Computing the integral of trig function under square root How can we solve this integral
$$\int \sqrt{\csc^2x -2} 
 \mathrm{d}x$$
My idea was substituting $\csc^2x=2\csc^2\theta$. Then the integral became $$\sqrt{2}\int \frac{\csc^2\theta-1}{\sqrt{2\csc^2\theta-1}} \mathrm{d}\theta$$ after a few simplifications. I don't know how to go about at this point. If i try breaking the numerator,still i would be left with an equivalent question of $$\int \sqrt{2\csc^2\theta-1}\mathrm{d}\theta$$ and $$\int \frac{\mathrm{d}\theta}{\sqrt{2\csc^2\theta-1}}$$
Could someone provide a cleaner approach to this problem or give new insights on how to further simply the above integrals?

Letting $u=\sqrt{\csc ^{2} x-2}$ transform the integral into
$$\begin{aligned}
I&=-\frac{1}{2} \int \frac{u^{2} d u}{\left(u^{2}+2\right) \sqrt{u^{2}+1}}\\& = \frac{1}{2} \underbrace{\int \frac{d u}{\sqrt{u^{2}+1}}}_{\sinh ^{-1} u+C_1}  + \underbrace{ \int \frac{d u}{\left(u^{2}+2\right) \sqrt{u^{2}+1}}}_{J}\end{aligned}
$$
For the integral $J$, letting $t=\frac{u}{\sqrt{u^{2}+1}} $ gives $u d u=\frac{t d t}{\left(1-t^{2}\right)^{2}}$ and
$$
\begin{aligned}
I &=\int \frac{d t}{\frac{2-t^2}{1-t^{2}}\left(1-t^{2}\right)} \\
&=\int \frac{d t}{2-t^{2}} \\
&=\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{t}{\sqrt{2}}\right)+C_{2}\\&= \frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{u}{\sqrt{2} \sqrt{u^{2}+1}}\right)+C_2
\end{aligned}
$$
Now we can conclude that
$$
I=-\frac{1}{2} \sinh ^{-1} \left(\sqrt{\csc ^{2} x-2}\right) +\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{\sqrt{\csc ^{2} x-2}}{\sqrt{2} \cot x}\right)+C
$$
