# calculate this line integral

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.

• You have a mistake in differentiation of $\cos\frac{t}{2}$ (you skipped the 1/2 in $\sin$). – orion Jan 19 at 23:43
• @orion it’s $2\cos\frac{t}{2}$, so he doesn’t. – cluelessatthis Jan 20 at 0:45
• It should be $-\sin \frac{t}{2}$, not $-\sin t$. The half doesn't disappear from inside the trigonometric function after differentiation. – orion Jan 20 at 0:52

$$r(t)=(\cos{t},\sin{t}-1,2\cos{\frac{t}{2}})$$ , $$t=[0,3\pi]$$

$$r'(t)=(-\sin{t},\cos{t},-\sin{\frac{t}{2}})$$ <- 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 $$u = sin^2(\frac{t}{2}) + 1$$

$$du = \frac{\sin{t}}{2} dt$$

$$t = 0, u = 1. t = 3\pi,u = 2.$$

$$I = 2\int_{1}^{2}\sqrt{u}\textrm{ }du = \frac{4}{3}(2^\frac{3}{2}-1^\frac{3}{2})=\frac{4}{3}\times(2\sqrt{2}-1)=\frac{8\sqrt{2}-4}{3}$$

• why is It $du=\frac{sint}{2}$ ? how did you get $\frac{sint}{2}$? – NPLS Jan 20 at 8:44
• $u = sin^2(\frac{t}{2}); du = 2sin(\frac{t}{2})cos(\frac{t}{2})\times \frac{1}{2}$ By Chain Rule. Simplify. – Gareth Ma Jan 21 at 9:05