Ask Me Help Desk - Integration

Please integrate this o to 2010pi (1/(tanx)^2011 +1)

Is this what you mean:

$\int_{0}^{2010\pi}\frac{1}{tan^{2011}(x)+1}dx$

If so, may I ask from where you got this crazy integration problem?

I would venture to say that this is not meant to be integrated by brute force. It is difficult to even get a computer to do it.

If we look at the graph, half of the interval is enclosed by the rectangles of width about 1.6 and height 1. The other half is at y=0, so no area

So, as an educated guess, I would say the area is about 3157.

Galactus, that integral is indeed crazy. However, if I read literally what the OP wrote, the 1 isn't in the denominator. Hence the integral is

$\int_0^{2010 \pi} {(\frac1{\tan^{2011}x}+1)} \;dx \;=\; \int_0^{2010 \pi} {(\cot^{2011}x})\;dx \; + \; \int_0^{2010 \pi} 1 \;dx$

This would be horribly painful to do as an indefinite integral, but as I'm sure you see, it's trivial as this particular definite integral.

To the OP:

The function cot(x) is odd-symmetric and periodic. The definite integral of cot(x) over any interval which is a multiple of $\pi$ is always 0. Because of the odd symmetry, raising the function cot(x) to any odd power always results in another function with odd symmetry over the same period. Hence, the integral of $\cot^3 x$ , $\cot^5 x$ , or even $\cot^{2011} x$ is also zero over an interval which is a multiple of $\pi$ (which is the case for your question).

Hence, the answer to your problem is simply

$\int_0^{2010 \pi} {(\frac1{\tan^{2011}x}+1)} \;dx \;=\; 0 \;+\;\int_0^{2010 \pi} 1 \;dx\;=\;2010 \pi$

Very good JC. I figured this problem had something to do with observation. This horrific integrals like this normally indicate this. I misread. Either way, good show. Now, let's see if the OP even comes back to appreciate your insight.

Thanks for your solution and you got it correct.but I think there can be a simple answer for this problem.because I found this from a mathematics paper.therefore please try it using a substitution.your attention for this will be highly appreciate.

I assume then that my interpretation was correct.

The solution is $\frac{2010\pi}{2}=1005\pi$

Try it with other powers of tan. You will see the solution comes out to half of the upper limit of integration.

$\int_{0}^{(n-1)\pi}\frac{1}{tan^{2n}(x)+1}dx=\frac{n\pi}{2}-\frac{\pi}{2}=\frac{(n-1)\pi}{2}$

Starting with a tan power of 2, we get $\frac{\pi}{2}$

$n=4, \;\ \frac{3\pi}{2}$

This is only valid for even powers of tan. At odd values it is undefined.

The integrand has period Pi, so try solving:

$\int_{0}^{\pi}\frac{1}{tan^{2n}(x)+1}dx$

Try letting $x=tan^{-1}(u), \;\ dx=\frac{1}{1+u^{2}}du$