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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. –  Ron Gordon Dec 23 '12 at 11:32
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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. –  AndreasS Dec 23 '12 at 12:02
    
@AndreasS Looking at a picture 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 $\tan (x)$. –  Matt N. Dec 25 '12 at 9:41
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2 Answers

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 i fully understand the above method in order to work out integrals as it was one of the first ways i was taught, however the question im currently doing asks for me to work the above integration using Riemanns Integral –  jill Dec 23 '12 at 11:37
    
@jill The integral I am using is the Riemann Integral –  Nameless Dec 23 '12 at 11:40
    
i understand the approach you took, however, that doesn't explain how you've worked it out using riemann's integral? as you've used th fundamental theorem and just in putted the answer into the riemanns summation –  jill Dec 23 '12 at 11:46
    
The indefinite integral is not the Riemann Integral. The definite one is however, and is related to the indefinite one via the 2nd fundumental theorem of calculus –  Nameless Dec 23 '12 at 11:48
    
ohhhhh okay now i see, so to compute this simply using riemanns theory would not be possible? as we'd have to use the rules of fundamental calculus? –  jill Dec 23 '12 at 11:51
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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 $0=x_0<x_1<\ldots< x_N={\pi\over3}$ 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 $$\lim_{N\to\infty} R_N=\lim_{N\to\infty}{2^{1/N}-1\over 1/N}=\log 2\ .$$

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