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The question is $\int_0^{\pi/2} \sin^n x \ \mathrm{d}x$. We know that this can be solve with a reduction formula if n is definite. And easily guess it results to 0 if $ n \rightarrow +\infty $.

This article says there is another way:

$\int_0^{\pi/2} \sin^n x \ \mathrm{d}x = \int_0^{\frac \pi 2 -\epsilon} \sin^n x \ \mathrm{d}x \ + \ \int_{\frac \pi 2 -\epsilon}^\frac \pi 2 \sin^n x \ \mathrm{d}x$

, where the two terms on the right yields 0.

My question is, I can't get the trick. The article didn't detail the steps after breaking the interval. I wish someone could write a deduction, or point out which formula/principle should be used. Thanks.

cuter44
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2 Answers2

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  • For the first term, recall that $q^n \rightarrow 0$ when $q \in (-1, 1)$.
  • For the second term, use the boundedness of $\sin$: $$ \left| \int_{\frac{\pi}{2} - \varepsilon}^{\frac{\pi}{2}} \sin^n x dx \right| \leq \int_{\frac{\pi}{2} - \varepsilon}^{\frac{\pi}{2}} 1 dx = \varepsilon $$
user7440
  • 832
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In the first term $\sin x \leq \sin(\frac {\pi} 2 -\epsilon)$ so the term is bounded by $(\frac {\pi} 2 -\epsilon)r^{n}$ where $r = \sin(\frac {\pi} 2 -\epsilon)$. Since $0<r<1$ the first term tends to $0$. The second term is bounded by $\int_{\frac {\pi} 2 -\epsilon }^{\frac {\pi} 2} 1 \, dx =\epsilon$.

Kavi Rama Murthy
  • 311,013