Indefinite Integration
Integration involving Secant / Trigonometric Substitution
nta_pyq_2023_jan
Grade 12
Question:
If $\displaystyle\int \sqrt{\sec 2x - 1}\,dx = \alpha \log_e\left|\cos 2x + \beta + \sqrt{\cos 2x\left(1+\cos\dfrac{1}{\beta}x\right)}\right| + \text{constant}$, then $\beta - \alpha$ is equal to ______.
Step-by-Step Solution
Key Concept: Write $\sec 2x - 1 = \frac{1-\cos 2x}{\cos 2x}$; use $\cos x = t$ substitution to reduce to $\int \frac{dt}{\sqrt{2t^2-1}}$.
$\int\sqrt{\sec 2x-1}\,dx = -\ln|\sqrt{2}\cos x+\sqrt{\cos 2x}|+c$. Comparing: $\alpha=-1/2$, $\beta=1/2$. $\beta-\alpha=1$.
Correct Answer: 1