it, u got (2 + tan 1/2 t)^2 right ? now, substitute for highest interval (t=0) and then subtract with for t=-pi/4 (low interval)

    • one year ago
  34. roselin Group Title
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    okay,

    • one year ago
  35. tanjung Group Title
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    do u have the options ?

    • one year ago
  36. roselin Group Title
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    options?

    • one year ago
  37. tanjung Group Title
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    multiple choices ?

    • one year ago
  38. roselin Group Title
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    no, but i used the wolfram and shows the ans is 5

    • one year ago
  39. roselin Group Title
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    i have another question, how would I put this problem in wolframaalpha? |dw:1355919813662:dw|

    • one year ago
  40. tanjung Group Title
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    sorry, i cant do it in wolfram :)

    • one year ago
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