Integral of sin^2l(x): Ideas Needed

[cragar]

<br /> \int(sin(x))^{2l}dx<br />
I tried using Eulers formula to rewrite sin(x) in terms of exponentials and thought about using trig manipulations but does any have any ideas.
where L is a positive unknown constant integer.
sinx is raised to the 2l power , not sure if my latex is working

 

[DivisionByZro]

Is this for school? If so, what class?

The solution cannot be expressed in terms of elementary functions as far as I know. One of the solutions can be found by using part of the hypergeometric function.

Also, the solution is kind of messy!

Good luck.

 

[coelho]

Well... if you want the final result be written in terms of sines and/or cosines, you could try the so-called reduction formulae, but i don't know how much useful they could be, as they usually write integrals of
sin(x)^{N}
as functions of some integral of
sin(x)^{N-2}.

If the final result doesn't need to be written in terms of sines and/or cosines, I think writing sin(x) as a function of exponentials is an way to solve it. The final result looks a bit ugly, and maybe it doesn't look very useful, but its a answer anyway.

First, write:
sin(x)=\frac{e^{ix}-e^{-ix}}{2i}

Then expand it using:

(a+b)^n=\sum^{n}_{k=0} C^{n}_{k} a^{k} b^{n-k}

As the summation and the integral are interchangeable, the integral becomes:

\int (sin(x))^{2L} = \frac{1}{(2i)^{2L}} \sum^{2L}_{k=0} C^{2L}_{k} (-1)^{2L-k} \int e^{2i(k-L)x} dx

Be careful when k=L, because in this case the integrand is 1.
Anyway, the final answer is:

\int (sin(x))^{2L} = \frac{1}{(2i)^{2L}} \left[ (-1)^L C^{2L}_{L} x + \sum^{2L}_{k=0, k\neq L} C^{2L}_{k} (-1)^{2L-k} \frac{e^{(2ik-2iL)x}}{2i(k-L)} \right]

I hope it be useful.

 

[cragar]

wow thanks for your answer

 

[coelho]

Youre welcome

Anyway... i think this formula, even looking impressive, is pretty much useless, with so many complex exponentials and such. So i worked more on it, expanded the exponentials and re-grouped them in sines, which gives this:

\int (sin(x))^{2L} dx = \frac{1}{(-4)^{L}} \left[ (-1)^{L} C^{2L}_{L} x + \sum^{L-1}_{k=0} (-1)^{k} C^{2L}_k \frac{sin[2(L-k) x]}{L-k} \right]

I hope this one be more useful than the previous.

 

[cragar]

you are like a hardcore mathematician. A couple steps above, why is it ok to intechange the summation and the integral sign. Is that because L is an integer. I was thinking it had to be a Riemann sum.

 

[coelho]

you are like a hardcore mathematician.

It could be taken as flattery, or as a subtle way of saying "your formulas are way too complicated"... I will choose the first one just kiddin!

A couple steps above, why is it ok to intechange the summation and the integral sign. Is that because L is an integer. I was thinking it had to be a Riemann sum.

It isn't because L is an integer, its simpler than this.

Remember that:

\int f_1 + f_2 + f_3 + f_4 + ... dx = \int f_1 dx + \int f_2 dx+ \int f_3 dx+ \int f_4 dx + ...

Writing this as a summation:

\sum_k f_k = f_1 + f_2 + f_3 + f_4 + ...

We get:

\int \sum_k f_k dx = \int f_1 + f_2 + f_3 + f_4 +... dx = \int f_1 dx + \int f_2 dx+ \int f_3 dx+ \int f_4 dx + ... = \sum_k \int f_k dx

Or:

\int \sum_k f_k dx = \sum_k \int f_k dx

Thats why we can interchange the summation and the integral.

 

[cragar]

oh ok i see, thanks for explaining . How far along are you in your math career?

 

[loosely]

change the sin function into taylor polynomial, then you are able to integrate it.

 

[Mute]

Remember that:

\int f_1 + f_2 + f_3 + f_4 + ... dx = \int f_1 dx + \int f_2 dx+ \int f_3 dx+ \int f_4 dx + ...

Writing this as a summation:

\sum_k f_k = f_1 + f_2 + f_3 + f_4 + ...

We get:

\int \sum_k f_k dx = \int f_1 + f_2 + f_3 + f_4 +... dx = \int f_1 dx + \int f_2 dx+ \int f_3 dx+ \int f_4 dx + ... = \sum_k \int f_k dx

Or:

\int \sum_k f_k dx = \sum_k \int f_k dx

Thats why we can interchange the summation and the integral.

To clarify: this result is not always valid. It is always valid for sums with a finite number of terms, as in the calculation in this thread. However, if the sum has an infinite number of terms, then there are cases in which exchanging the order of integration and summation is invalid.

change the sin function into taylor polynomial, then you are able to integrate it.

No, that will not help very much:

\int dx~(\sin(x))^{2L} = \int dx~\left(\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\x^{2k+1}\right)^{2L} = \int dx~\sum_{k_1=0}^\infty\sum_{k_2=0}^\infty \dots \sum_{k_{2L}=0}^\infty \frac{(-1)^{k_1+k_2+\dots k_{2L}}}{(2k_1+1)!(2k_2+1)!\dots(2k_{2L}+1)!} x^{2k_1+1}x^{2k_2+1}\dots x^{2k_{2L}+1}

which is less easy than what coelho suggested above! (2L infinite sums versus one finite sum!)

 

[gb7nash]

<br /> \int(sin(x))^{2l}dx<br />
I tried using Eulers formula to rewrite sin(x) in terms of exponentials and thought about using trig manipulations but does any have any ideas.
where L is a positive unknown constant integer.
sinx is raised to the 2l power , not sure if my latex is working

Hi cragar,

As far as I know there's not a closed formula for this integral. However, you can make use of the appropriate power reduction formula as many times as you need and take the integral after you only have powers of 1. For example:

\sin^4 x = (\sin^2 x)^2 = (\frac{1-\cos 2x}{2})^2 = \frac{3 - 4\cos 2x + \cos 4x}{8}

Edit: I didn't see coelho's formula. If you want to be general, then you can take his formula and integrate it.