Solve path integral

[Guest]

Good morning, I'm hesitating with the following exercise:

Let [tex]\sigma[/tex] be the path:

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

Evaluate the integral

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

So that's the statement. Now, what I've done is changing the variables:

[tex]\mathrm{d}x=-60\sin t\cos^{59}t\, \mathrm{d}t[/tex]

for the three x, y and z. Then the integral becomes

[tex]\int^{59π/2}_{0}−60\cos^{59}t\sin^{59}tdt+\int^{59π/2}_{0}60\sin^{59}t\cos^{59}tdt+\int^{59π/2}_{0}\cos^2t\sin^2tdt[/tex]

Is that procedure correct? I mean, would the solution to the sum of the three integrals be the solution of the first integral with [tex]\sigma[/tex]?

Thanks in advance

[Guest]

Yes, that is correct. And since the first two integrals have odd powers of both sine and cosine, you can factor out one to use with differential. For example to integrate [tex]sin^{59}(t)cos^{59}(t) dt[/tex], write it as [tex]sin^{59}(t)cos^{58}(t)(cos(t)dt)= sin^{59}(x)(cos^2(t))^{29}(cos(t)dt)= sin^{59}(t)(1- sin^2(t))^{29}(cos(t)dt)[/tex] and use the substitution [tex]u= sin(t)[/tex] so that [tex]du= cos(t)dt[/tex] and the integrand becomes [tex]u^{59}(1- u^2)^{29}du[/tex]. That is not at all complicated but it does look tedious!