# Yet another interesting integral

1. Mar 5, 2005

### msmith12

this integral was on a test I took recently--any thoughts or ideas on how to attack it?

$$\int_{0}^{1} \frac{x\sin{x}}{1+\cos^2{x}}dx$$

~matt

2. Mar 5, 2005

### PBRMEASAP

Dern, that is a nasty 'un. I can't find an antiderivative for it, so I'm hoping you don't need to. Is it acceptable to express the result as a series? If so, the 1/(1 + cos^2(x)) part can be expressed as a geometric series on the half-open interval (0, 1]. Then you can integrate term by term using integration by parts. You can also show that the integral on [0, epsilon] is bounded by some power of epsilon, so it is arbitrarily small.

Tentatively, it looks like I'm getting $$\sum_{j=0}^\infty (-1)^j \ \{ \frac{\cos^{2j+1}(1)}{2j + 1} \ + \ \frac{1}{2j + 1} \sum_{k=0}^j (-1)^k \ \left(\begin{array}{cc}j \\ k \end{array}\right) \ \frac{\sin^{2k+1}(1)}{2k+1} \}$$

Is there a direct way to do it? Seems like there should be.

Last edited: Mar 5, 2005
3. Mar 5, 2005

### dextercioby

Here's the solution.

Daniel.

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4. Mar 5, 2005

### BobG

The only CAS I have to work on it is Mupad and that does nothing with it.

I'd be tempted to change it to:

$$\int_{1}^{0} x \frac{-sin x dx}{1 + cos^2 x}$$

$$\int_{1}^{0} x \frac{-sin x dx}{1 + cos^2 x} = x tan^{-1}(cos x) - \int tan^{-1} (cos x)$$

$$\int_{1}^{0} x \frac{-sin x dx}{1 + cos^2 x} = x tan^{-1}(cos x) - x tan^{-1} (cos x) + \frac{1}{2} ln (1 + cos^2 x) \vert_{1}^{0}$$

$$\int_{1}^{0} x \frac{-sin x dx}{1 + cos^2 x} =\frac{1}{2} ln (1 + cos^2 x) \vert_{1}^{0}$$

5. Mar 5, 2005

### TheDestroyer

Give me 1-2 days, i'll try finding an antiderivative for it :P, and if not, i'll try transforming to Real sum :P heheheh

6. Mar 5, 2005

Daniel.

7. Mar 5, 2005

### TheDestroyer

BobG, You're integration isn't correct, You did 2 opposite partitions in 1 integral, the integration of $$arctan(cos(x))$$ isn't correct, and it's not possible to be integrated

8. Mar 5, 2005

### dextercioby

Oooops,yes,Bob,you did it.It should have been 1/2 ln of the denominator from the very first part integration.

And yes,the last integral is not that pretty.

Daniel.

9. Mar 5, 2005

### msmith12

i found the solution of it...

the hint that I got was--

$$\int_{0}^{1} xf(\sin{x})dx = \frac{\pi}{2} \int_{0}^{1} f(\sin{x})dx$$

from there, it isnt to bad...

10. Mar 5, 2005

### dextercioby

I think it is.And how did u get that $$\frac{\pi}{2}$$...?

Daniel.

11. Mar 5, 2005

### BobG

You're right. I should have written it on paper. First sub shows the error. (It works if the x before sine x were cosine x, instead).

Free, which is why I've been wondering how good it works. (At least, it has a free download version - Mupad light or some such)

12. Mar 5, 2005

### dextercioby

Daniel.

13. Mar 5, 2005

### msmith12

first of all, i just realized that i have the bounds wrong...
they should be from 0 -> pi... sorry about that, but

to get the $$\frac{\pi}{2}$$ part

take the equality... and subtract one from the other...

$$\int_{0}^{\pi} xf(\sin{x})dx = \frac{\pi}{2} \int_{0}^{\pi} f(\sin{x})dx$$
$$\int_{0}^{\pi} xf(\sin{x}) - \frac{\pi}{2}f(\sin{x})dx = 0$$
$$\int_{0}^{\pi} (x-\frac{\pi}{2})(f(\sin{x})) dx = 0$$
calling the integrand Z, we have

$$\int_{0}^{\frac{\pi}{2}} Z dx + \int_{\frac{\pi}{2}}^{\pi} Zdx = 0$$

but, due to the symmetry of both parts of Z about pi/2, we can rewrite this as...

$$\int_{0}^{\frac{\pi}{2}} Z dx - \int_{0}^{\frac{\pi}{2} Z dx = 0$$

which, is trivial...

and so for the original integral, you can rewrite as...

$$\int_{0}^{\pi} \frac {x\sin{x}}{2-\sin^{2}{x}}$$

which can be rewritten using the above identity as

$$\frac{\pi}{2}\int_{0}^{\pi} \frac {\sin{x}}{2-\sin^{2}{x}}$$

from here, rewrite the denominator in terms of cosine, and use a simple u substitution-- so you arrive at
$$\frac{\pi^{2}}{4}$$

whew...

14. Mar 5, 2005

### BobG

On the "downloads" page, MuPad Pro (which you have to pay for) is at the top. You have to scroll down to get to the free versions for which ever operating system you use.

This was on a test? My first impression was that this has to be an easy problem if you just look at it the right way. I keep thinking you definitely want the arctangent, arcosine, or arcsine since your limits of integration are 0 to 1 (they would work out very nicely). But the problem just doesn't work nicely at all. Did the teacher make a typo?

My calculator gives a numerical answer (which matches the answer dextercioby posted).

15. Mar 5, 2005

### dextercioby

Thank you,Bob.I have an ancient version of Maple (5,or something like that) which is implemented into SWP 2.5.Basically,there's no programming involved.So,if i'll need something better,i'll consider your suggestion.

Daniel.