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anonymous

  • 5 years ago

how do i put ln[ sectheta + tantheta] in terms of x

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  1. anonymous
    • 5 years ago
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    a. There's no "x" in this expression. b. In order to put it in terms of a variable; it has to be in an equation form.

  2. anonymous
    • 5 years ago
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    the ogrinal equation from solving it was intergal of dx/1-x^2 if that helps

  3. anonymous
    • 5 years ago
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    true, isn't there a given x np?

  4. anonymous
    • 5 years ago
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    HaHaHA (breaking down due to extreme lough)

  5. anonymous
    • 5 years ago
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    lol, what's so funny angoo?

  6. anonymous
    • 5 years ago
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    I don't know

  7. anonymous
    • 5 years ago
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    Ahhh, so you used a trig substitution. Now, go back to whatever value you used for x to solve the integral (in terms of theta), solve for sec(theta) and tan(theta) by drawing out the triangles, and put those values - in terms of x - in your final expression.

  8. anonymous
    • 5 years ago
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    >_> seriously AG

  9. anonymous
    • 5 years ago
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    just wrote that to make you smile

  10. anonymous
    • 5 years ago
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    did you understand what Quantum said np65? ^_^

  11. anonymous
    • 5 years ago
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    and it made me smile :)

  12. anonymous
    • 5 years ago
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    so i have the triangle would that put me at 1-x^2+x then?

  13. anonymous
    • 5 years ago
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    for the answer

  14. anonymous
    • 5 years ago
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    what was your value of x when you integrated?

  15. anonymous
    • 5 years ago
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    What trig substitution for x did you use to solve the integral?

  16. anonymous
    • 5 years ago
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    i set x=sin

  17. anonymous
    • 5 years ago
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    how did you set x = sin?

  18. anonymous
    • 5 years ago
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    alright i have the intergal of dx1-x^2 i set u =sintheta and du = costheta dtheta i then subed in u and du

  19. anonymous
    • 5 years ago
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    following that i got dtheta/costheta

  20. anonymous
    • 5 years ago
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    oh I got it!\[x = \sin \theta\] and you know that sin x = opposite/hyp your opposite in this case = x hyp = 1 draw the triangle and you'll be able to continue from here ^_^

  21. anonymous
    • 5 years ago
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    hence :\[\sec \theta= 1/\cos \theta\] \[\tan \theta = \sin \theta/ \cos \theta\] you have sin (theta), now find cos (theta) :) is it clear now?

  22. anonymous
    • 5 years ago
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    so then the other side would be 1-x^2 or use the p-thag therom to get the last side making it the sqrroot(-x^2+1)

  23. anonymous
    • 5 years ago
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    use the pyth.theorem :) that's it. you just don't have x, find it by applying the theorem , then find cos (theta)

  24. anonymous
    • 5 years ago
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    so the last side will be : \[\sqrt{1- x^2}\] ^_^

  25. anonymous
    • 5 years ago
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    awesome got it now

  26. anonymous
    • 5 years ago
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    thanks

  27. anonymous
    • 5 years ago
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    np :)

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