Prove $\lim_{n\to +\infty}\int_0^{\frac{\pi}{2}}\sin^n(x)\,dx=0$ Prove $$\lim_{n\to +\infty}\int_0^{\frac{\pi}{2}}\sin^n(x)\,dx=0$$
My attempt:
$$I_n=\int_0^{\frac{\pi}{2}}\sin^n(x)\,dx = \left.- \int_0^{\frac{\pi}{2}} \sin^{n-1} x \,d(\cos x) = -\sin^{n-1} x\cos x \right|_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x\,d(\sin^{n-1} x)$$
As $\left.-\sin^{n-1} x\cos x \vphantom{\dfrac11} \right|_0^{\frac{\pi}{2}} = 0$, hence
$$\int_0^{\frac{\pi}{2}}\sin^n(x)\,dx = (n-1) \int_0^{\frac{\pi}{2}}\sin^{n-1}(x) \, dx - (n-1)\int_0^{\frac{\pi}{2}}\sin^n(x) \, dx \Rightarrow I_n = \frac{(n-1)I_{n-2}}{n}$$ hence if $n=2k$: $$I_{2k}=\frac{(2k-1)!!}{(2k)!!}\cdot\frac{\pi}{2}$$ 
if $n=2k+1$:
$$I_{2k+1}=\frac{(2k)!!}{(2k+1)!!}$$
Got that by induction and because of $I_1=\int_0^{\frac{\pi}{2}}\sin x\,dx= \left.-\cos x \vphantom{\dfrac11} \right|_0^{\frac{\pi}{2}} = 1$.
So I have to prove that if $k \to +\infty \Rightarrow I_{2k} \to 0 \text{ and } I_{2k+1} \to 0$. I don't know how to do that. Please help
 A: This is a direct consequence of the Dominated Convergence Theorem. Note $\forall x,n$, $\sin^n(x) \le 1$ which is integrable over $[0,\frac{\pi}{2}]$ and $\sin^n(x) \to 0 \hspace{1mm} \forall x \in (0,\frac{\pi}{2})$.
EDIT: I felt kinda bad just leaving my solution like this, for those who aren't familiar with DCT. The idea is very simple (although what I'm about to say slightly differs from DCT). Since $\sin^n(x) \to 0$ uniformly on $[0,1]$, for any $\epsilon > 0$, for $n$ large enough, $\sin^n(x) \le \epsilon$. Therefore, $|\int_0^{2\pi} \sin^n(x)| \le 2\pi \epsilon$ for $n$ large enough. Since this holds for all $\epsilon > 0$, this limit must be $0$.
A: You can use the formule
$$(2k)!! =2^k k!, \quad (2k-1)!!=\frac{(2k)!}{2^k k!} $$
to obtain
$$I_{2k} =\frac{\pi}{2} \frac{(2k)!}{2^k k!} \frac{1}{2^k k!} = \frac{\pi}{2}\frac{(2k)(2k-1) \cdots (k+1)}{2^{2k} k!} \le \frac{\pi}{2}\frac{1}{2^k} \to 0. $$
$I_{2k+1}$ is similarly dealt with.
A: A stronger result
Let $\displaystyle{W_n=\int_0^{\pi/2}\sin^n(x)\,dx}$.
Integration by parts leads to the formula :
$$\forall n\ge2,\,nW_n=(n-1)W_{n-2}\tag{1}$$
which implies :
$$\forall n\ge1,\,nW_nW_{n-1}=\frac\pi 2$$
On the other hand, it's easy to see that the sequence $(W_n)_{n\ge0}$ is decreasing, and so :
$$W_{n+1}\le W_n\le W_{n-1}$$
After division by $W_{n-1}$ and combining with (1), we get :
$$\lim_{n\to\infty}\frac{W_n}{W_{n-1}}=1$$
Finally :
$$nW_n^2=\left(nW_nW_{n-1}\right)\left(\frac{W_n}{W_{n-1}}\right)\underset{n\to\infty}{\longrightarrow}\frac\pi2$$
We have proved the (very classical) result :

$$\int_0^{\pi/2}\sin^n(x)\,dx\sim\sqrt{\frac\pi{2n}}$$

In particular, we have $\displaystyle{\lim_{n\to\infty}W_n=0}$
A: $$\color{red}{\int_{0}^{\pi/2}\sin(x)^n\,dx} = \int_{0}^{\pi/2}\cos(x)^n\,dx \color{red}{\leq} \int_{0}^{\pi/2}e^{-nx^2/2}\,dx\leq \int_{0}^{+\infty}e^{-nx^2/2}\,dx=\color{red}{\sqrt{\frac{\pi}{2n}}}.$$
The crucial inequality $\cos(x)\leq e^{-x^2/2}$ over $(0,\pi/2)$ is equivalent to $\log\cos(x)\leq -\frac{x^2}{2}$, that is a consequence of $\tan(x)\geq x$ (just integrate both sides).
A: Different approach: For any $a\in [0,\pi/2),$ we have
$$\int_0^{\pi/2} \sin^n(x)\, dx = \int_0^{a} \sin^n(x)\, dx + \int_a^{\pi/2} \sin^n(x)\, dx \le a\cdot\sin^n(a) + 1\cdot (\pi/2-a).$$
Now the first term on the right $\to 0.$ Taking the $\limsup$ then shows
$$\tag 1 \limsup_{n\to \infty} \int_0^{\pi/2} \sin^n(x)\, dx \le 0 +(\pi/2-a).$$
Since $a$ is arbitrarily close to $\pi/2,$ we see the $\limsup$ is $0,$ which is the desired result.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\underline{\mrm{Laplace\ Method}}$:

\begin{align}
\lim_{n \to \infty}\int_{0}^{\pi/2}\sin^{n}\pars{x}\,\dd x & =
\lim_{n \to \infty}\int_{0}^{\pi/2}\exp\pars{n\ln\pars{\cos\pars{x}}}\,\dd x =
\lim_{n \to \infty}\int_{0}^{\infty}\expo{-nx^{2}/2}
\pars{1 - {n \over 12}\,x^{4}}\,\dd x
\\[5mm] & =
\root{\pi \over 2}\lim_{n \to \infty}
\pars{{1 \over n^{1/2}} - {1 \over 4n^{3/2}}} = \bbx{\ds{\large 0}}
\end{align}
