Can someone give me a few hints about how to solve this problem?

Question

Evaluate the line integral

$$zdx+xdy+xydz $$

where $C$ is the path of the helix

$$r(t)=4\cos t \,\textbf{i} +4\sin t \,\textbf{j}+3t\,\textbf{k} \;,\;\; 0\le t \le 2\pi$$

I've already posted this question wondering whether if my work was write. I solved it and plugged it into lon capa but it is incorrect. Can someone just give me a hint as to how i should start this problem?

http://i1317.photobucket.com/albums/t638/ayoshnav/Snapshot_20140330_5_zps4e599465.jpg

I took the integral from 0 to 2(pi) of -12tsin(t) + 16(cost(t))^2 + 48cos(t)sin(t). I broke it up into three separate integrals. The first one was -12tsin(t) from 0 to 2(pi). I evaluated this by computing uv - the integral of vdu. I took u to be -12t and du = -12. I let dv= sin(t) and v= -cost. So from that I got 12tcos(t) - the integral from 0 to 2(pi) of 12cos(t).

Then I looked at the second integral. I know the (cost)^2 is (1 + cos(2t))/2. Evaluating that integral I get 8t + 4sin(2t) from 0 to 2(pi).

Finally for the last one I got (48(sint)^3)/2 from 0 to 2(pi).

Adding those three and plugging 2(pi) and 0 in I get 12tcos(t) + 16(pi)

Answer 1

Hint: develop the integral

$$I:=\int\limits_0^{2\pi}\left( 3t(-4\sin t)+4\cos t(4\cos t)+16\cos t\sin t\cdot (3)\right)dt$$

Added:

$$I:=\int\limits_0^{2\pi}\left(-12t\cos t+16\cos^2t+\overbrace{24\sin 2t}^{=48\sin t\cos t}\right)dt=$$

$$=\left.12t\cos t\right|_0^{2\pi}-12\;\overbrace{\int\limits_0^{2\pi}\cos t\,dt}^{=0}+8\left(t+\sin t\cos t\right)_0^{2\pi}-\left.12\cos 2t\right|_0^{2\pi}=$$

$$=24\pi+8(2\pi)-12(1-1)=40\pi$$