Another Integral Limit

$\displaystyle I= \lim_{n\to \infty } \int_0^{\frac{\pi}{2}} \frac{1-\cos^{\frac{1}{n}}x }{\frac{1}{n}} dx$

Use L'Hospital Rule:

$\displaystyle I = \lim_{n\to \infty } \int_0^{\frac{\pi}{2}} \frac{-(\frac{1}{n} )' \cos^{\frac{1}{n}}x \ln(\cos x)}{(\frac{1}{n})'} dx$

$\displaystyle = -\lim_{n\to \infty } \int_0^{\frac{\pi}{2}} \cos^{\frac{1}{n}}x \ln(\cos x) dx$

$\displaystyle I = -\int_0^{\frac{\pi}{2}} \ln(\cos x) dx$

$\displaystyle I = -\int_0^{\frac{\pi}{2}} \ln(\sin x) dx ( {\color{#3D99F6}{x \to (\frac{\pi}{2} -x )}}$

$\displaystyle => 2I = -\int_0^{\frac{\pi}{2}} \ln(\frac{\sin 2x}{2}) dx$

$\displaystyle = -\int_0^{\frac{\pi}{2}} \ln(\sin 2x) dx + \int_0^{\frac{\pi}{2}} \ln(2)dx$

$\displaystyle = -\frac{1}{2} \int_0^{\pi} \ln(\sin x) dx + \frac{\pi}{2} \ln 2$

(used ${\color{#3D99F6}{x \to \frac{x}{2} }}$ )

$\displaystyle = -\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(\cos x) dx + \frac{\pi}{2} \ln 2$

(used ${\color{#3D99F6}{x \to (\frac{\pi}{2} -x ) }}$ )

$\displaystyle = I + \frac{\pi}{2} \ln 2$

$\displaystyle => \boxed{I = \frac{\pi}{2} \ln 2}$

Well, we can also solve this using the Beta Function and then using Stirling's approximation........!!!