# Trigonometry integration

1. Jul 1, 2006

### Orion1

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?

$$\int \sin(\cos x) dx$$

Last edited: Jul 1, 2006
2. Jul 1, 2006

### HallsofIvy

Staff Emeritus
It cannot be integrated in terms of elementary functions.

3. Jul 7, 2006

### Plastic Photon

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?

4. Jul 8, 2006

### d_leet

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.

5. Jul 8, 2006

### Robokapp

integral of sin(cos(x)) well i have a fancy calculator that says...

I take that as a no-no.

6. Jul 8, 2006

### Orion1

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

$$\int_a^b \sin(\cos x) dx = \lim_{n \rightarrow \infty} \sum_{k=1}^n \sin(\cos x_k) \Delta x_k$$

Last edited: Jul 8, 2006
7. Jul 8, 2006

### matt grime

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.'

8. Jul 13, 2006

### Orion1

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

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

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

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

$$\int_a^b \sin (\cos x) dx = \sin (\sin b) - \sin (\sin a)$$

9. Jul 13, 2006

### arildno

It doesn't bother you the least that what you've written is totally wrong?

Last edited: Jul 14, 2006
10. Jul 13, 2006

### shmoe

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.

11. Jul 15, 2006

### Orion1

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)?
$$f(x) = \sin(\cos x)$$
$$\int \sin(\cos x) dx = F(x) + C$$

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
12. Jul 16, 2006

### TD

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
13. Jul 16, 2006

### Integral

Staff Emeritus
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?

14. Jul 17, 2006

### Orion1

Last edited: Jul 17, 2006
15. Jul 17, 2006

### Office_Shredder

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

Orion, this is wrong:

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

The second summation should be inside of the first, not simply multiplying the first.

16. Jul 22, 2006

### Orion1

$$\sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n + 1}$$

$$\cos x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}$$

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

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

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

Reference:
http://en.wikipedia.org/wiki/Taylor_series

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

### HallsofIvy

Staff Emeritus
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.

18. Jul 23, 2006

### matt grime

Stop using the same letter for all summation indices.

Last edited: Jul 23, 2006
19. Feb 9, 2007

### xerobeatz

integration

hi, im new here, can i ask how 2 integrate "tan x"

20. Feb 9, 2007

### bomba923

Welcome to PF

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

Generally, such questions belong in this subforum :)

Last edited: Feb 9, 2007