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$\int_C xds$ where $x=\frac{3}{4}sin(2t)$ $y=\cos^3 t$ $z=\sin^3 t$ How do I solve this type of integrals?

I don't understand how integrate because $ds=\sqrt{dx^2+dy^2+dz^2}dt$ but I can't see how factorize properly to simplify the integral. Any help?

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\begin{eqnarray} ds &=& \sqrt{dx^2 + dy^2 + dz^2} \\ &=& \left[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2\right]^{1/2}dt \\ &=& \left[\frac{9}{4}\cos^2(2 t) + 9 \cos^4t \sin^2t + 9 \cos^2t \sin^4t \right]^{1/2}dt \\ &=&\left[\frac{9}{4}\cos^2(2 t) + 9 \cos^2t \sin^2t\underbrace{(\cos^2t + \sin^2 t)}_{\color{blue}{1}} \right]^{1/2} dt \\ &=& \frac{3}{2} \left[\cos^4t -2\cos^2t\sin^2t +\sin^2t + 4 \cos^2\cos^2t\right]^{1/2}dt \\&=& \frac{3}{2}[(\cos^2 t + \sin^2)^2]^{1/2} dt\\ &=& \frac{3}{2}dt \end{eqnarray}

The integral then becomes

$$ \int_C x~ds = \frac{3}{2}\int x(t)dt $$

Can you take it from here?

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  • $\begingroup$ Thanks, I can't see the factorization before, really thanks. $\endgroup$ – Ragnar1204 Sep 18 '17 at 12:16
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Hint: $\int_C f(x,y,z) ds =\int_a^bf(x(t),y(t),z(t))\sqrt{x'(t)^2+y'(t)^2+z'(t)^2} dt$ where $a\leq t \leq b$

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