Combine these two integrals into a single integral $$\int\frac{\textrm{cosec}^2x}{\sqrt{\textrm{cot}^2x-1}}dx-2\int\frac{\textrm{sin}x}{\sqrt{1-2\textrm{sin}^2x}}dx=$$

$\int\sqrt{\textrm{cosec}^2x-2}\:\:dx$


$\int\frac{\sqrt{\textrm{cos}2x}}{2\textrm{sin}x}\:\:dx$


$\int\frac{\sqrt{\textrm{cos}2x}}{\textrm{sin}x}\:\:dx$


$\int\sqrt{\textrm{cosec}^2x-1}\:\:dx$

I have calculated both integrals separately both nothing useful came out. Just to show my work the value of both integrals are as follows
$$\int\frac{\textrm{cosec}^2x}{\sqrt{\textrm{cot}^2x-1}}dx=-\ln|\sqrt{\textrm{cot}^2x-1}+\textrm{cot}x|+c$$
and
$$2\int\frac{\textrm{sin}x}{\sqrt{1-2\textrm{sin}^2x}}dx=-\sqrt{2}\:\: \cdot\ln|\sqrt{2\textrm{cos}^2x-1}+\sqrt{2}\:\:\textrm{cos}x|+c$$
Now I know that I can calculate value of all individual integrals and then check if that is equal to the difference of the original integrals. But I guess, that process is not the perfect one as that way is very tedious and time-taking. So I think there should be another more appropriate method.
Any help is greatly appreciated.
 A: First of all, notice that:
$$\int\frac{\textrm{cosec}^2x}{\sqrt{\textrm{cot}^2x-1}}dx-2\int\frac{\textrm{sin}x}{\sqrt{1-2\textrm{sin}^2x}}dx=\int\left(\frac{\textrm{cosec}^2x}{\sqrt{\textrm{cot}^2x-1}}-2\frac{\textrm{sin}x}{\sqrt{1-2\textrm{sin}^2x}}\right)dx.$$
Furthermore, we have that:
$$\frac{\textrm{cosec}^2x}{\sqrt{\textrm{cot}^2x-1}}-2\frac{\textrm{sin}x}{\sqrt{1-2\textrm{sin}^2x}} = \\
=\frac{\frac{1}{\textrm{sin}^2x}}{\sqrt{\frac{\textrm{cos}^2x}{\textrm{sin}^2x}-1}}-2\frac{\textrm{sin}x}{\sqrt{1-2\textrm{sin}^2x}} = \\
=\frac{\frac{1}{\textrm{sin}x}}{\sqrt{\textrm{cos}^2x-\textrm{sin}^2x}}-2\frac{\textrm{sin}x}{\sqrt{1-2\textrm{sin}^2x}} = \\
=\frac{\frac{1}{\textrm{sin}x}}{\sqrt{1 - 2\textrm{sin}^2x}}-2\frac{\textrm{sin}x}{\sqrt{1-2\textrm{sin}^2x}} = \\
=\frac{1}{\sqrt{1 - 2\textrm{sin}^2x}}\left(\frac{1}{\textrm{sin}x}-2\textrm{sin}x\right) = \\
=\frac{1}{\sqrt{1 - 2\textrm{sin}^2x}}\left(\frac{1 - 2\textrm{sin}^2x}{\textrm{sin}x}\right) = \\
=\frac{\sqrt{1 - 2\textrm{sin}^2x}}{\textrm{sin}x} = \\
=\sqrt{\frac{1}{\textrm{sin}^2x} - 2} = \sqrt{\textrm{cosec}^2x - 2}.$$
A: If we allow ourselves the, admittedly rather inspired, substitution:
$$u^2=1-\tan^2 x$$
then we get:
$$\cot^2 x - 1=\frac{u^2}{1-u^2}$$
$$\sec^2 x =2-u^2$$
$$\csc^2 x=\frac{2-u^2}{1-u^2}$$
$$\sin x =\sqrt{\frac{1-u^2}{2-u^2}}$$
$$\cos x =\sqrt{\frac{1}{2-u^2}}$$
$$dx =\frac{-u\, du}{(2-u^2)\sqrt{1-u^2}}$$
and the required integral becomes
$$\int\frac{\csc^2x}{\sqrt{\cot^2x-1}}dx- 2\frac{\sin x}{\sqrt{1-2 \sin^2x}}dx=\int -\frac{1}{1-u^2}+\frac{2}{2-u^2} \, du$$
$$=\frac 12 \ln \frac{u+1}{u-1}+\sqrt2 \ln \frac{u+\sqrt 2}{u-\sqrt 2}$$
where
$$u=\sqrt{1-\tan^2 x}$$
A: A bit tricky
$$A=\frac{\csc ^2(x)}{\sqrt{\cot ^2(x)-1}}-\frac{2 \sin (x)}{\sqrt{1-2 \sin ^2(x)}}$$
Use the tangent half-angle substitution $x=2 \tan ^{-1}(t)$ and simplify to obtain
$$\color{blue}{A=\frac{\sqrt{t^4-6 t^2+1}}{2 t}}$$Now, back to $x$ using $t=\tan \left(\frac{x}{2}\right)$ and obtain
$$A=\csc (x)\sqrt{\cos (2 x)} $$
