Question
Use Riemann sums to evaluate the integral $$\int_0^{\frac {\pi} {3}} \sin x\ \mathrm{d}x\ .$$
Hint
Use the fact that $$\sum_{i = 1}^n \sin (i\theta) = \frac {\sin \frac {(n + 1)\theta} {2} \sin \frac {n\theta} {2}} {\sin \frac {\theta} {2}}.$$
My Working
So far, I have gotten to the step where $$\int_0^{\frac {\pi} {3}} \sin x\, \mathrm{d}x =
\lim_{n \to \infty}\ \frac {\pi} {3n}\sum_{i = 1}^n \sin \frac {i\pi} {3n}.$$
Here is where I am stuck. Using the hint given, can I simply let $\theta$ = $\frac {\pi} {3n}$ and continue? I am not sure if this is mathematically correct as I see that in the hint, $\theta$ does not depend on $n$. However, if I make such a substitution, then it will, but again, I am not sure if this is relevant.
Any help will be greatly appreciated!
Best Answer
Since the hint for the evaluation of $\sum_{i=1}^{n} sin(i\theta)$ is provided, it probably wouldn't hurt to try to substitute in your choice of $\theta$ to obtain a preliminary result, and see whether you are able to evaluate it nicely before proceeding with the full working. Starting from your last equation:
\begin{align*} \int_0^{\frac{\pi}{3}} \sin x\ dx &= \lim_{n \to \infty} \frac{\pi}{3n}\sum_{i=1}^{n} \sin \frac{i\pi}{3n} \\ &= \lim_{n \to \infty} \frac{\pi}{3n}\frac{\sin \left(\frac{n+1}{2}\cdot\frac{\pi}{3n}\right) \cdot \sin \frac{\pi}{6}}{\sin \frac{\pi}{6n}} \\ &= \lim_{n \to \infty} \frac{\pi}{3n}\frac{\sin \left(\frac{\pi}{6} + \frac{\pi}{6n}\right) \cdot \sin \frac{\pi}{6}}{\sin \frac{\pi}{6n}} \end{align*}
and the rest should be manageable.
Hint: If you're still stuck in 'removing' terms with denominator $n$, consider how you may use the known limit $\lim_{x \to 0}\frac{\sin x}{x} = 1$.