# Evaluating: $\lim_{n\to\infty} \frac{\sqrt n}{\sqrt {2}^{n}}\int_{0}^{\frac{\pi}{2}} (\sin x+\cos x)^n dx$

I'm supposed to compute the following limit:

$$\lim_{n\to\infty} \frac{\sqrt n}{\sqrt {2}^{n}}\int_{0}^{\frac{\pi}{2}} (\sin x+\cos x)^n dx$$

I'm looking for a resonable approach in this case, if possible. Thanks.

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Laplace's method yields $$\frac{\sqrt n}{\sqrt {2}^{n}}\int_{0}^{\pi/2} (\sin x+\cos x)^n \mathrm dx=\int_{-\pi\sqrt{n}/4}^{\pi\sqrt{n}/4}\left(\cos(t/\sqrt{n})\right)^n\mathrm dt\to\int_{-\infty}^{+\infty}\mathrm e^{-t^2/2}\mathrm dt=\sqrt{2\pi}.$$

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this was a shot directly in the problem's heart. Great solution. Thanks! – user 1618033 Jun 16 '12 at 11:34

Since $\sin x + \cos x = \sqrt{2} \sin (x+\pi/4)$ we have $$\frac{\sqrt{n}}{ \sqrt{2}^n }\int^{\pi/2}_0 (\sin x+ \cos x)^n dx = \sqrt{n} \int^{3\pi/4}_{\pi/4} \sin^n x dx.$$

Since $\int^{\pi/4}_0\sin^n x dx \leq \frac{\pi}{4} \left( \frac{1}{\sqrt{2}} \right)^n \to 0$ our limit is the same as the limit of $$2\sqrt{n} \int^{\pi/2}_0 \sin^n x dx.$$

The integral can be evaluated by integration by parts or converting it into a Beta function, and Stirling's approximation shows the limit is $\sqrt{2\pi}.$

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if i rewrite the last integral as $\sqrt{n} \int^{\pi}_0 \sin^n x dx$, then i could use the integral generating the Wallis product, right? – user 1618033 Jun 16 '12 at 11:37
@Chris Yes, the Wallis product is often derived by integrating that integral by parts. – Ragib Zaman Jun 16 '12 at 11:46
thanks for your solution! – user 1618033 Jun 16 '12 at 11:47