Definite Integration
Indefinite Integration
Grade Class 12
Question:
If $\int \sqrt{\sec 2x - 1} dx = \alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x \left( 1 + \frac{\cos 2x}{\beta} \right)} \right| + \text{constant}$, then $\beta - \alpha$ is equal to ____.
Step-by-Step Solution
Key Concept: Use trigonometric identities to simplify the integrand: sqrt(sec 2x - 1) = sqrt((1-cos 2x)/cos 2x) = sqrt(2 sin^2 x / cos 2x). Substitute and integrate.
The integral is $\int \sqrt{\frac{1-\cos 2x}{\cos 2x}} dx = \int \sqrt{\frac{2\sin^2 x}{\cos 2x}} dx = \sqrt{2} \int \frac{\sin x}{\sqrt{\cos 2x}} dx$. Let $\cos x = t$, then $-\sin x dx = dt$. The integral becomes $-\sqrt{2} \int \frac{dt}{\sqrt{2t^2-1}} = -\int \frac{dt}{\sqrt{t^2 - 1/2}} = -\ln|t + \sqrt{t^2 - 1/2}| + C = -\ln|\cos x + \sqrt{\cos^2 x - 1/2}| + C$. Further manipulation leads to the form given in the question.
Correct Answer: 1