I was solving the following question.
Find the following limit. $$\lim_{n\to \infty}\dfrac1n \left(\dfrac{1}{1 + \sin\left(\dfrac{\pi}{2n}\right)} + \dfrac{1}{1 + \sin\left(\dfrac{2\pi}{2n}\right)} + \dfrac{1}{1 + \sin\left(\dfrac{3\pi}{2n}\right)} + ... +\dfrac{1}{1 + \sin\left(\dfrac{n\pi}{2n}\right)}\right)$$
Here's my method: $$\lim_{n\to \infty}\dfrac1n \left(\dfrac{1}{1 + \sin\left(\dfrac{\pi}{2n}\right)} + \dfrac{1}{1 + \sin\left(\dfrac{2\pi}{2n}\right)} + \dfrac{1}{1 + \sin\left(\dfrac{3\pi}{2n}\right)} + ...+ \dfrac{1}{1 + \sin\left(\dfrac{n\pi}{2n}\right)}\right)$$ $$ = \lim_{n\to \infty}\dfrac1n \left(\sum_{k=1}^n\dfrac{1}{1 + \sin\left(\dfrac{k\pi}{2n}\right)}\right)$$
Now from the Riemann sum the limit is equal to, $$\int\dfrac{1}{1 + \sin\left(\dfrac{x\pi}{2n}\right)} dx$$
Is it right? I don't know how to determine the limits of the integral as I'm new to this topic. Can anyone help me in this?
Remark:$$\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n} f(\frac{k}{n})=\int_0^1 f(x)dx$$ $$= \lim_{n\to \infty}\dfrac1n \left(\sum_{k=1}^n\dfrac{1}{1 + \sin\left(\dfrac{k\pi}{2n}\right)}\right) =\\ \lim_{n\to \infty}\dfrac1n \left(-1+\sum_{k=0}^n\dfrac{1}{1 + \sin\left(\dfrac{k\pi}{2n}\right)}\right)=\\ \lim_{n\to \infty}\dfrac1n \left(\sum_{k=0}^n\dfrac{1}{1 + \sin\left(\dfrac{k\pi}{2n}\right)}\right)-\lim_{n\to \infty }\frac1n=\\ \lim_{n\to \infty}\dfrac1n \left(\sum_{k=0}^nf(\frac{k}{n})\right)-\lim_{n\to \infty }\frac1n=\\\int_{0}^{1}\dfrac{1}{1 + \sin\left(\dfrac{x\pi}{2}\right)} dx -\lim_{n\to \infty }\frac1n=\\\int_{0}^{1}\dfrac{1}{1 + \sin\left(\dfrac{x\pi}{2}\right)} dx -0 $$ then you can use substitution $u=\frac{x\pi}{2}$ to solve the integral $$\int \frac 1{1+\sin u}du=\int \frac{\sec^2(\frac u2)}{(\tan(\frac u2)+1)^2}du =\int \frac{f'}{f^2}du=\\-2\frac{1}{f}=\\\frac{-2}{\tan(\frac u2)+1}+c$$