The length $\;L\;$ of the curve C given by
$\displaystyle \;\;x=3\cos\!\left(6t\right), \;\;y=18t+3\sin\!\left(6t\right), \, \;\; \displaystyle \;\; 0 \le t \le \frac{\pi }{6}\;\;$
is found by evaluating an integral of the form $\;\;{\displaystyle \int_{\alpha}^{\beta} \! J(t) \, dt }\;\;$ for a suitable integrand $\;J(t)\;$ and constants $\;\alpha\;$ and $\;\beta$.
So I found the derivative of:
x' = $18 \sin(6t)$
y' = $18 + 18 \cos(6t)$
Length of the curve is found by formula:
$L(c) = \;\;{\int_{\alpha}^{\beta} \! \sqrt{x'^2 + y'^2} \, dt }\;\;$
And I get this inside the integral:
$\int_{0}^{\pi/6} \sqrt{324sin^2(6t) + (324 + 648cos(6t) + 324cos^2(6t)} dt $
Then simplifying the integral:
$18 \cdot \int_{0}^{\pi/6} \sqrt{1+1+2 \cos(6t)} dt$
$ 18 \sqrt{2} \int_{0}^{\pi/6} \sqrt{1+\cos(6t) \cdot \frac{1-\cos(6t)}{1-\cos(6t)}} = \sqrt{\frac{\sin^2(6t)}{1- \cos(6t)} } dt$
Then doing u-substitution = $1-\cos(6t)$ I end up with:
$-3 \sqrt{2} \int_{0}^{2} u^{-1/2} du$
Evaluates to:
$-6 \sqrt{2} \cdot (1-\cos(6 \cdot \frac{\pi}{6})^{1/2} - 0$
But it's wrong, why?
I also just tried by leaving u, and get -12, but it's not correct.
You have erroneously introduced a minus sign when you have done your u-substitution. You were doing:
$$18\sqrt{2}\int_0^{\frac{\pi}{6}}\frac{\sin 6t}{\sqrt{1-\cos 6t}}dt$$
Let $u=1-\cos 6t$ so $du=6\sin 6t$
So your integral becomes:
$$3\sqrt{2}\int_0^2u^{-\frac{1}{2}}du$$