Solving the riemann sum of a trig function

Question

I know the function $f$ in this case will be $cos\frac{4k\pi}{5n}$. I'm quite sure how to compute the function. I tried U substitution but it's kind of messy, is there a simpler way to do this?

Answer 1

Expressing the integral using a Right-Hand Sum, you have $$\int_0^{4\pi/5}f(t)dt = \lim_{n\rightarrow\infty}\sum_{k=1}^n f\left(0+\frac{4k\pi/5}{n}\right)\left(\frac{4\pi/5}{n}\right)= \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n \left(\frac{4\pi}{5}\right)f\left(\frac{4k\pi}{5n}\right). $$ This matches what you're given if $$\cos\left(\frac{4k\pi}{5n}\right) = \left(\frac{4\pi}{5}\right)f\left(\frac{4k\pi}{5n}\right), $$ i.e., if $$f\left(\frac{4k\pi}{5n}\right) = \left(\frac{5}{4\pi}\right)\cos\left(\frac{4k\pi}{5n}\right). $$ So let $$f\left(t\right) = \left(\frac{5}{4\pi}\right)\cos\left(t\right). $$ From here you can just compute the integral.