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.
