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Trigonometric Integral

  1. May 23, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    If [itex]f(x)=cotx+tanx[/itex] then find [itex]\int{f(x)}dx[/itex]

    3. The attempt at a solution
    I was able to manipulate (not sure if in the right direction though) the function and resulted with [tex]f(x)=\frac{2}{sin(2x)}[/tex]

    I've also tried re-arranging things in a different way, but came up with nothing useful (I think). So then I'm stuck, any ideas?
     
  2. jcsd
  3. May 23, 2009 #2

    danago

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    You could possibly use the fact that (sin x)' = cos x, along with the idea that (ln f(x))' = f'(x) / f(x).

    Much easier to integrate term by term rather than combine the expressions into a single trig term.
     
  4. May 23, 2009 #3

    Mentallic

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    I'm not getting any form of f'(x)/f(x) and substituting u=sinx and du/dx=cosx doesn't seem like it gave me much either...
     
  5. May 23, 2009 #4

    danago

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    Look at the cot x term --

    cot x = cos x / sin x, which is exactly of the form f'(x) / f(x), since (sin x)' = cos x. A very similar idea applies to the tan x term.
     
  6. May 23, 2009 #5

    HallsofIvy

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    That wasn't what was suggested.

    [tex]cot(x)+ tan(x)= \frac{cos(x)}{sin(x)}+ \frac{sin(x)}{cos(x)}[/tex]

    Integrate by letting u= sin(x) in the first fraction and v= cos(x) in the second.
     
  7. May 23, 2009 #6

    Mentallic

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    Ahh thanks for all the help!

    Sorry, I kept looking at my manipulated expression.
     
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