TITLE: How find this limit $I=\lim_{n\to\infty}n^a\left(\int_{0}^{\pi/2}\sin{(nx)}\cos^n{x}dx\right)=b$
QUESTION [6 upvotes]: If the constant $a,b\neq 0$ such 
$$
I=\lim_{n \to \infty}\left[%
n^{a}\int_{0}^{\pi/2}\sin\left(nx\right)\cos^{n}\left(x\right)\,{\rm d}x
\right] = b
$$
find $a,b$
My idea: since
$$\sin{(nx)}=\dfrac{e^{inx}-e^{-inx}}{2i}$$
$$\cos{x}=\dfrac{1}{2}(e^{ix}+e^{-ix})$$
so
$$(e^{ix}+e^{-ix})^n=\sum_{k=0}^{n}\binom{n}{k}e^{i(n-k)x}e^{-ikx}$$
so
$$\sin{(nx)}\cos^n{x}=\dfrac{1}{2i}\cdot\dfrac{1}{2^n}(e^{inx}-e^{-inx})(e^{ix}+e^{-ix})^n=\dfrac{1}{2^{n+1}\cdot i}\sum_{k=0}^{n}\left(\binom{n}{k}(e^{i(2n-2k)x}-e^{-2ikx}\right)$$
then I can't.Thank you

REPLY [9 votes]: May I give it a try in one line? Since $\displaystyle I_k=\frac{I_{k-1}}{2}+\frac{1}{2k}\Rightarrow 2^kI_k=2^{k-1}I_{k-1}+\frac{2^{k-1}}{k}$, then 
$$\lim_{n\to\infty}n\int_0^{\pi/2}\sin(nx)\cos^n(x)\,\mathrm{d}x=\lim_{n\to\infty}\frac{n}{2^{n+1}}\sum_{k=1}^n\frac{2^k}{k}=1$$
as a consequence of the Cesaro-Stolz theorem.