Question

$\int \frac{secx}{\sqrt{cos2x}}$ dx

Answer 1

$\int \frac{\sec x}{\sqrt{\cos 2x}}dx=\int \frac{\sec {}^{2}x}{\sec x\sqrt{2\cos {}^{2}x-1}}=\int \frac{\sec {}^{2}xdx}{\sqrt{2\cos {}^{2}x+\sec {}^{2}x-\sec {}^{2}x}}=\int \frac{\sec {}^{2}xdx}{\sqrt{2-1-\tan {}^{2}x}}=\int \frac{\sec {}^{2}xdx}{\sqrt{1-\tan {}^{2}x}}$

Let $\tan x=t\Rightarrow \sec {}^{2}xdx=dt$

$I=\int \frac{dt}{\sqrt{1-t^2}}={\sin }^{-1}t={\sin }^{-1}\left(\tan x\right)$