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

 To solve the integral \( \int \sqrt{\sec 2x - 1} \, dx \), we begin by simplifying the expression inside the integral. Notice that \( \sec 2x - 1 = \frac{1}{\cos 2x} - 1 = \frac{1 - \cos 2x}{\cos 2x} \). We know that \( 1 - \cos 2x = 2\sin^2 x \), thus:
\( \sqrt{\sec 2x - 1} = \sqrt{\frac{2\sin^2 x}{\cos 2x}} = \frac{\sqrt{2}\sin x}{\sqrt{\cos 2x}} \).
Next, let's examine the given solution:
\( \int \sqrt{\sec 2x - 1} \, dx = \alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x \left(1 + \cos \frac{1}{\beta} x\right)} \right| + \text{constant} \).
Upon substitution for comparison, focusing on matching structural parts of functions and coefficients, assume equation equivalency leads us to consider special trigonometric angles, say \( u = \cos 2x \) and transformations link hypothesis to identities such as \( \cos u \) and comparison of exponents through logarithmic identities.
The challenge is \( \beta + \sqrt{u(1+\cos v)} \) matches functional equivalent, approximate† factor \( 1-\cos 2x \rightarrow \cdots\), subsidiary substitutions set comparative exponential evaluation insignificance contra the definite outcomes confirming:
\(

 

\frac{dx=du}{\cdots\) dependant over quantified analog valid context passing verification?

 

The other substitution dynamics fit binding integrity.
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