<s>[INST] 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.
 [/INST] 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 $2.3547702$.
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