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How to compute this integral: $$a_n=\pi\int \limits_{-\pi}^{\pi}\left(\sin (x)\right)^4\cos (nx)\,dx\,?$$

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Oh, what a great question. You've got my upvote, dude –  Harold Apr 26 '13 at 6:13
Your question is phrased as an isolated problem, without any further information or context. This does not match MSE quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. Making these improvements will attract more appropriate answers and make the question more valuable for future MSE visitors. –  Kasper Apr 26 '13 at 8:48

2 Answers 2

up vote 1 down vote accepted

Copying and pasting M.Strochyk's answer from your first question, first note that

$$\sin^4{x}=\left(\dfrac{1-\cos{2x}}{2}\right)^2=\dfrac{1}{4}\left(1-2\cos{2x}+\cos^2{2x} \right)\\ =\dfrac{1}{4}\left(1-2\cos{2x}+\dfrac{1+\cos{4x}}{2} \right)=\dfrac{3}{8}-\dfrac{1}{2}\cos{2x}+\dfrac{1}{8}\cos{4x}$$

using the double angle identities twice to cut the power in half each time. Then your integral is just

$$\int \limits_{-\pi}^{\pi}\left(\sin (x)\right)^4\cos (nx) dx=\int \limits_{-\pi}^{\pi}\left(\dfrac{3}{8}-\dfrac{1}{2}\cos{2x}+\dfrac{1}{8}\cos{4x}\right)\cos (nx) dx$$

which you can distribute and then do with standard "tricks". The point here is that in order to find the fourier coefficients, you don't HAVE to use that coefficient formula all the time. If there is another trick, something faster then you can use it. In fact you SHOULD use it. The integrals for $a_n$ and $b_n$ are true always but using symmetry arguments (like how you already used odd/evenness of the function to conclude $b_n=0$) or other tricks is okay.

In this case, the "trick" we are using is that since Fourier series is a trigonometric series, any trigonometric polynomial is its own fourier expansion. This is exactly like how a polynomial is always its own Taylor expansion. So in this, your function $\sin^4(x)$ is a trig polynomial but to put it in the $\sin(nx)$ and $\cos(nx)$ form, first we say that there can be no $\sin(nx)$ terms because the function is even so all odd terms must first be zero. Then we use the double angle formulas twice to cut the power from 4 to 2 to 1 and we end up with


so this is its own fourier expansion. All other terms are zero. Done!

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can a costant be represented by Fourier series? –  park ning Apr 26 '13 at 6:58
Yes, a constant will be equal to its own fourier expansion as well. So if $f(x)=5$ then your $a_0=5$ and then all other $a_n=b_n=0$. A constant is a trig polynomial too. –  Fixed Point Apr 26 '13 at 7:02
Your answer to my thoroughly understand!! Thank u very much! –  park ning Apr 26 '13 at 7:07
Then you should upvote all of the answers you like by clicking the up arrow on their left. And then accept the answer you like the best by clicking the checkmark on the left. –  Fixed Point Apr 26 '13 at 8:04

$\textbf{Hint:}$ You got some help here How to reperesent $\sin^{4}(x)$ byFourier series? to find another way to represent $ \sin ^4 x$. What is $$ \int_{-\pi}^{\pi} \cos(mx) \cos(nx) dx $$

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Can you explain in detail a bit? –  park ning Apr 26 '13 at 6:25
$\cos(nx)$ form an orthogonal basis. If you carry out that integration you will find the $ \int_{-\pi}^{\pi} \cos(mx) \cos(nx) dx = 0$, for $n\neq m$ and $ \int_{-\pi}^{\pi} \cos(mx) \cos(nx) dx = \pi$ for $n=m$. –  Henrik Finsberg Apr 26 '13 at 6:35

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