Arc length using Integration In Thomas's calculus $10^{th}$ edition, chapter $5$, exercises $5.3$, problem $9$, I am asked to find the length of an arc with the equation:
$$x=\int^y_0\sqrt{\sec^4(t)-1} dt$$
and $\displaystyle {\frac{-\pi}{4}} \leq y \leq {\frac{\pi}{4}}$.
The problem is, I'm not entirely sure what the equation is supposed to be... The previous problems were relatively straight forward, with simple $x=f(y)$ equations. The second part, the parameters of $y$, is confusing me-- is this a parametric of some sort (considering the "$t$")? I've looked around the chapter and can't seem to find another example like it...
If someone could point me in the right direction, that'd be really helpful.
 A: Each value of $y$ gives a particular value of $x$, so that $x$ is a function of $y$. The graph of this function, from $y=-\frac{\pi}{4}$ to $y=\frac{\pi}{4}$ is what you are meant to  be computing.
The reason they are giving you this complicated-looking function is that it will make the integral for computing the arc length very easy. Remember that if $x=f(y)$ is a function, then the arc length of the graph of $x=f(y)$ from $y=a$ to $y=b$ is given by
$$\mathrm{Length} = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy.$$
If $$x(y) = \int_0^y\sqrt{\sec^4(t)-1}\,dt$$
then by the Fundamental Theorem of Calculus, what is $\frac{dx}{dy}$?
A: In general, the length $s$ is given by:
$$
s = \int_{t_0}^{t_1} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \, dt
$$
In your exercise, $x$ is a function of $y$. Therefore:
$$
s = \int_{y_0}^{y_1} \sqrt{\left(\frac{dx}{dy}\right)^{2} + 1} \, dy
$$
Now:
$$
x = \int^y_0 \sqrt{\sec^4(t) - 1} \, dt = F(y) - F(0)
$$
Where $F$ is an antiderivative of $\sqrt{\sec^4(t) - 1}$.
Therefore:
$$
\frac{dx}{dy} = F^{\prime}(y) = \sqrt{\sec^4(y) - 1}
$$
Can you take it from here?
