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. 
 A: HINT: prove that $$\int_{0}^\pi(\sin(x))^ndx=\frac{\sqrt{\pi}\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}$$
A: A simpler approach than Dr Graubner's:
Your limit can be rewritten as:
$$
\lim_{n\to \infty} \frac{\int_0^\pi \sin^n(x) d x}{\sqrt{\frac{1}{n}}}.
$$
Note that both the numerator and denominator go to zero, so you can use L'Hôpital's rule. I am sure you can differentiate the denominator. For the numerator, differentiating under the integral sign with respect to $n$ gives:
$$\int_0^\pi \cos x \sin^{n-1} x dx.$$
I am sure you can complete the argument...
A: 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 $\sqrt{2\pi}$.
