Should be easy, but I got stuck!

[siaosi]

Completely stuck with this:

\int_0^\infty sin^{2008} x dx


Any help on how to tackle this please?

[Hurkyl]

What have you tried? (And are you sure you wrote it correctly?)

[siaosi]

Thanks for you prompt reply,

Yes it's written correctly. I've tried a formula that I found in a book, that says it is:

1*3*5*...*(n-1)/2*4*6*...*(n) * pi

But it just can't be as simple as that... I'm sure this is a tricky question.

[Defennder]

Hmm, I have to say that doesn't seem particular plausible. Think of the area under the graph of \sin^{2008} x from 0 to \infty. Did you mistype the integral limits?

[siaosi]

Ok, I have re-checked it, I had it wrong, it is from 0 to pi. Sorry guys.



\int_0^\pi sin^{2008} x dx


Still buffed though!

[snipez90]

Hmm well you could apply the usual methods for starting these integrals. Consider breaking down the integral into two (one with limits of integration of 0 to pi/2 and another of pi/2 to pi). Then consider the symmetry of sin(x) to realize that these two integrals are equal. Then try a substitution involving the identity cos(x) = sin(pi/2 - x).

But then again in this case it might complicate the problem. Is there an easy way to factor sin2008(x) + cos2008(x)

[Pere Callahan]

I would suggest you try integration by parts to derive a recursion formula for


I_n=\int_0^\pi{dx \sin^nx}



Then you can solve this recursion formula and plug in n=2008.

[snipez90]

I would suggest you try integration by parts to derive a recursion formula for


I_n=\int_0^\pi{dx \sin^nx}



Then you can solve this recursion formula and plug in n=2008.

I have seen that recursion formula before and I can tell you without remembering the specifics that 2008 is a high enough number that this approach will probably be not very pleasant.

[Pere Callahan]

I have seen that recursion formula before and I can tell you without remembering the specifics that 2008 is a high enough number that this approach will probably be not very pleasant.



I can tell you that the result is not very pleasant:smile:

[snipez90]

Haha, well I didn't mean to be rude, since my approach seemed to get no where. I forgot that a^n + b^n cannot really be factored readily.

[tiny-tim]

(2n-1)\int_0^{\pi}\,sin^{2n}x\,dx

=\ (2n-1)\int_0^{\pi}\,sin^{2n-2}x\,dx\ -\ \int_0^{\pi}\,(2n-1)\,sin^{2n-2}x\,cos^2xdx

=\ (2n-1)\int_0^{\pi}\,sin^{2n-2}x\,dx\ -\ [sin^{2n-1}x\,cosx]_0^{\pi}\ -\ \int_0^{\pi}\,sin^{2n}x\,dx

So \int_0^{\pi}\,sin^{2n}x\,dx\ =\ \frac{2n-1}{2n}\,\int_0^{\pi}\,sin^{2n-2}x\,dx :smile:

[Pere Callahan]

Yes, and the same works if you take n instead of 2n.:smile:

[tiny-tim]

Is \frac{1}{\pi}\,\int_0^\pi sin^{2008} x dx

the probability of tossing a coin 2008 times :rolleyes: and getting exactly 1004 heads?

If so, why … ? :confused:

(\int_0^\pi sin^m x dx is rational for odd m, and a rational multiple of π for even m)

[NoMoreExams]

How can it be? Tossing a coin is a discrete even -> discrete probability distribution function but integrals are analagous to continous ones. Or am I misunderstanding your question tiny-tim?

[Pere Callahan]

Well, maybe, Tiny-Tim just observed that the two quantities - the integral on the one hand, and the describes probability - are the same.

The probability of having k heads in 2k tosses is


p_{2k}=\left(\stackrel{2k}{k}\right)\left(\frac{1}{2}\right)^{2k}=\frac{(2k!)}{k!k!}\left(\frac{1}{2}\right)^{2k}


On the other hand, from the recursion formula described above one finds (using I_0=\pi)


\frac{1}{\pi}I_{2k}=\frac{(2k-1)!!}{(2k)!!}


Now one can use the identites (2k)!!=2^k k! and (2k-1)!!=\frac{(2k)!}{2^k k!} to find


\frac{1}{\pi}I_{2k} = \frac{(2k)!}{2^kk!2^kk!}=p_{2k}


If the question were not about 2008 but some odd number, something similar would probably be true.

[tiny-tim]

\frac{1}{\pi}I_{2k} = \frac{(2k)!}{2^kk!2^kk!}=p_{2k}

Yes … but why?

:cry: what does it all mean? :cry:

[Pere Callahan]

I don't know ....

[siaosi]

Wow, thanks for all the answers!

(2n-1)\int_0^{\pi}\,sin^{2n}x\,dx

=\ (2n-1)\int_0^{\pi}\,sin^{2n-2}x\,dx\ -\ \int_0^{\pi}\,(2n-1)\,sin^{2n-2}x\,cos^2xdx

=\ (2n-1)\int_0^{\pi}\,sin^{2n-2}x\,dx\ -\ [sin^{2n-1}x\,cosx]_0^{\pi}\ -\ \int_0^{\pi}\,sin^{2n}x\,dx

So \int_0^{\pi}\,sin^{2n}x\,dx\ =\ \frac{2n-1}{2n}\,\int_0^{\pi}\,sin^{2n-2}x\,dx :smile:

I have done my research and from various books I have found the same answer as this. This is all fine, but I have a loooooooong way to reach the number 2008... So I am still confused.

[Pere Callahan]

See posts #7 and #15.

[tiny-tim]

I have done my research and from various books I have found the same answer as this. This is all fine, but I have a loooooooong way to reach the number 2008... So I am still confused.

Hi siaosi! :smile:

The best thing to do, if you're confused by complicated equations, is to give things short names so that equatins look really simple.

In this case, define P2n = ∫0π sin2nx dx.

Then that equation was P2n =
P2(n-1) (2n-1)/2n,

which you can then see is
P2(n-2) (2n-1)(2n-3)/2n(2n-2),

and so on … until you reach P0, which you can easily work out! :smile:

Now can you see that there's a pattern … so you don't have to worry about making thousands of calculations! :wink: