Average value of a function

In order to approximate the average value, I take equally spaced values across the interval $[0,\frac{\pi}{3}]$ , substitute them into the function, and then take the average. In order to produce equal spacing across the entire interval, the values I take from the interval must take the form $\frac{\pi}{3}\frac{k}{n}$ for each integer $k$ such that $0\leq k\leq n$ . (Note that equal spacing is not required, but it works all the same.) As the number of points approaches infinity, the idea is that the approximation will get closer to the true value. The formula for this average is $AM=\sum_{k=0}^n \frac{\tan{(\frac{\pi}{3}\frac{k}{n})}}{n}$ . Next I compare this to the Riemann sum $R=\sum_{k=0}^n\tan{(\frac{\pi}{3}\frac{k}{n}})\frac{\pi/3}{n}$ , which describes the area under the curve. There is a simple relationship between these two formulas, namely, $AM=\frac{3}{\pi}R$ . The Riemann sum can be calculated using the Fundamental Theorem of Calculus (in other words you can use the antiderivative of the function to calculate the area/Riemann sum). The solution then follows as: $\Rightarrow AM=\frac{3}{\pi}\int_0^{\pi/3} \tan{x}dx=\frac{3}{\pi} \int_0^{\pi/3} \frac{\sin{x}}{\cos{x}}dx=\frac{3}{\pi} \int_0^{\pi/3} \frac{\sin{x}}{\cos{x}}\frac{dx}{d(\cos{x})}d(\cos{x})=\frac{3}{\pi} \int_0^{\pi/3} \frac{\sin{x}}{\cos{x}}\Big(\frac{d(\cos{x})}{dx}\Big)^{-1}d(\cos{x})=\frac{3}{\pi} \int_0^{\pi/3} \frac{\sin{x}}{\cos{x}}\Big(-\sin{x}\Big)^{-1}d(\cos{x})$ $=-\frac{3}{\pi} \int_0^{\pi/3} \frac{1}{\cos{x}}d(\cos{x})=-\frac{3}{\pi}\ln{(\cos{x})}\Big|_0^{\pi/3}=-\frac{3}{\pi}(\ln{(\cos{\frac{\pi}{3}})}-\ln{(\cos{0})})=-\frac{3}{\pi}(\ln{(\frac{1}{2})}-\ln{(1)})=-\frac{3}{\pi}(\ln{(\frac{1}{2})})=\frac{3}{\pi}\ln{\Big(\Big(\frac{1}{2}\Big)^{-1}\Big)}=\frac{3}{\pi}\ln{2}$ .