Calculus II, Curve length question. Find the length of the curve
$x= \int_0^y\sqrt{\sec ^4(3 t)-1}dt, \quad 0\le y\le 9$
A bit stumped, without the 'y' in the upper limit it'd make a lot more sense to me.
Advice or solutions with explanation would be very appreciated.

$$\frac{dx}{dy} = \sqrt{\sec^4{3 y}-1}$$
Arc length is then
$$\begin{align}\int_0^9 dy \sqrt{1+\left ( \frac{dx}{dy} \right )^2}  &= \int_0^9 dy\, \sec^2{3 y} \\ &= \frac13 \tan{27} \end{align}$$
