# Approximation of $n^a \int_{0}^{\pi/n}\sin^b(t)dt$

How can I find an approximation of $$n^a \int_{0}^{\pi/n}\sin^b(t)dt$$ when $n\rightarrow \infty$, $(a,b>0)$ ?

-
Thank you very much! –  Chon Oct 25 '11 at 12:59
Since $\sin(t)\sim t$ when $t\to0$, the integral is equivalent to $$\left[(b+1)^{-1}t^{b+1}\right]_0^{\pi/n}=(b+1)^{-1}(\pi/n)^{b+1},$$ and the equivalent you are looking for is $\pi^{b+1}(b+1)^{-1}n^{a-b-1}$.