So my components are $$f(x)= 3\cos t, \ \text g(y)=4\sin t ,\ \text h(z)= t$$
$$f'(x)=-3\sin t, \ \text g'(y)=4\cos t,\ \text h'(z)=1$$
I found the derivatives of each component and plugin to the arc length and have this
$$\int_ 0 ^{4\pi} \sqrt{9\sin^2t+16\cos^2t+1} dt $$
After this I'm stuck. I saw examples on YouTube where they used trig identities $$\sin^2t + \cos^2t = 1$$ but the coefficient are not the same so I'm not able to factor out the coefficient so what now? Please help.
On
the integral $$\int_ 0 ^{4\pi} \sqrt{9\sin^2t+16\cos^2t+1} dt $$ can be rewrited as $$\int_ 0 ^{4\pi} \sqrt{17-7\sin^2t} dt $$ and then you factor out the sqrt17 leaving you with :
$$\int_ 0 ^{4\pi} \sqrt{17}\sqrt{1-(7/17)\sin^2t} dt $$ which is a special integral, an incomplete elliptic integral of the second kind, which result is given by 8sqrt(17)E(7/17)
you can find more about these types of integrals and how to solve them here https://en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_second_kind
Actually the integral $$ \int_0^{4\pi} \sqrt{9\sin^2t+16\cos^2t+1} dt = 8\int_0^{\pi/2} \sqrt{9\sin^2t+16\cos^2t+1} dt $$ cannot be solved by elementary methods. The answer involves elliptic integrals of the second kind:
In other words let $$ E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2 x}dx $$ and solve it in terms of $E$. You can use $16\cos^2t=16(1-\sin^2t)$ to get started.