Compute the limit. Compute the limit. $$\lim _{n \to \infty} \left(\sqrt{n} \int_{0}^{\pi} (\sin x)^n dx\right)$$ I have no clue where to start with this problem so any help is greatly appreciated. 

Another possibility besides the explicit calculation of the integral (which one can do recursively, for example) is to use Laplace's method to approximate
$$
\int_0^\pi\sin^n(x)\,dx=\int_0^\pi e^{n\ln\sin(x)}\,dx=\int_0^\pi e^{n\,f(x)}\,dx
$$
where (in Wiki notations) $f(x)=\ln\sin(x)$ and $x_0=\frac{\pi}{2}$. We have $f''(x_0)=-\frac{1}{\sin^2(\pi/2)}=-1$, so the methods says that
$$
\int_0^\pi\sin^n(x)\,dx\sim e^{n\ln\sin(\pi/2)}\sqrt{\frac{2\pi}{n\cdot 1}}=\frac{\sqrt{2\pi}}{\sqrt{n}}
$$
which gives the limit to be $\sqrt{2\pi}$.
