Key Concepts and Formulas

• Trigonometric Identities: $\sec x = \frac{1}{\cos x}$, $1 - \cos 2x = 2\sin^2 x$, $\cos 2x = 2\cos^2 x - 1$
• Integration: $\int \frac{1}{\sqrt{x^2 - a^2}} dx = \ln|x + \sqrt{x^2 - a^2}| + C$

Step-by-Step Solution

Step 1: Rewrite the integral using trigonometric identities.

We start by rewriting $\sec 2x$ in terms of $\cos 2x$ and then use the identity $1 - \cos 2x = 2\sin^2 x$.

$I = \int \sqrt{\sec 2x - 1} dx = \int \sqrt{\frac{1}{\cos 2x} - 1} dx = \int \sqrt{\frac{1 - \cos 2x}{\cos 2x}} dx = \int \sqrt{\frac{2\sin^2 x}{\cos 2x}} dx$

Step 2: Introduce a substitution to simplify the integral.

Let $t = \sqrt{2} \cos x$. Then, $dt = -\sqrt{2} \sin x \, dx$. We aim to express the integral in terms of $t$ and $dt$. From $t = \sqrt{2}\cos x$, we have $\cos x = \frac{t}{\sqrt{2}}$, so $\cos 2x = 2\cos^2 x - 1 = 2\left(\frac{t}{\sqrt{2}}\right)^2 - 1 = t^2 - 1$. Also, $\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \frac{t^2}{2}} = \frac{\sqrt{2 - t^2}}{\sqrt{2}}$. Then, $\int \sqrt{\frac{2\sin^2 x}{\cos 2x}} dx = \int \frac{\sqrt{2}\sin x}{\sqrt{\cos 2x}}dx = \int \frac{\sqrt{2}\frac{\sqrt{2-t^2}}{\sqrt{2}}}{\sqrt{t^2-1}} dx = \int \frac{\sqrt{2-t^2}}{\sqrt{t^2-1}} dx$. Since $dt = -\sqrt{2}\sin x dx$, we have $dx = \frac{dt}{-\sqrt{2}\sin x} = \frac{dt}{-\sqrt{2}\frac{\sqrt{2-t^2}}{\sqrt{2}}} = \frac{-dt}{\sqrt{2-t^2}}$. Then, $\int \frac{\sqrt{2-t^2}}{\sqrt{t^2-1}}\frac{-dt}{\sqrt{2-t^2}} = \int \frac{-1}{\sqrt{t^2-1}} dt = -\ln|t + \sqrt{t^2-1}| + C$

Step 3: Substitute back to get the integral in terms of $x$.

Now we substitute $t = \sqrt{2}\cos x$ back into the expression: $-\ln|\sqrt{2}\cos x + \sqrt{2\cos^2 x - 1}| + C = -\ln|\sqrt{2}\cos x + \sqrt{\cos 2x}| + C$

Step 4: Manipulate the logarithmic expression.

We want to get the expression in the form $\alpha \log_e |\cos 2x + \beta + \sqrt{\cos 2x(1 + \cos 2x)}| + C$. Let's try to rewrite our current result: $-\ln|\sqrt{2}\cos x + \sqrt{\cos 2x}| + C = -\frac{1}{2}\ln|(\sqrt{2}\cos x + \sqrt{\cos 2x})^2| + C = -\frac{1}{2}\ln|(2\cos^2 x + \cos 2x + 2\sqrt{2}\cos x \sqrt{\cos 2x})| + C$ Since $2\cos^2 x = \cos 2x + 1$, we have $-\frac{1}{2}\ln|(\cos 2x + 1 + \cos 2x + 2\sqrt{\cos 2x(\cos 2x + 1)})| + C = -\frac{1}{2}\ln|(2\cos 2x + 1 + 2\sqrt{\cos 2x(\cos 2x + 1)})| + C$ $= -\frac{1}{2}\ln|2(\cos 2x + \frac{1}{2} + \sqrt{\cos 2x(\cos 2x + 1)})| + C = -\frac{1}{2}\ln 2 -\frac{1}{2}\ln|(\cos 2x + \frac{1}{2} + \sqrt{\cos 2x(\cos 2x + 1)})| + C$ $= -\frac{1}{2}\ln|(\cos 2x + \frac{1}{2} + \sqrt{\cos 2x(1 + \cos 2x)})| + C'$

Step 5: Identify $\alpha$ and $\beta$ and calculate $\beta - \alpha$.

Comparing this with the given form $\alpha \log_e |\cos 2x + \beta + \sqrt{\cos 2x(1 + \cos 2x)}| + C$, we have $\alpha = -\frac{1}{2}$ and $\beta = \frac{1}{2}$. Therefore, $\beta - \alpha = \frac{1}{2} - \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1$.

Common Mistakes & Tips

• Be careful with the signs when substituting back.
• Remember trigonometric identities to simplify the expression.
• When manipulating the logarithmic expression, remember the properties of logarithms.

Summary

We started by simplifying the integral using trigonometric identities and then introduced a substitution. After integrating and substituting back, we manipulated the expression to match the given form and identified the values of $\alpha$ and $\beta$. Finally, we calculated $\beta - \alpha$.

The final answer is \boxed{1}.