TITLE: Integral and its limit
QUESTION [6 upvotes]: Evaluate: $$ \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$
Rewriting this as $$I(n)= \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$ and then Differentiating under the Integral Sign with respect to $n$. I was unable to go much further with this.
$$$$ Original problem: $$ \large\lim_{n \to \infty} \int_{0}^{\pi/2}n \left(1-\sqrt[n]{\cos x}\right) \mathrm{d}x$$

REPLY [3 votes]: I think you should solve the actual problem by solving the limit first.
The limit can be taken into the integral, so now the problem becomes
$$\int_{0}^{\pi /2}\lim_{n\rightarrow \infty }n(1-\sqrt[n]{\cos x})dx$$
By L'Hôpital's rule,
$$\lim_{n \to \infty }n(1-\sqrt[n]{\cos x})=\lim_{n \to \infty }\frac{-\cos^{\frac 1n}x\cdot \ln \cos x\cdot \frac {-1}{n^2}}{-\frac{1}{n^2}}=-\ln \cos x$$
Hence we have 
$$\lim_{n \to \infty }\int_{0}^{\pi /2}n(1-\sqrt[n]{\cos x})dx=-\int_{0}^{\pi /2}\ln \cos x dx$$
Now the integral on the right side is $-\frac{\pi }{2}\ln2$. You can refer to here.
Thus
$$\lim_{n \to \infty }\int_{0}^{\pi /2}n(1-\sqrt[n]{\cos x})dx=\frac{\pi }{2}\ln2$$