Integral and its limit

Expand $\cos^{\frac1n}x$ in powers of $\frac{1}{n}$ using that $$\cos^{\varepsilon}x=1+\varepsilon\ln\cos x+O\left(\varepsilon^2\right).$$ Now the limit becomes equal to $$\lim_{n\to \infty}n\int_{0}^{\pi/2}\left(1-\cos^{\frac1n}x\right)dx=-\int_0^{\pi/2}\ln\cos x\,dx=\frac{\pi\ln 2}{2}.$$

P.S. The integral can also be evaluated in terms of gamma functions (see here), but this is of little help for evaluating the limit.

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

$$\lim_{n \to \infty }\int_{0}^{\pi /2}n(1-\sqrt[n]{\cos x})dx=\frac{\pi }{2}\ln2$$