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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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    Not sure if this is the best way to go about computing this integral. You really want to consider the antiderivative of $\tan{x}$ and use the Fundamental Theorem of Calculus.2012-12-23
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    When I look at this (and the answer using the FToC) - and especially the result - I asked myself if one could compare this series to the alternating harmonic series. For example bound it from above or below by the alternating series. I don't know if this could work, but it would seem interesting. At least the way using the FToC is far more efficient and usable.2012-12-23
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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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