$$\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx$$
$$\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}dx=\int^{\frac{\pi}{4}}_{0} \frac{\cos 2x-1}{\cos 2x+1}\cdot\frac{\cos 2x-1}{\cos 2x-1}dx=\int^{\frac{\pi}{4}}_{0} \frac{\cos^2 2x-2\cos2x+1}{\cos^22x-1}dx$$
$$=\int^{\frac{\pi}{4}}_{0} \frac{\cos^2 2x-2\cos2x+1}{-\sin^22x}dx$$
$u=\cos2x$
$du=-2\sin2x$
Is this substitution is ok? or do dx must be in the numerator?