I would like to find an equivalent of the sequence $u_{n}$ where
$ u_{n}=\int_0^{\pi/2} \cos\left(\frac{\pi}{2}\sin(x)\right)^n \mathrm dx $
The substitution $x\rightarrow \frac{\pi}{2}\sin(x)$ gives:
$ u_{n}=\frac{2}{\pi}\int_0^{\pi/2} \frac{\cos(x)^n}{\sqrt{1-(\frac{2x}{\pi})^2}} \mathrm dx$
Numerical values seem to show that:
$ \int_0^{\pi/2} \frac{\cos(x)^n}{\sqrt{1-(\frac{2x}{\pi})^2}} \mathrm dx \sim_{n\rightarrow \infty} \int_0^{\pi/2} \cos(x)^n \mathrm dx \sim_{n\rightarrow \infty} \sqrt{\frac{\pi}{2n}}$
And $ u_{n} \sim_{n\rightarrow \infty} \sqrt{\frac{2}{\pi n}}$
So my question is:
How can I show that
$ \int_0^{\pi/2} {\cos(x)^n}\left(\frac{1}{\sqrt{1-(\frac{2x}{\pi})^2}}-1 \right) \mathrm dx=_{n\rightarrow \infty} o(1/\sqrt{n}) $ ?