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  1. J

    Integration problem

    I didn't really take a look at your solution But in order to solve the integral ,, just substitute >>> u = 8 + x^2 it's straight forward ;)
  2. J

    Find the arc length of f(x) (x^(5/4))/5

    just make the integral like this >>> int of (1/4)Sqrt(16+ Sqrt(x)) dx then you can substitute >> u = 16 + sqrt(x) it'll be easy ;)
  3. J

    A bit tricky integral

    Ahaaaa ,, that's entirely awesome :D thanks soooooo much ,, to all of you guys :) now ,, i can go to rest :)
  4. J

    A bit tricky integral

    But i can't factor (x^4)+1 i mean it's (x^4)+1 ,, not (x^4)-1 i don't know how to use partial fractions with that !!
  5. J

    A bit tricky integral

    hey guys ,, i managed to make the integral look like this [2][int{1/(1+(u^4)) du}] would that be any good !!
  6. J

    A bit tricky integral

    at what step exactly?! and how can i do that ?! would you please show me how :) :blushing:
  7. J

    A bit tricky integral

    it tried it ,, it would also give me : (constant) times int{Sqrt(tan w) dw}
  8. J

    A bit tricky integral

    are you sure about the partial fractions? ,, cuz i didn't enjoy any !! i got the Integral of (2(y^2) / (1 + y^4))dy !!
  9. J

    A bit tricky integral

    Umm ,,, i tried what you said ,, i think that will turn the integral back to itself :S except i'll get Integral of Sqrt(tan(x))dx I'm really stuck now :biggrin: Thanks by the way I appreciate it :D and if you get any ideas i'll be glad to try some :)
  10. J

    A bit tricky integral

    I think that won't work ,, bcoz i'll get Integral of (Sqrt(u)/(1+u^2))du then what the next step?!! i also tried integration by parts after the step u mentioned ,, but it just gets more complicated :confused: More Help Is Appreciated :)
  11. J

    Finding the area of a circle using integration

    You can always get the radius from circle equations !! However, all circle equations are integrated by trigonometric substitution and it can also be done by integration by parts but that is a bit tricky!!
  12. J

    A bit tricky integral

    a bit tricky integral!! Homework Statement Integral of (sqrt(cot[x])dx) Homework Equations I just need a hint :) The Attempt at a Solution
  13. J

    Integrating a circle

    I'm really not sure about the difference of the two :S Does each one have a different method of integration? Thanks in advance :)
  14. J

    Integrating a circle

    thanx for the reply :smile: well ,, the thing is that i'm not really good at contour integration ,, i've been searching for a text to study contour integration for ages ,, and still can't find one with good details and examples and still don't know when to use contour integration!! and...
  15. J

    Integrating a circle

    Homework Statement integrating a circle ,, my main question is that, can we integrate it by contour integration technique ? and if yes ,, would you plz show me how :) or just give me a hint :D Thanks is advance :-) Homework Equations y^2 + x^2 = a^2 where a= r suppose that...
  16. J

    Simple Harmonic Motion an Object Attached to a Spring

    Thanks Alot :D you really helped me out and solved my question ,, i appreciate it :) about that side note :P ,, i only did it to get the equations u mentioned :D Happy Holidays :)
  17. J

    Simple Harmonic Motion an Object Attached to a Spring

    your explanation was really more than awesome :) ,, but there's only one thing i didn't really get it, that's when you said : "So let's examine a function of time, A\cos{(\omega t+\phi)}" ,, is it because the drawing of cos(x) function is similar to the oscillation of a spring ,, if so then i...
  18. J

    Simple Harmonic Motion an Object Attached to a Spring

    again and i'm sorry .. if omega = sqrt(k/m) then T= (2(pi))/(sqrt(k/m)) but we studied that T = 2(pi)(sqrt(k/m)) so the sqrt(k/m) is above not at the bottom .. i'm confused NOW :S
  19. J

    Simple Harmonic Motion an Object Attached to a Spring

    alright :) but what does omega represent?!
  20. J

    Simple Harmonic Motion an Object Attached to a Spring

    thanks a lot RoyalCat :) you just got right :D but isn't the left extreme the same as the point the whole system started from when the force of a hand was released? .. remember that the length of an oscillatin of a spring isn't determined by its mass ,, but by the place where you pushed it and...
  21. J

    Simple Harmonic Motion an Object Attached to a Spring

    guys to be honest ,, it's not a simple question at all !! as you may think that it's just a question you may face and solve in a minutes!! you know our teacher said that whoever solve it will get a full mark in all the exams without taking them ,, and since a week nothing happened :D and about...
  22. J

    Simple Harmonic Motion an Object Attached to a Spring

    ok :) i did think like finding the point where they have different velocities .. i just don't know how to begin ,, and how should i draw the velocity as a function of displacement?! thanx any way :D
  23. J

    Simple Harmonic Motion an Object Attached to a Spring

    thanks for the reply :) anyway ,, i think i understood like 50% of what you have said :( The thing is that the two masses (m1 & m2) .. starts accelerating from the left toward the equilibrium position where x=0 then the spring starts to decelerate and its speed will get slower .. from this...
  24. J

    Simple Harmonic Motion an Object Attached to a Spring

    Simple Harmonic Motion .. an Object Attached to a Spring!! Hi everybody :smile: .. i'm glad that i joined this forum so that i can help people and benefit at the same time :) .. i wanted to start with this problem right here ,, and hope that i get the hand from you :) it's a question i've...
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