Line integral with parametrization from (1,0,1) to (0,1,1)

[Ted123]

Homework Statement

[PLAIN]http://img690.imageshack.us/img690/4543/vectort.jpg

For part (iv) "starting at $(1,0)$ and ending at $(0,1)$" should read "starting at $(1,0,1)$ and ending at $(0,1,1)$".

Homework Equations

The Attempt at a Solution

I've done the 1st 3 parts (I think) - can anyone check whether they agree with my answers please. For part (iv) I've parametrised it as follows but can't see how to do the integration.

My answers for parts (i)-(iii):
(i) $\frac{69}{70}$

(ii) $\frac{29}{3}$

(iii) $\frac{1}{4}(96+\pi- 24\pi ^2 + \pi ^4)$

My attempt at part (iv) - how do I do the integration?

${\bf p}(t) = (\text{cos}^3\,t , \text{sin}^3\,t , 1)\;\;\;\;0\leq t \leq \frac{\pi}{2}$

${\bf p}'(t) = (-3\text{sin}\,t\,\text{cos}^2\,t, 3\text{sin}^2\,t\,\text{cos}\,t , 0)\;\;\;\;0\leq t \leq \frac{\pi}{2}$

And so the integral equals:

$\displaystyle \int^{\frac{\pi}{2}}_0 \left[ xy\frac{dx}{dt} + yz\frac{dy}{dt} + zx\frac{dz}{dt} \right]\;dt = \int^{\frac{\pi}{2}}_0 \left[ 3\text{cos}^5\,t\,\text{sin}^4\,t\, + 3\text{sin}^5\,t\,\text{cos}\,t \right]\;dt$

 

[hunt_mat]

I get the final integral to be:
$$-9\int_{0}^{\frac{\pi}{2}}\cos^{6}t(1-\cos^{2}t)^{2}\sin tdt$$
Use the substitution $$u=\cos t$$

 

[Ted123]

I get the final integral to be:
$$-9\int_{0}^{\frac{\pi}{2}}\cos^{6}t(1-\cos^{2}t)^{2}\sin tdt$$
Use the substitution $$u=\cos t$$

How does

$\displaystyle -9\int_{0}^{\frac{\pi}{2}}\cos^{6}t(1-\cos^{2}t)^{2}\sin tdt = \int^{\frac{\pi}{2}}_0 \left[ 3\text{sin}^5\,t\,\text{cos}\,t - 3\text{cos}^5\,t\,\text{sin}^4\,t\, \right]\;dt$ ?