Question

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

Answer 1

Correct Answer: 1

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

Answer 2

The integral can be solved as follows: \[ \int \sqrt{\sec 2x - 1} \, dx = \int \sqrt{1 - \cos 2x} \, dx \] This is equivalent to: \[ = \sqrt{2} \int \frac{\sin x}{\sqrt{2\cos^2 x - 1}} \, dx \] Let \( \cos x = t \), so \( -\sin x \, dx = dt \). The integral becomes: \[ = -\int \sqrt{2} \frac{dt}{\sqrt{2t^2 - 1}} \] This simplifies to: \[ = -\ln \left( \sqrt{2} \cos x + \sqrt{\cos 2x} \right) + c \] Therefore, we have: \[ = -\frac{1}{2} \ln \left( 2 \cos^2 x + \cos 2x + 2 \sqrt{2} \cos x \right) + c \] This leads to: \[ \beta = \frac{1}{2}, \quad \alpha = -\frac{1}{2} \quad \Rightarrow \quad \beta - \alpha = 1 \]