How find this limit $I=\lim_{n\to\infty}n^a\left(\int_{0}^{\pi/2}\sin{(nx)}\cos^n{x}dx\right)=b$ 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
 A: 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.
A: Let
\begin{align*}
  I_n &= \int_{0}^{\pi/2}\sin{(nx)}\cos^n{x}dx
\end{align*}
A reduction formula seems to be
\begin{align*}
  I_n &= \frac{1}{2}\left(I_{n-1}+\frac{1}{n}\right) \\
  I_0 &= 0
\end{align*}
Simplifying that gives:
\begin{align*}
  I_n &= \sum_{k=1}^n \frac{1}{2^k\left(n-k+1\right)}
\end{align*}
and the generating function for $I_n$ is 
\begin{align*}
  G(z) &= -\frac{\log{\left(1-z\right)}}{\left(2-z\right)}
\end{align*}
Now, we need to get the asymptotics for $[z^n]$
Update
For the asymptotic analysis, we may use the theorem VI.12, pp.434 in analytic combinatorics, which states:

Let $a(z) = \sum_n a_n z^n$ and $b(z) = \sum_n b_n z^n$ be two power series with radii of convergence $\alpha > \beta \ge 0$, respectively. Assume that
  $b(z)$ satisfies the ratio test,$$\frac{b_{n-1}}{b_n}\to \beta \; \; \text{ as } n\to\infty$$
Then, the coefficients of the product $f(z) = a(z)\cdot b(z)$ satisfy $$[z^n] f(z) \sim a\left(\beta\right)b_n\; \; \text{ as } n\to\infty$$ provided $a(\beta) \ne 0$

Hence, in our gf, let 
\begin{align*}
  a(z) &= \sum_n \frac{1}{2^n}z^n \\
  b(z) &= \sum_n \frac{1}{n} z^n \\
  \implies [z^n]G(z) &\sim \frac{1}{2}\cdot a(1)\cdot \frac{1}{n} \;\;\text{since $\frac{b_{n-1}}{b_n}\to 1$} \\
  [z^n]G(z) &\sim \frac{1}{n}
\end{align*}
Therefore,
\begin{align*}
  \lim_{n\to\infty} n \cdot I_n &= 1
\end{align*}
A: Continuing with what gar did, if you do partial fractions on the sum for $I_n$, you get
$$\frac{1}{2^k (n-k+1)} = \frac{2^{-n-1}}{n-k+1} + \frac{1-2^{-1-n}}{(n+1) 2^{k}}.$$
The sum over the second term is $O(1/n)$, and this estimate is sharp. The sum over the first term is $O(n 2^{-n})$, which you can obtain by replacing $k$ with $n$ (which can only increase the term). So the first term will drop out as $n \to \infty$ no matter what $a$ is, and all we need be concerned with is the second term. Moreover for $a<1$ it will converge to zero, for $a>1$ it will diverge. For $a=1$ we should have a finite limit, which is 
$$\lim_{n \to \infty} \frac{n \left (1-2^{-1-n} \right )}{n+1} \sum_{k=1}^n \frac{1}{2^k} = 1$$
A: From what you've done already, notice that since the LHS is real, the RHS is equal to its own real part. Because of the $1/i$, this means we need to compute the imaginary part of the sum. We get (for the sum):
$$\sum_{k=0}^n {n \choose k} \left (  \sin((2n-2k)x) - \sin(-2kx) \right )$$
This should be easy to integrate term by term, and because you're on $[0,\pi/2]$ and all the frequencies are even, all the definite integrals should have a very simple form.
You also should do your binomial expansion again, because I'm fairly sure there is a small error. You should have frequency $n+k-(n-k)=2k$ with a plus sign and frequency $-n+k-(n-k)=-2n+2k$ with a minus sign.
A: Here is an approach with very basic steps
$$
\begin{align}
&n\int_0^{\pi/2}\sin(nx)\cos^n(x)\,\mathrm{d}x\\
&=\frac{n}{2^n}\sum_{k=0}^n\binom{n}{k}\int_0^{\pi/2}\sin(nx)\cos((n-2k)x)\,\mathrm{d}x\tag{1}\\
&=\frac{n}{2^n}\sum_{k=0}^n\binom{n}{k}\int_0^{\pi/2}\frac12\left[\vphantom{\frac12}\sin(2(n-k)x)+\sin(2kx)\right]\,\mathrm{d}x\tag{2}\\
&=\frac{n}{2^n}\sum_{k=0}^n\binom{n}{k}\int_0^{\pi/2}\sin(2kx)\,\mathrm{d}x\tag{3}\\
&=\frac{n}{2^n}\sum_{k=1}^n\binom{n}{k}\frac1{2k}(1-\cos(k\pi))\tag{4}\\
&=\frac{n}{2^{n+1}}\sum_{k=1}^n\binom{n}{k}\frac{1^k-(-1)^k}{k}\tag{5}\\
&=\frac{n}{2^{n+1}}\int_{-1}^1\frac{(1+x)^n-1}{x}\,\mathrm{d}x\tag{6}\\
&=\frac{n}{2^{n+1}}\int_0^2\frac{x^n-1}{x-1}\,\mathrm{d}x\tag{7}\\
&=\frac{n}{2^{n+1}}\int_0^2\sum_{k=1}^nx^{k-1}\,\mathrm{d}x\tag{8}\\
&=\frac{n}{2^{n+1}}\sum_{k=1}^n\frac{2^k}{k}\tag{9}\\
&=\frac{n}{2^{n+1}}\sum_{k=1}^n\frac{2^{n+1-k}}{n+1-k}\tag{10}\\
&=\sum_{k=1}^n\frac{2^{-k}}{1-(k-1)/n}\tag{11}\\
&\to\sum_{k=1}^\infty2^{-k}\tag{12}\\[6pt]
&=1\tag{13}
\end{align}
$$
Explanation:
$\ \:(1):$ $\cos^n(x)=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^n=\frac1{2^n}\sum\limits_{k=0}^n\binom{n}{k}e^{i(n-2k)x}=\frac1{2^n}\sum\limits_{k=0}^n\binom{n}{k}\cos((n-2k)x)$
$\ \:(2):$ $\sin(A)\cos(B)=\frac12\left[\vphantom{\frac12}\sin(A+B)+\sin(A-B)\right]$
$\ \:(3):$ combine identical terms
$\ \:(4):$ integrate
$\ \:(5):$ evaluate
$\ \:(6):$ expand and integrate to get $(5)$
$\ \:(7):$ substitute $x\mapsto x-1$
$\ \:(8):$ expand quotient
$\ \:(9):$ integrate
$(10):$ reindex $k\mapsto n+1-k$
$(11):$ distribute over sum
$(12):$ dominated convergence: dominated by $k2^{-k}$
$(13):$ geometric sum

Proof and Use of a Nice Recursion
Using a trigonometric identity and integrating by parts, we see that
$$
\begin{align}
I_n
&=\int_0^{\pi/2}\sin(nx)\cos^n(x)\,\mathrm{d}x\\
&=\int_0^{\pi/2}\left[\vphantom{\frac12}\sin((n+1)x)\cos(x)-\cos((n+1)x)\sin(x)\right]\cos^n(x)\,\mathrm{d}x\\
&=\int_0^{\pi/2}\sin((n+1)x)\cos^{n+1}(x)\,\mathrm{d}x+\int_0^{\pi/2}\cos((n+1)x)\frac{\mathrm{d}\cos^{n+1}(x)}{n+1}\\
&=\int_0^{\pi/2}\sin((n+1)x)\cos^{n+1}(x)\,\mathrm{d}x-\frac1{n+1}+\int_0^{\pi/2}\sin((n+1)x)\cos^{n+1}(x)\,\mathrm{d}x\\
&=2I_{n+1}-\frac1{n+1}
\end{align}
$$
Substituting $n\mapsto n-1$ and solving for $I_n$ yields the recursion mentioned by gar and Chris's sis:
$$
I_n=\frac12\left(I_{n-1}+\frac1n\right)
$$
Multiplying by $2^n$ gives
$$
2^nI_n=2^{n-1}I_{n-1}+\frac{2^{n-1}}{n}
$$
and since $I_0=0$, we get
$$
2^nI_n=\sum_{k=1}^n\frac{2^{k-1}}{k}
$$
Therefore,
$$
\begin{align}
n\,I_n
&=n\sum_{k=1}^n\frac{2^{k-n-1}}{k}\\
&=n\sum_{k=1}^n\frac{2^{-k}}{n+1-k}\\
&=\sum_{k=1}^n\frac{2^{-k}}{1-(k-1)/n}\\
\end{align}
$$
which is $(11)$ above.
