Question

\(\int \text{cosec}^{-1}(\sqrt{\frac{a+x}{x}})\,\text{d}x =\)

• A. \(aθtan^2θ-atanθ-aθ+c\)  \((\text{where }θ = tan^{-1}(\sqrt{\frac{x}{a}}))\) and c is the constant of integration
• B. \(aθtan^2θ-atanθ+aθ+c\)  \((\text{where }θ = tan^{-1}(\sqrt{\frac{x}{a}}))\) and c is the constant of integration
• C. \(aθtan^2θ+atanθ-aθ+c\)  \((\text{where }θ = tan^{-1}(\sqrt{\frac{x}{a}}))\) and c is the constant of integration
• D. \(aθtan^2θ+atanθ+aθ+c\)  \((\text{where }θ = tan^{-1}(\sqrt{\frac{x}{a}}))\) and c is the constant of integration

Hint:

Substitute x = a tan^2 theta so the inverse cosecant collapses to theta, then integrate by parts.

Answer 1

The Correct Option is B

Step 1: Substitution
$x = a\tan^2\theta$ makes $\sqrt{(a+x)/x} = \csc\theta$, so the integrand is $\theta$.

Step 2: Differential
$dx = 2a\tan\theta\sec^2\theta\,d\theta = d(a\tan^2\theta)$.

Step 3: By parts
$\int\theta\,d(a\tan^2\theta) = a\theta\tan^2\theta - a\int\tan^2\theta\,d\theta = a\theta\tan^2\theta - a\tan\theta + a\theta + c$. Option (B).

Final Answer:
Option (B). \[ \boxed{a\theta\tan^2\theta - a\tan\theta + a\theta + c} \]