\int\frac{dx}{\sin x(2\cos^2x-1)}=\int\frac{-\sin x\,dx}{-\sin^2x(2\cos^2x-1)}=\int\frac{d(\cos x)}{(\cos^2x-1)(2\cos^2 x-1)}=\int\frac{dt}{(t^2-1)(2t^2-1)}