<s>[INST] How to obtain the asymptote formula of $\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx$? I want to obtain the formula in the form as follows:
$$ \int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx=A+\frac{B}{n}+\frac{C}{n^2}+\frac{D}{n^3}+o\left(\frac{1}{n^3}\right).$$
At least, it holds that $$\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx = \int_{0}^{\pi/2}\sqrt[n]{\sin x}\,dx = \int_{0}^{1}\frac{u^{1/n}}{\sqrt{1-u^2}}\,du = \frac{1}{2}\int_{0}^{1}(1-t)^{-\frac{1}{2}}t^{-\frac{1}{2}+\frac{1}{2n}}\,dt=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)}{2\,\Gamma\left(1+\frac{1}{2n}\right)},$$
but how to go on with this?
 [/INST] You properly wrote
$$I_n=\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx =\frac{\sqrt \pi}2\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)}{\Gamma\left(1+\frac{1}{2n}\right)}$$ Now, take logarithms for the gamma functions, useStirling approximation of the gamma function and continue with Taylor expansions to get
$$\log \left(\frac{\Gamma \left(\frac{1}{2}+\frac{1}{2n}\right)}{\Gamma \left(1+\frac{1}{2
   n}\right)}\right)=\frac{\log (\pi )}{2}-\frac{\log (2)}{n}+\frac{\pi ^2}{24 n^2}-\frac{\zeta (3)}{4
   n^3}+O\left(\frac{1}{n^4}\right)$$
Take the exponential
$$\frac{\Gamma \left(\frac{1}{2}+\frac{1}{2n}\right)}{\Gamma \left(1+\frac{1}{2
   n}\right)}=\sqrt{\pi }-\frac{\sqrt{\pi } \log (2)}{n}+\frac{\sqrt{\pi } \left(\pi ^2+12 \log
   ^2(2)\right)}{24 n^2}-\frac{\sqrt{\pi } \left(6 \zeta (3)+4 \log ^3(2)+\pi ^2
   \log (2)\right)}{24 n^3}+O\left(\frac{1}{n^4}\right)$$
$$I_n=\frac{\pi }{2}-\frac{\pi  \log (2)}{2 n}+\frac{\pi ^3+12 \pi  \log ^2(2)}{48
   n^2}-\frac{6 \pi  \zeta (3)+4 \pi  \log ^3(2)+\pi ^3 \log (2)}{48
   n^3}+O\left(\frac{1}{n^4}\right)$$
</s>