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Solve x'=sinx

  1. Sep 14, 2008 #1
    i'm having trouble with this.

    i first need to solve x'=sinx. the answer is given as t=ln|(csc(x0)+cot(x0))/(csc(x)+cot(x))|]

    i'm not sure where those x0's came in from.

    here's what i did:

    Int(1/sinx)dx=Int(1)dt
    Int(csc(x))dx=Int(1)dt
    Int((-1)[(-1)csc(x)(cot(x)+csc(x))]/(cot(x)+csc(x)))dx=Int(1)dt (this is the same as the line above but then lets you change Int(f'(x)/f(x)) to ln(f(x)))
    -ln(cot(x)+csc(x))=t

    so as you can see this answer isn't the same as the answer that was provided. what am i doing wrong and where do those x0's come in from?

    any help is appreciated.
     
  2. jcsd
  3. Sep 14, 2008 #2

    HallsofIvy

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    Re: x'=sinx

    What happened to your constant of integration?

    If [itex]\int f(x)dx= \int g(x)dx[/itex] then the most you can say is that f(x)= g(x)+ C where C is some constant- and if you are calling your function "x" then "x0" is a good a name as "C".

    Your last line should be -ln|cos(x)+ csc(x)|+ C= t. Of course, - ln|cos(x)+ csc(x)|= ln|1/(cos(x)+ csc(x))| and, of course, - ln|cos(x)+ csc(x)|+ C= ln|1/(cos(x)+ csc(x))|+ C= ln|D/(cos(x)+ csc(x)| where D= eC is just another constant. The numerator above, csc(x0)+cot(x0), is just a number like "D". I presume they have some reason for writing it that way but it is exactly the same thing.
     
  4. Sep 14, 2008 #3
    Re: x'=sinx

    ahhhhh. i see. thank you.
     
  5. Sep 14, 2008 #4
    Re: x'=sinx

    my next problem is that it says show that for x0=pi/4 you can solve x=2arctan((e^t)/(1+sqrt(2))

    here's what i have so far, working from the previous equation:

    t=ln|((csc(pi/4)+cot(pi/4))/(csc(x)+cot(x))|
    t=ln|((2/sqrt(2)+1)/(csc(x)+cot(x))|
    e^t=(2/sqrt(2)+1)/(csc(x)+cot(x))
    csc(x)+cot(x)=(2/sqrt(2)+1)/e^t

    then i'm stuck. i don't see how to isolate x and where an arctan would come from.

    thank you in advance.
     
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