How to solve this integral and have arccos(...) as a result?

Question

$$\int {\sqrt{\csc^{2}x -1}} \, d(\cos^2x)$$

I need to solve this integral in order to arrive to a solution that looks like $x= \arccos(...)$ The main substitution is already done, I don't know how to put this equation in a software so I need a help.

Answer 1

$$\begin{align}\int \sqrt{\csc^{2}x -1} \, d(\cos^2x) &= \int\frac {\cos x} {\sin x}\left( -2\cos x \sin x \right)\, dx\\ &=-\int(1+\cos 2x)\, dx\\ &=-x+\frac12 \sin 2x+c\end{align}$$

is this what you want?

Answer 2

$$\int \sqrt{\csc^2 x-1}\,d(\cos^2 x)\\ =\int \sqrt{\cot^2 x} \cdot 2\cos x\cdot(-\sin x)\,dx\\ =-\int \frac{\cos x}{\sin x}\cdot 2\cos x\cdot \sin x\,dx\\ =-2\int \cos^2 x\, dx$$

Can you continue from here?