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The Question i have is: Calculate the following Riemann Integral
$\int_0^\frac{\pi}3 \tan(x) \,dx.$

I know that $\int_a^b f(x) \, dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*) \Delta X$
and so I've worked out $\Delta X = \frac {b-a} n = \frac {\frac \pi 3} n = \frac \pi {3n}$
and also $ x_i^* = a+ (\Delta X)i = 0 + (\frac \pi {3n})i$.

So for my question I know that the $\int_0^ \frac\pi 3 tan(x) \, dx = \lim_{n\to\infty} \sum_{i=1}^n \tan((\frac \pi {3n})i) \times \frac \pi {3n} $

but I am not 100% sure where to go from here.

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    @AndreasS Looking at a [picture](http://i.stack.imgur.com/SQXdk.png) suggests that the tangent curve is the same as the $1/x$ curve flipped horizontally at an axis near $1$ and slightly shifted upwards. Perhaps one can figure out how to center the point of intersection of both functions in $[0,\pi/3]$ so that one can compute the integral of $\tilde{f}(x)$ obtained from $1/x$ instead of $\tan (x)$.2012-12-25

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Calculating integrals this way can be very hard... That's why we have the Funcdumental Theorem of Calculus:

$\int \tan x\, dx=\int\frac{\sin x}{\cos x}\, dx$ Substituting $u=\cos x $ yields $\int \tan x\, dx=\int\frac{-1}{u}\, dx=-\ln\left|u\right|+c=-\ln\left|\cos x\right|+c$ Therefore, by the 2nd Fundumental Theorem of calculus, $\int_0^\frac{\pi}3 \tan(x)\, dx=-\ln\frac12+\ln1=\ln 2$ This also implies that $\lim_{n\to\infty} \sum_{i=1}^n\frac{\pi}{3n} \tan((\frac \pi {3n})i)=\ln 2 $

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    @rlgordonma im assuming the question isn't asking for the evaluation as it simple asks "calculate the following Riemann integrals" and then gives a couple of questions2012-12-23
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The integral in question can indeed be computed as a limit of Riemann sums.

We consider Riemann sums $R_N:=\sum_{k=1}^N \tan(\xi_k)(x_k-x_{k-1})\qquad(1)$ where the partition $0=x_0 is chosen as follows: $x_k:=\arccos\bigl(2^{-k/N}\bigr)\qquad(0\leq k\leq N)\ ;$ and the sampling points $\xi_k \in [x_{k-1},x_k]$ are chosen later.

Fix $k$ for the moment. Then $x_k-x_{k-1}=\arccos'(\tau)\bigl(2^{-k/N}-2^{-(k-1)/N}\bigr)\qquad(2)$ for some $\tau\in\bigl[2^{-k/N},\>2^{-(k-1)/N}\bigr]$. Now

$\arccos'(\tau)={1\over\cos'(\arccos\tau)}=-{1\over \sin\xi}\ ,\qquad(3)$ where $\cos\xi=\tau$. It follows that $2^{(k-1)/N}\leq{1\over\cos\xi}\leq 2^{k/N}$ or ${1\over\cos\xi}=2^{k/N}\cdot 2^{-\Theta/N}$ for some $\Theta\in[0,1]$. Now chose this $\xi$ as the $\xi_k$ in $(1)$. Then we get, using $(2)$ and $(3)$: $R_N=\sum_{k=1}^N{\sin\xi_k\over \cos\xi_k}{1\over\sin\xi_k}\bigl(2^{-(k-1)/N}-2^{-k/N}\bigr)=\sum_{k=1}^N 2^{-\Theta_k/N}(2^{1/N}-1)\ .$ For large $N$ the factors $2^{-\Theta_k/N}$ are arbitrarily close to $1$. Therefore the last sum essentially consists of $N$ terms of equal size $2^{1/N}-1$. (The obvious squeezing argument can be supplied by the reader.) It follows that $\lim_{N\to\infty} R_N=\lim_{N\to\infty}{2^{1/N}-1\over 1/N}=\log 2\ .$