Analyzing a Smooth Curve for -π < t < π

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Homework Statement



Determine where r(t) is a smooth curve for -pi <t<pi
R(t)= (x(t),y(t))=(4sin^3(t), 4cos^3(t))

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The Attempt at a Solution



To be honest I have no idea where to start. I know what a smooth function is but my understanding is that the sin(t) and cos(t) functions over all of t are smooth. No corners.
Any starting help would be appreciated.
 
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Yes, but you are NOT asked if sine and cosine are smooth- you are asked if F is smooth. What happens if the denominator of a fraction goes to 0? What fraction is involved here?
 
The function is not continuous at that particular point that makes the denominator go to zero.
Perhaps we could rewrite in the complex plane?
 
No, it is not necessary to work with the complex plane. What is the definition of "smooth curve"?
 
As someone on mathstackexchange said, a smooth curve is a curve with no stubble, like this: :bugeye:
 

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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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