help integrating I'm really struggling to integrate this.
$$\int_0^t \sqrt{\cos^2\left(1\right)\sinh^2\left(u\right)+4\sin^2\left(\dfrac{1}{2}\right)\cosh^2\left(\dfrac{u}{2}\right)} \,\mathrm{d}u$$
This is to find the arc length of a plane curve. An online integral calculator gives me such a complicated result with $\ln$ and I don't know how they got to that result.
I also then need to use the result to find the inverse.
 A: Hint:
Set $a={{\cos }^{2}}\left( 1 \right),b={{\sin }^{2}}\left( \frac{1}{2} \right)$
and using   $\sinh \left( u \right)=2\sinh \left( \frac{u}{2} \right)\cosh \left( \frac{u}{2} \right)$ the integral becomes:
$$\begin{align}
  & =\int{\sqrt{4a{{\sinh }^{2}}\left( \frac{u}{2} \right){{\cosh }^{2}}\left( \frac{u}{2} \right)+4b{{\cosh }^{2}}\left( \frac{u}{2} \right)}du} \\ 
 & \ =\int{\sqrt{4{{\cosh }^{2}}\left( \frac{u}{2} \right)\left( a{{\sinh }^{2}}\left( \frac{u}{2} \right)+b \right)}du} \\ 
 & =\int{2\cosh \left( \frac{u}{2} \right)\sqrt{a{{\sinh }^{2}}\left( \frac{u}{2} \right)+b}du}\ \quad ,x=\sinh \left( \frac{u}{2} \right) \\ 
 & =4\int{\sqrt{a{{x}^{2}}+b}\ dx} \\ 
\end{align}$$
A: So I followed your hint and got
$s(t)=2\sqrt{b}\sinh\left(\dfrac{u}{2}\right)\sqrt{\dfrac{a\sinh^2\left(\frac{u}{2}\right)}{b}+1}+\dfrac{2b\ln\left(\left|\sqrt{a\sinh^2\left(\frac{u}{2}\right)+b}+\sqrt{a}\sinh\left(\frac{u}{2}\right)\right|\right)}{\sqrt{a}}$
After substituting the values of $a$ and $b$ into the equation and then calculating the definite integral from lower bound $0$ and upper bound $t$ I will need to find the inverse.
Any ideas how to go about this with a $ln$ in the result??? I need to make $t$ the subject
