# Homework Help: Should be easy, but I got stuck!

1. Aug 25, 2008

### siaosi

Completely stuck with this:

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

Any help on how to tackle this please?

2. Aug 25, 2008

### Hurkyl

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

3. Aug 25, 2008

### siaosi

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.

4. Aug 25, 2008

### 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?

5. Aug 25, 2008

### 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!

6. Aug 25, 2008

### 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)

7. Aug 26, 2008

### 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.

8. Aug 26, 2008

### snipez90

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.

9. Aug 26, 2008

### Pere Callahan

I can tell you that the result is not very pleasant

10. Aug 26, 2008

### 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.

11. Aug 26, 2008

### 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 $$\int_0^{\pi}\,sin^{2n}x\,dx\ =\ \frac{2n-1}{2n}\,\int_0^{\pi}\,sin^{2n-2}x\,dx$$

12. Aug 27, 2008

### Pere Callahan

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

13. Aug 27, 2008

### tiny-tim

probability …

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

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

If so, why … ?

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

14. Aug 27, 2008

### 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?

15. Aug 27, 2008

### 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.

16. Aug 27, 2008

### tiny-tim

Yes … but why?

what does it all mean?

17. Aug 27, 2008

### Pere Callahan

I don't know ....

18. Aug 27, 2008

### siaosi

Wow, thanks for all the answers!

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.

19. Aug 27, 2008

### Pere Callahan

See posts #7 and #15.

20. Aug 27, 2008

### tiny-tim

Hi siaosi!

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!

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