Question

Evaluate : \(\int \frac{dx}{\sqrt{x^2 - a^2}}\).

Hint:

Memorizing the three standard integrals involving square roots is crucial for competitive exams: \(\int \frac{dx}{\sqrt{a^2 - x^2}}\), \(\int \frac{dx}{\sqrt{x^2 + a^2}}\), and \(\int \frac{dx}{\sqrt{x^2 - a^2}}\). This saves significant time compared to deriving them from scratch.

Answer 1

Step 1: Understanding the Concept:
This question asks for the evaluation of a standard indefinite integral. The integrand is of a specific form whose result is a well-known formula in calculus.
Step 2: Key Formula or Approach:
The integral is a standard form:
\[ \int \frac{dx}{\sqrt{x^2 - a^2}} = \ln |x + \sqrt{x^2 - a^2}| + C \] where \(C\) is the constant of integration.
This formula can be derived using trigonometric substitution by setting \(x = a \sec \theta\).
Step 3: Detailed Explanation:
This is a direct application of a standard integration formula.
Given the integral: \[ I = \int \frac{dx}{\sqrt{x^2 - a^2}} \] By comparing this with the standard formula \(\int \frac{du}{\sqrt{u^2 - a^2}} = \ln |u + \sqrt{u^2 - a^2}| + C\), where \(u = x\), we can directly write the result.
\[ I = \ln |x + \sqrt{x^2 - a^2}| + C \] Step 4: Final Answer:
The evaluation of the integral is \(\ln |x + \sqrt{x^2 - a^2}| + C\).