Riemann Sum Question: Is my Answer Correct?

Checkfate
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Hello, just going through some Riemann sum problems before I hit integrals and I am like 99% sure that this answer from my text is wrong but I want to make sure. It's not really an important question so if you have better things to do, help the next guy :) But checking this over would be appreciated!

http://img178.imageshack.us/img178/875/incorrectyp5.jpg

I am trying to find an approximation of the integral using a Riemann sum.

Here is the question.

http://img216.imageshack.us/img216/2113/incorrect2jj8.jpg

Why are they using 32 as the denominators for their endpoints? I have the same numerators but my denominator is 12 since \Delta x=\frac{\pi}{6}. And for my first midpoint I would use half that.. \frac{\pi}{12}, and then add delta x again n times, right?
 
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It's sec(x/3).
 
Sigh, how dumb of me... thanks, arg!

If I had f()'s around my fractions I would have been fine, forgot about that. :( Thanks though, and sorry.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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