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Trigonometry integration

  1. Jul 1, 2006 #1

    Is it possible to integrate a trigonometry function within a function, or is this a formula that cannot be integrated in terms of elementary functions?

    [tex]\int \sin(\cos x) dx[/tex]
    Last edited: Jul 1, 2006
  2. jcsd
  3. Jul 1, 2006 #2


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    It cannot be integrated in terms of elementary functions.
  4. Jul 7, 2006 #3
    just to clear up

    If sin(x)=y and arcsin(y)=x , why can sin(arcsiny) not be integrated with elementary function? It is equivalent to integrating sin(x) which is -cosx + C, correct? So why is sin(cosx) any different, does this not apply to inverse trig functions?
  5. Jul 8, 2006 #4
    Actually sin(arcsin(y)) = y in the interval from -pi/2 to pi/2, and so in that interval it would be equivalent to integrating y which would be y2/2 + C. sin(cos(x)) is a very different stroy however, and as has already been stated in this thread cannot be integrated in terms of elementary functions.
  6. Jul 8, 2006 #5
    integral of sin(cos(x)) well i have a fancy calculator that says...

    Error. Sorry, Wcalc is unable to complete your requested computation. Please verify your data.

    I take that as a no-no.
  7. Jul 8, 2006 #6

    Is it possible to solve the area under the curve using the Riemann sum?
    [tex]\int_a^b f(x) dx = \lim_{n \rightarrow \infty} \sum_{k=1}^n f(x_k) \Delta x_k[/tex]

    [tex]\int_a^b \sin(\cos x) dx = \lim_{n \rightarrow \infty} \sum_{k=1}^n \sin(\cos x_k) \Delta x_k[/tex]
    Last edited: Jul 8, 2006
  8. Jul 8, 2006 #7

    matt grime

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    If by solve you mean 'numerically approximate' then obviously the answer is yes. The function is continuous and Riemann integrable. If you mean is it possible to find the exact answer in some explicit form (where explicit is to be determined by your own opinion) between any two limits then the answer is 'no, not for all a and b under most definitions of explicit.'
  9. Jul 13, 2006 #8

    Here is the theorem that I developed for any trig function within a function:

    [tex]\int u[v(x)] dx = u [ \int v(x) dx ] + C[/tex]

    [tex]\int \sin (\cos x) dx = \sin [ \int \cos x \; dx ] = \sin (\sin x) + C[/tex]

    [tex]\int_a^b u[v(x)] dx = u [ \int v(x) dx ]_b - u [ \int v(x) dx ]_a[/tex]

    [tex]\int_a^b \sin (\cos x) dx = \sin (\sin b) - \sin (\sin a)[/tex]
  10. Jul 13, 2006 #9


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    It doesn't bother you the least that what you've written is totally wrong?
    Last edited: Jul 14, 2006
  11. Jul 13, 2006 #10


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    It might be better if you don't call this a theorem. Try a few examples where u and v are polynomials and you should see how very, very wrong this is.
  12. Jul 15, 2006 #11

    I retract my term 'theorem', perhaps I should have used the term pseudo-developmental philosophy...

    Please post a graph of what F(x) appears like versus f(x)?
    [tex]f(x) = \sin(\cos x)[/tex]
    [tex]\int \sin(\cos x) dx = F(x) + C[/tex]

    I would post my graphs for f(x) = Sin (Cos x) and F(x) = Sin (Sin x), however these functions have been refuted here.

    The graph for F(x) can be drawn or graphed manually, can it not?
    Last edited: Jul 15, 2006
  13. Jul 16, 2006 #12


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    In blue, f(x); in green, a possible F(x) with c taken 0.

    http://img132.imageshack.us/img132/1402/intle6.jpg" [Broken]
    Last edited by a moderator: May 2, 2017
  14. Jul 16, 2006 #13


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    What do you mean by "refuted"? Did you read Matt Grimes post? Perhaps if you did not understand it you should make an effort to. What is means is that your function is fine, and an integral exists. But as others have said the integral cannot be expressed in terms of elementary functions.

    Consider this for small x; Sin(x) = x. Now look at the first terms of Taylor series expansion of cos(x):
    Cos(x) ~ x + (x^2)/2 + ...

    You can see that your function will differ from Sin(x) only in when Cos(x) is large. These functions can easily be plotted and it indeed tracks sin(x) very nicely except that the maximum values.

    Now again Where do you see that anyone has "refuted" these functions?
  15. Jul 17, 2006 #14
    Last edited: Jul 17, 2006
  16. Jul 17, 2006 #15


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    If by Cos(x) ~ x you mean Cos(x) ~ 1, then sure.

    Orion, this is wrong:

    [tex]\sin (\cos x) = \left(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \right) \left( \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n} \right)^{2n + 1}[/tex]

    The second summation should be inside of the first, not simply multiplying the first.
  17. Jul 22, 2006 #16
    [tex]\sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n + 1}[/tex]

    [tex]\cos x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}[/tex]

    [tex]\sin (\cos x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \left( \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n} \right)^{2n + 1}[/tex]

    [tex]\int \sin ( \cos x) dx = \int \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \left( \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n} \right)^{2n + 1} dx[/tex]

    [Color = Blue]
    Would it require a full page of proof in order to prove this equation?

    Last edited: Jul 23, 2006
  18. Jul 23, 2006 #17


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    To prove what? You've just replaced the trig functions with their Taylor's series. What is there to prove? Of course, you still haven't done the integral which was the whole point. Now, if you had moved the integral inside the sums, that would have been something to prove- but you wouldn't need a whole sum- just note that power series converge uniformly inside their radius of convergence and that the power series for sine and cosine have infinite radius of convergence.
  19. Jul 23, 2006 #18

    matt grime

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    Stop using the same letter for all summation indices.
    Last edited: Jul 23, 2006
  20. Feb 9, 2007 #19

    hi, im new here, can i ask how 2 integrate "tan x"
  21. Feb 9, 2007 #20
    Welcome to PF

    [tex]\int {\tan x\,dx} = \int {\frac{{\sin x}}
    {{\cos x}}\,dx} = - \int {\frac{{d\left( {\cos x} \right)}}
    {{\cos x}}} = - \log \left| {\cos x} \right| + C = \log \left| {\sec x} \right| + C[/tex]

    Generally, such questions belong in this subforum :)
    Last edited: Feb 9, 2007
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