Evaluate $\int_a^b \sqrt{(1+ m^2\cosh^2x)}dx$ $$\int_a^b \sqrt{(1+ m^2\cosh^2x)}dx$$ where $a = 0$ and $b= \ln 2$
I just need a little hint to start. Please don't answer it completely. Just a hint.
Edit : Originally the question was to find the arc length of the curve $f(x) = m \sinh x$ on the interval $[0, \ln 2]$. But I cannot evaluate this integral at all.
 A: $$ I = \int_0^{\ln(2)}  \sqrt{(1+ m^2\cosh^2x)}dx$$
As other user pointed out in the comments this integral is an elliptic integral and there is no way to deal with it out of the theory of elliptic integrals if you want to express it in a closed form.
Do the following change of variable $w = ix$, then :
$$ I = i \int_0^{-i\ln(2)} \sqrt{1+ m^2\cos^2 w}dw  = i \int_0^{-i\ln(2)} \sqrt{1+ m^2(1-\sin^2 w)}dw =  i \int_0^{-i\ln(2)} \sqrt{1+ m^2-m^2\sin^2 w}dw = i\sqrt{1+m^2}\int_0^{-i\ln(2)} \sqrt{1-\frac{m^2}{1+m^2}\sin^2 w} $$
Recall the definiton of the incomplete elliptic integral of the second kind:
$$E(p,\phi) = \int_{0}^{\phi} \sqrt{1-p^2\sin^2 \theta}d\theta \quad q^2=1-p^2$$
where $p$ is the modulus and $q$ is the complementary modulus. $\phi$ is  named the amplitude
Hence,
in your case $p^2 = \frac{m^2}{1+m^2}$ is the modulus, $\phi = -i\ln(2)$ is the amplitude.
$$I = i\sqrt{1+m^2}E\left(\sqrt{\frac{m^2}{1+m^2}},-i\ln(2) \right) $$
In principle, $p$ is real and $|p|<1$ but the theory can be extended to consider complex modulus.
Same thing with the amplitude: the basic theory states that $0\leq \phi \leq \frac{\pi}{2}$. However, the theory can be extended to imaginary amplitudes, as is the case with your integral.
For imaginary amplitude, the incomplete elliptic integral are themselves imaginary and involve the guadermanian function as well as hyperbolic functions:
$$E(p,i\phi) = i\left[F(q,\operatorname{gd}(\phi))-E(q,\operatorname{gd}(\phi)) + \tanh(\phi)\sqrt{1+p^2\sinh^2(\phi)}\right]$$
where $F(\cdot,\cdot)$ is the incomplete elliptic integral of the first kind:
$$F(p,\phi) = \int_{0}^{\phi} \frac{d\phi}{\sqrt{1-p^2\sin^2 \theta}}d\theta$$
Hence, your integral expressed in clsed form is:
$$I = \sqrt{1+m^2}\left[F\left(\frac{1}{\sqrt{1+m^2}},\operatorname{gd}(-\ln(2))\right)-E\left(\frac{1}{\sqrt{1+m^2}},\operatorname{gd}(-\ln(2))\right)-\frac{3}{5}\sqrt{1+\frac{m^2}{1+m^2}\frac{9}{16}}\right]$$
where
$$ q = \sqrt{1-p^2 }= \sqrt{1-\frac{m^2}{1+m^2}} = \frac{1}{\sqrt{1+m^2}}$$
Wolfram evaluates $\operatorname{gd}(-\ln(2)) = 2\cot^{-1}(2)-\frac{\pi}{2}$
Take care when evaluating ellptic integrals with software given that someties they use a slight different notation.
A: If you do not know yet about elliptic integrals, you do not have much choice.
Start using
$$\cosh^2(x)=\frac 1 2(1+\cosh(2x))$$ which makes
$$1+m^2\cosh^2(x)=\frac{1}{2} m^2 \cosh (2 x)+\frac{m^2}{2}+1=\frac{1}{2} m^2\Big[1+\frac{2}{m^2}+\cosh (2 x) \Big]$$
Now, use the series expansion of $\cosh(2x)$. This will give
$$1+m^2\cosh^2(x)=\left(m^2+1\right)+m^2 x^2+\frac{m^2}{3} x^4+\frac{2 m^2 }{45}x^6+\frac{m^2  }{315} x^8+O\left(x^{10}\right)$$
Use the binomial theorem
$$\sqrt{1+m^2\cosh^2(x)}=\sqrt{m^2+1}+\frac{m^2 x^2}{2 \sqrt{m^2+1}}+\frac{m^2 \left(m^2+4\right) x^4}{24
   \left(m^2+1\right)^{3/2}}+\frac{m^2 \left(m^4-28 m^2+16\right) x^6}{720
   \left(m^2+1\right)^{5/2}}+\frac{m^2 \left(m^6+696 m^4-816 m^2+64\right) x^8}{40320
   \left(m^2+1\right)^{7/2}}+O\left(x^{10}\right)$$ Integrate termwise to have an approximate antiderivative. Use the bounds.
Suppose that, with your given bounds, we use this truncated series for $m=3$. This would give as a numerical result $2.3547717$ while using the elliptic integral, the result would be $2.3547702$.
A: 
Define the function $\mathcal{I}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{I}{\left(s,\omega\right)}:=\int_{0}^{\omega}\mathrm{d}\nu\,\sqrt{1+s^{2}\cosh^{2}{\left(\nu\right)}}.$$
As @zinoviev's answer illustrates, the integral actually becomes pretty straightforward once you are sufficiently well-versed in the properties and relations of elliptic integrals. But the involvement of complex variables is a layer of complication I'd rather avoid for elliptic integral newbies.
The following approach may be more longwinded, but it has the appeal of sticking to real variables.

Given $0<s\land0<\omega$ and setting $z:=\cosh{\left(\omega\right)}>1$,
$$\begin{align}
\mathcal{I}{\left(s,\omega\right)}
&=\int_{0}^{\omega}\mathrm{d}\nu\,\sqrt{1+s^{2}\cosh^{2}{\left(\nu\right)}}\\
&=\int_{\cosh{\left(0\right)}}^{\cosh{\left(\omega\right)}}\mathrm{d}x\,\frac{\sqrt{1+s^{2}x^{2}}}{\sqrt{x^{2}-1}};~~~\small{\left[\nu=\ln{\left(x+\sqrt{x^{2}-1}\right)}\right]}\\
&=\int_{1}^{z}\mathrm{d}x\,\frac{\sqrt{1+s^{2}x^{2}}}{\sqrt{x^{2}-1}}\\
&=\int_{1}^{z}\mathrm{d}x\,\frac{1+s^{2}x^{2}}{\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}\\
&=\int_{1}^{z}\mathrm{d}x\,\frac{1}{\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}+\int_{1}^{z}\mathrm{d}x\,\frac{s^{2}x^{2}}{\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}.\\
\end{align}$$

Consider the following derivative:
$$\frac{d}{dx}\left[\frac{\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}{x}\right]=\frac{s^{2}x^{4}+1}{x^{2}\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}.$$
This leads to the integration formula
$$\int_{1}^{z}\mathrm{d}x\,\frac{s^{2}x^{2}}{\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}=\frac{\sqrt{\left(z^{2}-1\right)\left(1+s^{2}z^{2}\right)}}{z}-\int_{1}^{z}\mathrm{d}x\,\frac{1}{x^{2}\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}.$$
Then,
$$\begin{align}
\mathcal{I}{\left(s,\omega\right)}
&=\int_{1}^{z}\mathrm{d}x\,\frac{1}{\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}+\int_{1}^{z}\mathrm{d}x\,\frac{s^{2}x^{2}}{\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}\\
&=\int_{1}^{z}\mathrm{d}x\,\frac{1}{\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}\\
&~~~~~+\frac{\sqrt{\left(z^{2}-1\right)\left(1+s^{2}z^{2}\right)}}{z}-\int_{1}^{z}\mathrm{d}x\,\frac{1}{x^{2}\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}\\
&=\frac{\sqrt{\left(z^{2}-1\right)\left(1+s^{2}z^{2}\right)}}{z}+\int_{1}^{z}\mathrm{d}x\,\frac{1-x^{-2}}{\sqrt{\left(x^{2}-1\right)\left(1+s^{2}x^{2}\right)}}\\
&=\frac{\sqrt{\left(z^{2}-1\right)\left(1+s^{2}z^{2}\right)}}{z}\\
&~~~~~+\int_{0}^{\frac{\pi}{2}-\arcsin{\left(\frac{1}{z}\right)}}\mathrm{d}\varphi\,\frac{\tan{\left(\varphi\right)}\sec{\left(\varphi\right)}\left[\sec^{2}{\left(\varphi\right)}-1\right]}{\sec^{2}{\left(\varphi\right)}\sqrt{\left[\sec^{2}{\left(\varphi\right)}-1\right]\left[1+s^{2}\sec^{2}{\left(\varphi\right)}\right]}};~~~\small{\left[x=\sec{\left(\varphi\right)}\right]}\\
&=\frac{\sqrt{\left(z^{2}-1\right)\left(1+s^{2}z^{2}\right)}}{z}+\int_{0}^{\frac{\pi}{2}-\arcsin{\left(\frac{1}{z}\right)}}\mathrm{d}\varphi\,\frac{\sin^{2}{\left(\varphi\right)}}{\sqrt{1+s^{2}-\sin^{2}{\left(\varphi\right)}}}.\\
\end{align}$$
It should be clear how to proceed from this point, so I leave the rest as an exercise for the reader.

A: Put $ x\to ix$.
We get elliptic integrals in closed form as for example similar to computing circumference of an ellipse.
Now put back $ x \to - ix $ again to put result  back to hyperbolic functions.
