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use parseval's identity to evaluate the integral

\begin{equation} \int_{-\pi}^{\pi}(\sin x)^4dx\end{equation}

I'm familiar with Parseval's identity which states that for each piecewise continuous complex function $f$ we have the equality \begin{equation} \int_{-\pi}^{\pi}\left|f(x) \right|^{2}dx=\frac{|a_{0}|^{2}}{2}+\sum_{n=1}^{\infty}\left(|a_{n}|^{2}+|b_{n}|^{2} \right) \end{equation} where $a_{n}$ and $b_{n}$ are the Fourier coefficients of $f$.

I'm confused how evaluate $\sin^4x$

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  • $\begingroup$ Isn't it missing a $\frac{1}{\pi}$ besides that integral? $\endgroup$ – Aaron Maroja May 22 '15 at 20:25
  • $\begingroup$ no, I have written it correct $\endgroup$ – user155971 May 22 '15 at 20:29
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    $\begingroup$ I think it suffies to compute the fourier coefficients of $\sin^{2}(x)$ if you insist of using Parseval $\endgroup$ – TheOscillator May 22 '15 at 20:29
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We have: $$\sin^2(x) = \frac{1-\cos(2x)}{2}\tag{1}$$ hence Parseval's identity implies: $$ \int_{-\pi}^{\pi}\sin^4(x)\,dx = 2\pi\cdot\frac{1}{4}+\pi\cdot\frac{1}{4}=\color{red}{\frac{3\pi}{4}}.\tag{2}$$

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  • $\begingroup$ thank you for helping me but I need the steps to know how you solve it $\endgroup$ – user155971 May 22 '15 at 20:34
  • $\begingroup$ @user155971: the first line gives the Fourier coefficients of $\sin^2(x)$. The second line is the application of Parseval's identity. $\endgroup$ – Jack D'Aurizio May 22 '15 at 20:37
  • $\begingroup$ Call me crazy, but I would wrote that in reverse order, that is $\pi \cdot \frac{1}{4} + ...$, because $a_0=\frac{1}{2}$. I mean, your answer is correct, but... That just me. And oh, yes, answer is $3\pi/4$, or? $\endgroup$ – Cortizol May 22 '15 at 20:41
  • $\begingroup$ @Cortizol: yes, sure, $\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}$, my bad. $\endgroup$ – Jack D'Aurizio May 22 '15 at 20:55
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If you are going to use Parseval, then you need to evaluate $a_{n}=\frac{1}{\pi}\int_{-\pi }^{\pi }\sin ^{2}t\cos ntdt$. Since the integrand is even, it suffices to integrate from $0$ to $\pi $ and multiply by $2$.

$b_{n}$ is a similar integral with $\sin nt$.But the integrand in this case is odd, so these terms vanish.

So you are left with the integrals $a_{n}$, which are pretty easy to do.

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