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?

If you're asking how to simplify the bottom integral's integrand, you can note that $\csc{(x)} = \frac{1}{\sin{(x)}}$ and $1 - \cos^2{(x)} = \sin^2{(x)}$ to get
$$\frac{1}{\sqrt{2\csc^{2}\left(\theta\right)-1}} = -\frac{-\sin\left(\theta\right)}{\sqrt{\cos^{2}\left(\theta\right)+1}}.$$
I think you got it from here.
