The question is : Is $\sqrt 3$ the length of $$\Gamma =\{\gamma (t)=(t,\sin(t),\sqrt 2\cos(t))\mid t\in [0,1]\} \ \ ?$$ So it is $$\int_0^1\|\gamma '(t)\|dt=\int_0^1 \sqrt{1+2\sin^2(t)+\cos^2(t)}dt$$
I tries to do many substitution as $\cos^2(t)=\frac{1+\cos(2t)}{2}$, $\sin^2(t)=\frac{1-\cos(2t)}{2}$, or $1=\cos^2(t)+\sin^2(t)$ but I can't conclude. I'm sure there is a trick but I don't see it, could someone help ?
You are misreading the question that was posed to you. It asks if $\sqrt{3}$ can be the value of said integral. It is possible to avoid straight-forward integration by exploring the possibility that it may not. To this end you may be able to find an (easy) inequality to show this.
Notice that $2+\sin^2(t) < 3$ for every $t\in[0;1]$. Thus: $$||\Gamma||=\int\limits_0^1 \sqrt{2+\sin^2(t)} \mathrm{d}t < \int\limits_0^1\sqrt{3} \mathrm{d}t=\sqrt{3},$$ which answers your question.
$$1=\cos^2x+\sin^2x$$ $$2=1+\cos^2x+\sin^2x$$ $$2+\sin^2x=1+\cos^2x+2\sin^2x$$ so $$||\Gamma||=\int_0^1\sqrt{1+\cos^2x+2\sin^2x}\,\mathrm dx=\int_0^1\sqrt{2+\sin^2x}\,\mathrm dx$$ Then recall the definition of the incomplete elliptic integral of the second kind $$\mathrm E(\phi,k)=\int_0^\phi\sqrt{1-k\sin^2x}\,\mathrm dx$$ So we have that $$||\Gamma||=\sqrt{2}\,\mathrm E\bigg(1,-\frac12\bigg)\approx 1.064728654529...$$ Which might (but probably not) have a closed form