If int sqrt{sec 2x - 1} , dx = alpha log_e le | vedclass

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(B) We have $I = \int \sqrt{\sec 2x - 1} \, dx = \int \sqrt{\frac{1 - \cos 2x}{\cos 2x}} \, dx = \int \sqrt{\frac{2 \sin^2 x}{\cos 2x}} \, dx = \sqrt{

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(B) We have $I = \int \sqrt{\sec 2x - 1} \, dx = \int \sqrt{\frac{1 - \cos 2x}{\cos 2x}} \, dx = \int \sqrt{\frac{2 \sin^2 x}{\cos 2x}} \, dx = \sqrt{2} \int \frac{\sin x}{\sqrt{\cos 2x}} \, dx$.Let $\cos x = t$,then $-\sin x \, dx = dt$.$I = -\sqrt{2} \int \frac{dt}{\sqrt{2t^2 - 1}} = -\sqrt{2} \int \frac{dt}{\sqrt{2(t^2 - 1/2)}} = -\int \frac{dt}{\sqrt{t^2 - 1/2}}$.Using $\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln |x + \sqrt{x^2 - a^2}|$,we get $I = -\ln |t + \sqrt{t^2 - 1/2}| + C = -\ln |\cos x + \sqrt{\cos^2 x - 1/2}| + C$.$I = -\ln |\cos x + \sqrt{\frac{2 \cos^2 x - 1}{2}}| + C = -\ln |\cos x + \frac{1}{\sqrt{2}} \sqrt{\cos 2x}| + C$.Multiply and divide by $\sqrt{2}$ inside the log: $I = -\ln |\frac{\sqrt{2} \cos x + \sqrt{\cos 2x}}{\sqrt{2}}| + C = -\ln |\sqrt{2} \cos x + \sqrt{\cos 2x}| + C_1$.Squaring the term inside the log: $(\sqrt{2} \cos x + \sqrt{\cos 2x})^2 = 2 \cos^2 x + \cos 2x + 2\sqrt{2} \cos x \sqrt{\cos 2x} = (1 + \cos 2x) + \cos 2x + 2\sqrt{\cos 2x} \sqrt{2 \cos^2 x} = 2 \cos 2x + 1 + 2\sqrt{\cos 2x} \sqrt{1 + \cos 2x}$.Thus,$I = -\frac{1}{2} \ln |2 \cos 2x + 1 + 2\sqrt{\cos 2x (1 + \cos 2x)}| + C_2 = -\frac{1}{2} \ln |\cos 2x + 1/2 + \sqrt{\cos 2x (1 + \cos 2x)}| + C_3$.Comparing with the given form,$\alpha = -1/2$ and $\beta = 1/2$.Therefore,$\beta - \alpha = 1/2 - (-1/2) = 1$.

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

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