Proving $\int_0^{\pi/4}\sin^n(x)\,dx>\frac{1}{2^{n/2}(n+2)}$ I want to prove:$$\int_0^{\pi/4}\sin^n(x)\,dx>\frac{1}{2^{n/2}(n+2)}$$
This came up when I was working on this
question that only asked for elementary calculus solution. Some trivial lower bounds such as: 
$$ \sin(x)\geq\frac{2\sqrt{2}}{\pi}x$$
or even a slightly stronger one:
$$\sin(x)\geq \dfrac{3}{\pi}x\cdot 1_{[0,\pi/6]}+\left(\frac{6(\sqrt{2}-1)}{\pi}x+\dfrac{3-2\sqrt{2}}{2}\right)\cdot 1_{[\pi/6, \pi/4]}$$
were not tight enough to prove the assertion. 
Ideally, one could find the asymptotic expansion of:
$$n2^{n/2}\int_0^{\pi/4}\sin^n(x)\,dx$$
which can be seen to be converging to $1.$

I would (optionally) substitute $\sin x=t$ and integrate by parts twice:
\begin{align}\int_0^{1/\sqrt{2}}\frac{t^n\,dt}{(1-t^2)^{1/2}}&=\frac{1}{n+1}\left(\left.\frac{t^{n+1}}{(1-t^2)^{1/2}}\right|_0^{1/\sqrt{2}}-\int_0^{1/\sqrt{2}}\frac{t^{n+2}\,dt}{(1-t^2)^{3/2}}\right)\\&=\frac{1}{n+1}\left(2^{-n/2}-\frac{1}{n+3}\left.\frac{t^{n+3}}{(1-t^2)^{3/2}}\right|_0^{1/\sqrt{2}}+\ldots\right)\\&>\frac{2^{-n/2}}{n+1}\left(1-\frac{1}{n+3}\right)=2^{-n/2}\frac{n+2}{(n+2)^2-1}>\begin{bmatrix}\text{what}\\ \text{we}\\ \text{need }\end{bmatrix}\end{align}
