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 $1.$

Remark: @metamorphy gave a very nice solution by IBP. Can we use bounds for the integrand instead? I want to give some alternative solutions.
With the substitution $\sin x = t$, we have
$$\int_0^{\pi/4} \sin^n x \mathrm{d} x = \int_0^{1/\sqrt{2}} \frac{t^n}{\sqrt{1 - t^2}} \mathrm{d} t.$$
Since $\frac{1}{\sqrt{1 - t^2}} \ge 2t$ for all $t$ in $[0, 1/\sqrt{2}]$
(with equality if and only if $t = 1/\sqrt{2}$), we have
$$\int_0^{1/\sqrt{2}} \frac{t^n}{\sqrt{1 - t^2}} \mathrm{d} t > \int_0^{1/\sqrt{2}} 2t^{n+1} \mathrm{d} t
= \frac{1}{2^{n/2}(n + 2)}.$$
We are done.
Note: If we use better lower bounds for the integrand, we get better lower bounds for the integral.
About the lower bound $2t$: The Taylor expansion of $\frac{1}{\sqrt{1 - t^2}}$ around $t = \tfrac{1}{\sqrt{2}}$
is $\sqrt{2} + 2(t - \tfrac{1}{\sqrt{2}}) + \cdots$.
It is easy to prove that $\sqrt{2} + 2(t - \tfrac{1}{\sqrt{2}}) = 2t$ is a lower bound for $\frac{1}{\sqrt{1 - t^2}}$.
