<s>[INST] Find length of the arc of $y^2=x$ I am embarrassingly stuck on this example. My textbook provides the answer (picture below) and the steps, but I am unable to follow the math. I have been stuck for more then an hour.
Find the length of the arc of the parabola $y^2=x$ from (0,0) to (1,1). 
Since $x=y^2$, we have $\frac{dx}{dy} = 2y$.
$$\begin{align}
L &= \int_{0}^{1} \sqrt{1+(\frac{dx}{dy})^2}dy \\
&= \int_{0}^{1} \sqrt{1+(2y)^2}dy
\end{align}$$
Here is where I am getting stuck. I am supposed to make a trigonometric where $y = \frac{1}{2}\tan\theta$, but based off the trig substitution equation I have, $\sqrt{a^2+y^2} \rightarrow y = \tan\theta$ I am getting confused how I am supposed to get $y = \frac{1}{2}\tan\theta$. I keep thinking it should be $ y = \frac{1}{4}\tan\theta$.
Here is where I get my $y$ value from: 
$$\sqrt{a^2 + y^2} = \sqrt{1+ 2^2 y^2} \\
= \sqrt{\frac{1}{4} + y^2}$$ Which gives me $y = \frac{1}{4} tan\theta$
Here is a picture of the example in the textbook. I am not able to follow the math :(

 [/INST] If you have a sum under the square root sign ($\sqrt{a^2+x^2}=\sqrt{\frac{1}{a^2}\left(1+a^2x^2\right)}=\frac{1}{a}\sqrt{1+a^2x^2},\ a>0$), your trigonometric substitution would be $x=\frac{1}{a}\tan{\theta}$:
$$
\int\sqrt{1+4y^2}\,dy=\int\sqrt{1+2^2y^2}\,dy\rightarrow y=
\frac{1}{2}\tan{\theta}, dy=\frac{dy}{d\theta}d\theta=
\frac{1}{2}\sec^2\theta\,d\theta\\
\int\sqrt{1+4\left(\frac{1}{2}\tan{\theta}\right)^2}\frac{1}{2}\sec^2\theta\,d\theta=
\frac{1}{2}\int\sqrt{1+4\cdot\frac{1}{4}\tan^2{\theta}}\sec^2\theta\,d\theta=\\
\frac{1}{2}\int\sqrt{1+\tan^2{\theta}}\sec^2\theta\,d\theta=
\frac{1}{2}\int\sqrt{\sec^2\theta}\sec^2\theta\,d\theta=\\
\frac{1}{2}\int|\sec{\theta}|\sec^2\theta\,d\theta=
\frac{1}{2}\int\sec^3\theta\,d\theta.
$$
Here, $\sec\theta>0$, so we can drop the absolute value sign. That's because we're considering only an interval on which the secant function is positive (the interval of choice is typically $0<\theta<\frac{\pi}{2}$).
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