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Trigonometric Integral

  • Thread starter Mentallic
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  • #1
Mentallic
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Homework Statement


If [itex]f(x)=cotx+tanx[/itex] then find [itex]\int{f(x)}dx[/itex]

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?
 

Answers and Replies

  • #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.
 
  • #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...
 
  • #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.
 
  • #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.
 
  • #6
Mentallic
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Ahh thanks for all the help!

Sorry, I kept looking at my manipulated expression.
 

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