this is my curve : $r(t)=(\cos{t},\sin{t}-1,2\cos{\frac{t}{2}})$ , $t=[0,3\pi]$
$r'(t)=(-\sin{t},\cos{t},-\sin{t})$
$||r'(t)||=\sqrt{\sin^2{t}+1}$
I have to calculate: $\int{(y+1)}ds$
So I have : $\int_{0}^{3\pi}\sin t\sqrt{\sin^2{t}+1}dt$
Is this right? If it is, can you help me figure out how to compute this integral? I tried a lot of substitution but I can't get it any simpler.
$r(t)=(\cos{t},\sin{t}-1,2\cos{\frac{t}{2}})$ , $t=[0,3\pi]$
$r'(t)=(-\sin{t},\cos{t},-\sin{\frac{t}{2}})$ <- You forgot sin(t/2)
$||r'(t)||=\sqrt{\sin^2{\frac{t}{2}}+1}$
Calculate: $\int{(y+1)}ds$
-> $I = \int_{0}^{3\pi}\sin t\sqrt{\sin^2{\frac{t}{2}}+1}dt$
Apply subsitution $u = sin^2(\frac{t}{2}) + 1$
$du = \frac{\sin{t}}{2} dt$
$t = 0, u = 1. t = 3\pi,u = 2.$
$I = 2\int_{1}^{2}\sqrt{u}\textrm{ }du = \frac{4}{3}(2^\frac{3}{2}-1^\frac{3}{2})=\frac{4}{3}\times(2\sqrt{2}-1)=\frac{8\sqrt{2}-4}{3}$