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roselin

  • 2 years ago

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

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  1. RadEn
    • 2 years ago
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    there tan1/2 t right ?

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

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

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

  5. roselin
    • 2 years ago
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    where did you take the half from?

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

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

  8. Kainui
    • 2 years ago
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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.

  9. RadEn
    • 2 years ago
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    agreed ^

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

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

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

  13. roselin
    • 2 years ago
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    oh okay,

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

  15. RadEn
    • 2 years ago
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    with A is a constant

  16. roselin
    • 2 years ago
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    okay

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

  18. roselin
    • 2 years ago
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    ye

  19. RadEn
    • 2 years ago
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    ok, what's the result it ...?

  20. RadEn
    • 2 years ago
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    int 2u du = ... ?

  21. roselin
    • 2 years ago
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    sec^2 i/2 t^2/2

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

  23. roselin
    • 2 years ago
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    I am confused,

  24. roselin
    • 2 years ago
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    i got it,

  25. roselin
    • 2 years ago
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    |dw:1355891028471:dw|

  26. roselin
    • 2 years ago
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    yes, and then?

  27. Kainui
    • 2 years ago
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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".

  28. roselin
    • 2 years ago
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    so is that ( tan 1/2 t)^2 ?

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

  30. roselin
    • 2 years ago
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    tan of 0= 0 and tan of -pi/4 = -1

  31. tanjung
    • 2 years ago
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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 :)

  32. roselin
    • 2 years ago
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    and where do i put te integral sign ?

  33. tanjung
    • 2 years ago
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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)

  34. roselin
    • 2 years ago
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    okay,

  35. tanjung
    • 2 years ago
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    do u have the options ?

  36. roselin
    • 2 years ago
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    options?

  37. tanjung
    • 2 years ago
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    multiple choices ?

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

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

  40. tanjung
    • 2 years ago
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    sorry, i cant do it in wolfram :)

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