Show that a sine-power integral over a shortened interval tends to zero

Key takeaway: This is one of those limits where the interval endpoint does most of the work. Once the interval stops short of $\pi/2$, the power kills the integrand.

The statement is

$\lim_{n\to\infty}\int_{0}^{\frac{\pi}{2}-a}(\sin x)^n\,dx=0,$

with fixed $a>0$.

Because $x\le \frac{\pi}{2}-a$ on the interval, we have

$\sin x\le \sin\left(\frac{\pi}{2}-a\right)=\cos a<1.$

Hence

$0\le (\sin x)^n\le (\cos a)^n.$

Integrating both sides gives

\le \left(\frac{\pi}{2}-a\right)(\cos a)^n.$$ Now let $n\to\infty$. Since $0<\cos a<1$, $$ (\cos a)^n\to 0, $$ so the right-hand side goes to $0$. The squeeze theorem finishes the proof: $$\boxed{\lim_{n\to\infty}\int_0^{\frac{\pi}{2}-a}(\sin x)^n\,dx=0.}$$ If you want the intuition in one line: on this interval, $\sin x$ never gets close to 1, so repeated powers force it down to 0 very fast. --- **Pitfalls the pros know** 👇 Do not replace $\sin x$ by 1 just because the interval ends near $\pi/2$. The fixed gap $a$ is essential. Without that gap, the integrand can stay large near the endpoint and the limit changes. **What if the problem changes?** If the interval were $[0,\frac{\pi}{2}-a_n]$ with $a_n\to 0$, you would need more information about how fast $a_n$ shrinks. For very small gaps, the integral may no longer go to 0. The fixed positive $a$ is what makes the proof immediate. `Tags`: endpoint estimate, squeeze theorem, bounded integrand