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roselin

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
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    there tan1/2 t right ?

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

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

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

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

    • one year ago
  6. RadEn
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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
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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
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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
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    agreed ^

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

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

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

    • one year ago
  13. roselin
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    oh okay,

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

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

    • one year ago
  16. roselin
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    okay

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

    • one year ago
  18. roselin
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    ye

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

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

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

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

    • one year ago
  23. roselin
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    I am confused,

    • one year ago
  24. roselin
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    i got it,

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

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

    • one year ago
  27. Kainui
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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
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    so is that ( tan 1/2 t)^2 ?

    • one year ago
  29. RadEn
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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
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    tan of 0= 0 and tan of -pi/4 = -1

    • one year ago
  31. tanjung
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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
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    and where do i put te integral sign ?

    • one year ago
  33. tanjung
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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
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    okay,

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

    • one year ago
  36. roselin
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    options?

    • one year ago
  37. tanjung
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    multiple choices ?

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

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

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

    • one year ago
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