Q.$\int \dfrac{\sec x}{\sqrt{\cos 2x}}\,dx$ is:

Writing $\cos 2x = \cos^2x(1-\tan^2x)$ converts the integrand into $\sec^2x/\sqrt{1-\tan^2x}$, and substituting $t=\tan x$ gives a standard $\int dt/\sqrt{1-t^2}$ integral.

$\dfrac{\sec x}{\sqrt{\cos 2x}} = \dfrac{1}{\cos x\sqrt{\cos 2x}}$.

Use $\cos 2x = \cos^2x - \sin^2x = \cos^2x(1-\tan^2x)$, so $\sqrt{\cos 2x} = \cos x\sqrt{1-\tan^2x}$ (taking $\cos x>0$).

Then $\dfrac{1}{\cos x\cdot\cos x\sqrt{1-\tan^2x}} = \dfrac{\sec^2x}{\sqrt{1-\tan^2x}}$.

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