# Solve path integral

### Solve path integral

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

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

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

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

$$\mathrm{d}x=-60\sin t\cos^{59}t\, \mathrm{d}t$$

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

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

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

Guest

### Re: Solve path integral

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 $$sin^{59}(t)cos^{59}(t) dt$$, write it as $$sin^{59}(t)cos^{58}(t)(cos(t)dt)= sin^{59}(x)(cos^2(t))^{29}(cos(t)dt)= sin^{59}(t)(1- sin^2(t))^{29}(cos(t)dt)$$ and use the substitution $$u= sin(t)$$ so that $$du= cos(t)dt$$ and the integrand becomes $$u^{59}(1- u^2)^{29}du$$. That is not at all complicated but it does look tedious!
Guest