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In an answer here on Math.SE it is claimed that $$\int_{-a}^a \sin^{100}x\,\mathrm dx = \frac{99}{100} \int_{-a}^a \sin^{98}x\,\mathrm dx$$ but I don't understand how it could be.

Since $$ \begin{aligned} \int \sin^n x\,\mathrm dx &= -\cos x\sin^{n - 1} x + (n - 1) \int \cos^2 x\sin^{n - 2}\,\mathrm dx =\\ &= -\cos x\sin^{n - 1} x + (n - 1) \int(1 - \sin^2 x)\sin^{n - 2} x\,\mathrm dx =\\ &= -\cos x\sin^{n - 1} x + (n - 1)\int\sin^{n - 2} x\,\mathrm dx - (n - 1)\int\sin^n x\,\mathrm dx =\\ &= -\frac{\cos x\sin^{n - 1} x}n + \frac{n - 1}n \int\sin^{n - 2} x\,\mathrm dx, \end{aligned} $$ and for $n$ even $$-\frac{\cos x\sin^{n - 1} x}n$$ is odd, how can it disappear from the result? What am I missing?

EDIT: For reference, this is the answer in question.

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    $\begingroup$ The result is not correct unless $a$ is an integer multiple of $\pi/2$. $\endgroup$ – Winther Feb 25 '15 at 22:00
  • $\begingroup$ @Winther: Ah, that's what I was suspecting. Actually the author later considers the integral from $-\pi$ to $\pi$, so it's probably a typo. $\endgroup$ – rubik Feb 25 '15 at 22:01
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    $\begingroup$ I made some experiments with $\int_{-1}^1$ and it doesn't hold.. $\endgroup$ – Peter Franek Feb 25 '15 at 22:02

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