Solving path integral

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Let $\sigma$ be the path:

$$x=\cos^{60}t, \;\;\;\;y=\sin^{60}t,\;\;\;\;z=t, \;\;\;\;0 \le t \le \frac{59 \pi}{2}$$

Evaluate the integral

$$\int_{\sigma}\sin^{58}z\;dx+\cos^{58}z\;dy+(x^2y^2)^{\frac{1}{60}}dz$$

The first thing I've tried to do is to obtain the $f$ function in terms of $x,y$ and $z$ and then apply the path integral theorem:

$$\int_{\sigma}f \;ds = \int_{a}^{b} f(x(t),y(t),z(t))||\sigma'(t)||dt$$

And I reached that $f(x,y,z) = \sin^{58}z+cos^{58}z+(x^2y^2)^{\frac{1}{60}}$, but I'm not sure it is correct. Could you give me any hint?

Thanks in advance.

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Just do it as three integrals. In the first one we have $$x=cos^{60}z=cos^{60}t$$ so that $$\mathrm{d}x=-60\sin t\cos^{59}t\, \mathrm{d}t$$ Then the integral becomes$$-60\int_0^{59\pi/2}\sin^{59}t\cos^{59}t\,\mathrm{d}t$$