Find Area of Polar Curve: Cos(6ǿ)

gr3g1
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Hey guys,

I just wanted to know, if my polar curve is cos(6ǿ) and i have to find the length of one peddle... can i use the interval of 0 and Pi/12?

Thanks a lot!
 
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One peddle? Oh! One "petal"! cosine goes from 0 to 0 between \pi/2 and \pi/2[/tex]. 6\theta= \pi/2 when \theta= \pi/12.<br /> <br /> I would recommend using -\pi/12 to \pi/12. 0 to \pi/12 will only give half of a petal. Of course, if you are really clever, you could use 0 to \pi/12 and then multiply by 2.
 
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lol, sorry, i meant petal :P
Thanks a lot
 
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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