What is this problem asking? (Arc length) All of the previous questions are asking me to find the distance of a line in between a set of x-values, for example the distance of $y=x^2$ when $-1\leq x\leq 2$. The problem I don’t understand is asking for the length of $y=\int_0^x \tan t\,dt$ when $0\leq x\leq \frac{\pi}{6}$. How do I go about this?
The arc length of the curve given by  $y=\int_0^x \tan t\,dt$ when $0\leq x\leq \frac{\pi}{6}$ is 
$$
\begin{align*}
\text{ArcLength}(y)
&amp;= \int_0^{\pi/6} \sqrt{1+(y'(x))^2}dx \\
&amp;= \int_0^{\pi/6} \sqrt{1+\tan^2 x} \: dx \hspace{4mm} \mbox{ by the Fundamental Theorem of Calculus}\\ 
&amp;= \int_0^{\pi/6} \sqrt{\sec^2 x} \: dx \\
&amp;= \int_0^{\pi/6} |\sec x| \: dx \\ 
&amp;= \int_0^{\pi/6} \sec x \: dx   \hspace{4mm} \mbox{ since }\sec x \geq 0 \mbox{ on } [0,\pi/6]  \\ 
&amp;= \ln|\sec x + \tan x | \: \bigg|_0^{\pi/6} \\ 
&amp;= \ln \left|\sec \left(\frac{\pi}{6}\right) 
 + \tan \left(\frac{\pi}{6}\right) \right| - \ln|\sec 0 + \tan 0| \\
&amp;= \ln \left(\frac{2}{\sqrt{3}}+ \frac{1}{\sqrt{3}}\right) - \ln (1+0) \\
&amp;= \ln \left( \frac{3}{\sqrt{3}}\right) \\ 
&amp;= \ln\left( \sqrt{3} \right) \\  
&amp;= \boxed{\frac{\ln\left( 3\right)}{2}}. \\ 
\end{align*}  
$$
Hint By the Fundamental Theorem of Calculus,
$$y'= \tan(x)$$