Why the integral of $(\sin x)^n$ goes to zero before $\pi/2$

Key takeaway: The key idea is that on any interval ending before $\pi/2$, the sine function stays strictly below 1. Once you lock in that gap, the power $n$ forces the integrand down fast.

We want to prove

$\lim_{n\to\infty}\int_0^{\frac{\pi}{2}-a}(\sin x)^n\,dx=0,$

where $a>0$ is fixed.

Step 1: Find a uniform bound for $\sin x$

On the interval

$0\le x\le \frac{\pi}{2}-a,$

we have

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

Since $a>0$, we know

$0<\cos a<1.$

So for every $x$ in the interval,

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

Step 2: Bound the integral

Therefore,

\le \int_0^{\frac{\pi}{2}-a}(\cos a)^n\,dx.$$ Because $(\cos a)^n$ is constant with respect to $x$, $$\int_0^{\frac{\pi}{2}-a}(\cos a)^n\,dx =\left(\frac{\pi}{2}-a\right)(\cos a)^n.$$ So $$0\le \int_0^{\frac{\pi}{2}-a}(\sin x)^n\,dx \le \left(\frac{\pi}{2}-a\right)(\cos a)^n.$$ ### Step 3: Take the limit Since $0<\cos a<1$, $$\lim_{n\to\infty}(\cos a)^n=0.$$ Multiplying by the fixed constant $\frac{\pi}{2}-a$ does not change that, so by the squeeze theorem, $$\lim_{n\to\infty}\int_0^{\frac{\pi}{2}-a}(\sin x)^n\,dx=0.$$ That is exactly what we needed to show. --- **Pitfalls the pros know** 👇 The main trap is forgetting that $a$ is fixed. If $a$ depended on $n$ and drifted toward 0, then the bound $\cos a<1$ might not be enough. Here the interval ends a fixed distance before $\pi/2$, and that fixed gap is what makes the argument work. **What if the problem changes?** If the upper limit were exactly $\pi/2$, the answer would be different: the integral would no longer be trapped by a constant strictly less than 1 across the whole interval. In that case, the limit is not 0; it is asymptotic to a quantity of order $n^{-1/2}$. So the endpoint matters a lot. `Tags`: squeeze theorem, uniform bound, trigonometric limit