How to prove $$\int_a^{\pi/2}\cos^nxdx\le e^{-na^2/2}\int_0^{\pi/2}\cos^nxdx,$$ where $n\in\mathbb N$ and $a\in[0,\pi/2]$?
I noticed that if we can prove $$\cos^na\le nae^{-na^2/2}\int_0^{\pi/2}\cos^nxdx,$$ apply $\displaystyle\int_a^{\pi/2}$ to both side, the conclusion will follow. But unfortunately, this inequality above is not true. When $a=0$, $LHS=1>0=RHS$. Also, Wallis' formula can help us find $\displaystyle\int_0^{\pi/2}\cos^nxdx$. I'm not sure if it helps.
Let $0\leq x\leq \frac{\pi}{2}$ and $0\leq a\leq \frac{\pi}{2}$. Then we can see $$ \cos(x+a) = \cos x\cos a-\sin x\sin a\leq \cos a\cos x. $$ Let $$ f(t) = -\frac{t^2}{2} -\ln (\cos t), \quad 0\leq t<\frac{\pi}{2}. $$ Then, $f(0) = 0$ and $$ f'(t) = -t +\tan t\geq 0. $$ Thus we have $$ f(a) \geq 0, $$and $$ \cos a \leq e^{\frac{-a^2}{2}}. $$ This implies $$ \cos^n(x+a) \leq e^{-\frac{na^2}{2}}\cos^n x $$ and hence $$ \int_0^{\frac{\pi}{2}-a}\cos^n(x+a)\leq e^{-\frac{na^2}{2}}\int_0^{\frac{\pi}{2}-a}\cos^n x\leq e^{-\frac{na^2}{2}}\int_0^{\frac{\pi}{2}}\cos^n x. $$