Find the limit of $\Bigl[\sin\Bigl(\dfrac{n}{n^2+1^2}\Bigr)+\sin\Bigl(\dfrac{n}{n^2+2^2}\Bigr)...+\sin\Bigl(\dfrac{n}{n^2+n^2}\Bigr)\Bigr]$ using Riemann integrals of a suitable function.
$\Bigl[\sin\Bigl(\dfrac{n}{n^2+1^2}\Bigr)+\sin\Bigl(\dfrac{n}{n^2+2^2}\Bigr)...+\sin\Bigl(\dfrac{n}{n^2+n^2}\Bigr)\Bigr]=\dfrac{1}{n}\Bigl[\sin\Bigl(\dfrac{1}{1+(\dfrac{1}{n})^2}\Bigr)+\sin\Bigl(\dfrac{1}{1+(\dfrac{2}{n})^2}\Bigr)+...\Bigr]$
Hence the function is $\sin\Bigl(\dfrac{1}{1+x^2}\Bigr)$ and it goes from $0$ to $1$
$\displaystyle\int_0^1\sin\Bigl(\dfrac{1}{1+x^2}\Bigr)dx=?$
This is not an easy integral to compute, but the main question in this exercise is i think to convert the sum into an integral and not the computing, So maybe i did a mistake somewhere ?
In this case, as $n \to \infty$, the argument of the sine function is small so that we may make the approximation
$$\lim_{n\to\infty}\sum_{k=1}^n \sin{\left ( \frac{n}{n^2+k^2}\right )} = \lim_{n\to\infty}\frac1{n}\sum_{k=1}^n \frac1{1+\frac{k^2}{n^2}}$$
which is indeed a Riemann sum with limit
$$\int_0^1 \frac{dx}{1+x^2} = \frac{\pi}{4} $$