During an arc length calculation I reached the following integral and I am having hard time calculating it: $$\int\sqrt{36\sin^2(2t)+6\cos^2(t)}\,dt=\sqrt{6}\int\cos t \sqrt{24\sin^2(t)+1}\;dt$$
Where $t\in\left[{-\pi}/{2},{\pi}/{2}\right]$
I tried several techniques but couldn't manage to proceed that much.
Some hints will be appreciated.
$$\int_{-\pi/2}^{\pi/2}\cos t\sqrt{24\sin^2 t+1}dt\underset{u=\sin(t)}{=}\int_{-1}^{1} \sqrt{24u^2+1}du\underset{y=\sqrt{24}u}{=}\frac{1}{\sqrt{24}}\int_{-\sqrt{24}}^{\sqrt{24}}\sqrt{y^2+1}dy$$
Then you pose $y=\text{sinh}(v)$, you use the identity $\cosh^2(x)=1+\sinh^2(x)$ and you conclue.