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: Understanding the Concept
Let \(x = a\tan^2\theta\), so \(\theta = \tan^{-1}\sqrt{x/a}\). Then \(\dfrac{a + x}{x} = \dfrac{\sec^2\theta}{\tan^2\theta} = \dfrac{1}{\sin^2\theta}\).

Step 2: Simplify the integrand
\[ \text{cosec}^{-1}\sqrt{\frac{a + x}{x}} = \text{cosec}^{-1}\left(\frac{1}{\sin\theta}\right) = \theta \]
Also \(dx = 2a\tan\theta\sec^2\theta\,d\theta\). So the integral is \(\int 2a\,\theta\tan\theta\sec^2\theta\,d\theta\).

Step 3: Integrate by parts
Take \(u = \theta\), \(dv = 2a\tan\theta\sec^2\theta\,d\theta\), so \(v = a\tan^2\theta\).
\[ = a\theta\tan^2\theta - a\int\tan^2\theta\,d\theta = a\theta\tan^2\theta - a(\tan\theta - \theta) \]
\[ = a\theta\tan^2\theta - a\tan\theta + a\theta + c \]
This is option (B). The sign of the last term is plus because \(\int\tan^2\theta\,d\theta = \tan\theta - \theta\).

Final Answer:
The integral is \(a\theta\tan^2\theta - a\tan\theta + a\theta + c\), option (B). \[ \boxed{a\theta\tan^2\theta - a\tan\theta + a\theta + c} \]