Question

If $\int \sqrt{\sec 2x-1}dx=\alpha {\log }_e\left|\cos 2x+\beta +\sqrt{\cos 2x\left(1+\cos \frac{1}{\beta }x\right)}\right|+$ constant, then $\beta -\alpha$ is equal to_______

Answer 1

Answer: 1

$\int \sqrt{\sec 2x-1}dx=\int \sqrt{\frac{1-\cos 2x}{\cos 2x}}dx$
$=\sqrt{2}\int \frac{\sin x}{\sqrt{2{\cos }^{2}x-1}}dx$
$\text{ put }\cos x=t\Rightarrow -\sin xdx=dt$
$=-\sqrt{2}\int \frac{dt}{\sqrt{2t^2-1}}$
$=-\ln |\sqrt{2}\cos x+\sqrt{\cos 2x}|+c$
$=-\frac{1}{2}\ln \left|2\cos {}^{2}x+\cos 2x+2\sqrt{\cos 2x}\cdot \sqrt{2}\cos x\right|+c$
$=-\frac{1}{2}\ln \left|\cos 2x+\frac{1}{2}+\sqrt{\cos 2x}\cdot \sqrt{1+\cos 2x}\right|+c$
$\because \beta =\frac{1}{2},\alpha =-\frac{1}{2}$
$\Rightarrow \beta -\alpha =1$