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I am trying to calculate the definite integral $$\int_0^\pi \frac{\sin(\frac{21}{2}x)}{\sin(\frac{1}{2}x)} dx.$$ Wolfram Alpha says here that the answer is $\pi$. I replaced 21 by other constants and think that in general, $\int_0^\pi \frac{\sin(\frac{n}{2}x)}{\sin(\frac{1}{2}x)} dx = \pi$ for all odd $n \in \mathbb Z$.

However I have no idea how to approach this problem. I tried substituting $u = x/2$ to simplify the integral a bit to $$2\int_0^{\pi/2} \frac{\sin(nu)}{\sin(u)}.$$ Then I thought that I could maybe use the identity $\sin(nx) = \sin(x)\cos( (n-1)x) + \sin((n-1)x)\cos(x)$. For instance, since \begin{align*} \sin(3x) &= \sin(2x)\cos(x) + \cos(2x)\sin(x) \\ &= 2\sin(x) \cos^2(x) + \cos^2(x)\sin(x) - \sin^3(x) \\ &= \sin(x)(3\cos^2(x) - \sin^2(x)) \end{align*} the integral would become $$ 2\int_0^{\pi/2} 3\cos^2(x) -\sin^2(x) dx $$ which I am able to solve: \begin{align*} 2\int_0^{\pi/2} 3\cos^2(x) -\sin^2(x) dx &= 2[\frac{3}{2}(x+\sin(x)\cos(x)) - \frac{1}{2}(x - \sin(x)\cos(x))]^{\pi/2}_0 \\ &= [2x + 4\sin(x)\cos(x)]^{\pi/2}_0 \\ &= \pi. \end{align*} However I don't know how to generalize this approach because the expansion for $\sin(21x)$ would have lots of unwieldy terms. Is there another way to do this problem that I am missing?

aras
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2 Answers2

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Consider, from where you left off, $$I_n=2\int\frac{\sin n\theta}{\sin\theta} d\theta, $$ where $n$ is odd. Noting that $$\frac {\sin n\theta-\sin(n-2)\theta}{\sin\theta}=2\cos(n-1)\theta, $$

We have $$I_n=I_{n-2}+4\int^{\frac{\pi}{2}}_0\cos(n-1)\theta d\theta$$ $$=I_{n-2}+0$$ since $n-1$ is even.

Hence $$I_n=I_{n-2}=...=I_1$$ and the value of is last integral is $\pi$

David Quinn
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  • Why is the second equation true? I can't see how to proceed from $\sin(n\theta) - \sin( (n-2)\theta) = \sin^2(\theta)\cos( (n-2)\theta ) + \cos^2(\theta)\sin( (n-2)\theta ) = \sin^2(\theta) [\cos( (n-2)\theta) - \sin( (n-2)\theta) ]$ – aras Apr 02 '16 at 18:28
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    @aras $\sin A-\sin B=2\cos \frac {A+B}{2} \sin \frac {A-B}{2}$ – David Quinn Apr 02 '16 at 18:33
  • Ah, I didn't know that identity. Thank you! – aras Apr 02 '16 at 18:46
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This is exactly the closed form of the Dirichlet's Kernel when N = 10.

$\sum _{n=-N}^{N} $ $e^{-inx}$ = $\frac{sin(\frac{(2N+1)x}{2})}{sin(\frac{x}{2})}$

Integration the series term by term, the terms with positive values of n will cancel the terms with negative ones, and the value of the integral when $n=0$ is $ \int_0 ^\pi e^{0ix} = \pi $, thus $ \int _0 ^{\pi} \frac{sin(\frac{(2N+1)x}{2})}{sin(\frac{x}{2})} dx = \pi $ for any integer value of N.

Burrrrb
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