Find the length of $y=x^2$ from $x=0$ to $x=1$ $$L=\int_0^1 \sqrt{1+4x^2} dx = \frac{1}{4}\left(\theta+2\sqrt{5}\right)$$
It tells you to use $2x=\sinh\theta$.
I've tried the best I could, but I can never get the right side of the equation.
Can you help me?
 A: Hint
Use substitution as you wrote above:
$$
\int_0^1 \sqrt{1+4x^2} \;\mathrm{dx}= \left\vert
\begin{array}{c}
2x = \sinh\theta\\
4x^2 = \sinh^2\theta\\
dx = \frac{1}{2}\cosh\theta \;\mathrm{d\theta}
\end{array}
\right\vert = 
\int_{0}^{\log(2+\sqrt{5})}
\sqrt{1+\sinh^2\theta}\frac{1}{2}\cosh\theta\;\mathrm{d\theta}\\
$$
Now use the equality
$$
\cosh^2 \theta - \sinh^2\theta = 1.
$$
Note Do not forget to recalculate the boundaries, it is an easy mistake often done. For that you might want to use the definition of $\sinh \theta$:
$$
\begin{align*}
\sinh \theta &= \frac{e^\theta - e^{-\theta}}{2}\\
\cosh \theta &= \frac{e^\theta + e^{-\theta}}{2}.
\end{align*}
$$
A: Using $2x=\sinh\theta$ yield $x=\frac{1}{2}\sinh\theta$ and $dx=\frac{1}{2}\cosh\theta\,d\theta$. The limit of integral for $0<x<1$ equals $0<\theta<\sinh^{-1}2$. Hence
\begin{align}
\int_0^1 \sqrt{1+4x^2}\,dx&=\int_0^{\sinh^{-1}2} \sqrt{1+4\left(\frac{1}{2}\sinh\theta\right)^2}\frac{1}{2}\cosh\theta\,d\theta\\
&=\frac{1}{2}\int_0^{\sinh^{-1}2} \sqrt{1+\sinh^2\theta}\cosh\theta\,d\theta\\
&=\frac{1}{2}\int_0^{\sinh^{-1}2} \cosh^2\theta\,d\theta\\
&=\frac{1}{2}\int_0^{\sinh^{-1}2} \cosh^2\theta\,d\theta\\
&=\frac{1}{2}\int_0^{\sinh^{-1}2} \frac{1}{2}(1+\cosh2\theta)\,d\theta\\
&=\frac{1}{4}\left[\theta+\frac{1}{2}\sinh2\theta\right]_0^{\sinh^{-1}2}\\
&=\frac{1}{4}(\sinh^{-1}2+2\sqrt{5})
\end{align}
The last step can be calculated as follows, let $y=\sinh^{-1}2$ then $\sinh y=2$ and
$$
\cosh y=\sqrt{1+\sinh^2 y}=\sqrt{5}\qquad\rightarrow\qquad\text{take the principal root}
$$
then
$$
\sinh2\theta=2\sinh\theta\cosh\theta=2\times2\times\sqrt{5}=4\sqrt{5}
$$
