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$y = \sin(\pi x), 0 \le x \le 1$

$y' = \pi \cos(\pi x)$

$(y')^2 = \pi^2\cos^2(\pi x)$

$ds = \sqrt{1 + \pi^2 \cos^2(\pi x)}$

$r = y = \sin(\pi x)$

$S = \displaystyle\int_0^1 2 \pi\sin(\pi x)\sqrt{1 + \pi^2 \cos^2(\pi x)} dx$

I'm a little lost as to what to do next. Should I have simplified ds more or do I need to do a substitution with $u = \pi^2\cos^2(\pi x)$?

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Your questions can be much easier to read if you $\LaTeX$ them. If you right click and Show Source you can see how, then enclose it in dollar signs. – Ross Millikan Mar 7 '11 at 19:16
up vote 1 down vote accepted

Hint: I think a substitution of $u=\pi \cos(\pi x)$, will help as you have $du$ available.

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After doing that I get: the integral from 0 to 1 of [(-2/pi) sqr(1 + u^2pi) du]. Would I have to do a trig substitution? – Krysten Mar 7 '11 at 20:35
I don't think you should have a $\pi$ under the square root, and $\sqrt{1+u^2}du$ may be in your tables, or you can substitute $u=\tan(\theta)$ But I think the limits in $u$ are not 0 to 1. – Ross Millikan Mar 7 '11 at 20:49
ahh now i see! thanks – Krysten Mar 8 '11 at 0:47

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