Q62 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
$\int \text{cosec}^{-1}\left(\sqrt{\dfrac{a+x}{x}}\right)\, \text{d}x =$
- A$a\theta\tan^2\theta - a\tan\theta - a\theta + c$ $\left(\text{where } \theta = \tan^{-1}\left(\sqrt{\dfrac{x}{a}}\right)\right)$ and c is the constant of integration
- Correct answerB$a\theta\tan^2\theta - a\tan\theta + a\theta + c$ $\left(\text{where } \theta = \tan^{-1}\left(\sqrt{\dfrac{x}{a}}\right)\right)$ and c is the constant of integration
- C$a\theta\tan^2\theta + a\tan\theta - a\theta + c$ $\left(\text{where } \theta = \tan^{-1}\left(\sqrt{\dfrac{x}{a}}\right)\right)$ and c is the constant of integration
- D$a\theta\tan^2\theta + a\tan\theta + a\theta + c$ $\left(\text{where } \theta = \tan^{-1}\left(\sqrt{\dfrac{x}{a}}\right)\right)$ and c is the constant of integration
Solution
Let $I = \int \text{cosec}^{-1}\left(\sqrt{\dfrac{a+x}{x}}\right)\, \text{d}x$.
Substitute $x = a\tan^2\theta$, which gives $\theta = \tan^{-1}\left(\sqrt{\dfrac{x}{a}}\right)$.
Differentiating both sides, we get $\text{d}x = 2a\tan\theta\sec^2\theta\, \text{d}\theta$.
The term inside the inverse trigonometric function simplifies as:
$\sqrt{\dfrac{a+x}{x}} = \sqrt{\dfrac{a+a\tan^2\theta}{a\tan^2\theta}} = \sqrt{\dfrac{1+\tan^2\theta}{\tan^2\theta}} = \sqrt{\dfrac{\sec^2\theta}{\tan^2\theta}} = \text{cosec}\theta$
Therefore, the integrand becomes $\text{cosec}^{-1}(\text{cosec}\theta) = \theta$.
The integral transforms to:
$I = \int \theta (2a\tan\theta\sec^2\theta)\, \text{d}\theta$
Applying integration by parts, taking $\theta$ as the first function and $2a\tan\theta\sec^2\theta$ as the second function:
$I = \theta \int 2a\tan\theta\sec^2\theta\, \text{d}\theta - \int \left( \dfrac{\text{d}}{\text{d}\theta}(\theta) \int 2a\tan\theta\sec^2\theta\, \text{d}\theta \right) \text{d}\theta$
Since $\int 2a\tan\theta\sec^2\theta\, \text{d}\theta = a\tan^2\theta$, we get:
$I = \theta(a\tan^2\theta) - \int 1 \cdot (a\tan^2\theta)\, \text{d}\theta$
$I = a\theta\tan^2\theta - a\int (\sec^2\theta - 1)\, \text{d}\theta$
$I = a\theta\tan^2\theta - a(\tan\theta - \theta) + c$
$I = a\theta\tan^2\theta - a\tan\theta + a\theta + c$
Answer: $a\theta\tan^2\theta - a\tan\theta + a\theta + c$ $\left(\text{where } \theta = \tan^{-1}\left(\sqrt{\dfrac{x}{a}}\right)\right)$ and c is the constant of integration
Substitute $x = a\tan^2\theta$, which gives $\theta = \tan^{-1}\left(\sqrt{\dfrac{x}{a}}\right)$.
Differentiating both sides, we get $\text{d}x = 2a\tan\theta\sec^2\theta\, \text{d}\theta$.
The term inside the inverse trigonometric function simplifies as:
$\sqrt{\dfrac{a+x}{x}} = \sqrt{\dfrac{a+a\tan^2\theta}{a\tan^2\theta}} = \sqrt{\dfrac{1+\tan^2\theta}{\tan^2\theta}} = \sqrt{\dfrac{\sec^2\theta}{\tan^2\theta}} = \text{cosec}\theta$
Therefore, the integrand becomes $\text{cosec}^{-1}(\text{cosec}\theta) = \theta$.
The integral transforms to:
$I = \int \theta (2a\tan\theta\sec^2\theta)\, \text{d}\theta$
Applying integration by parts, taking $\theta$ as the first function and $2a\tan\theta\sec^2\theta$ as the second function:
$I = \theta \int 2a\tan\theta\sec^2\theta\, \text{d}\theta - \int \left( \dfrac{\text{d}}{\text{d}\theta}(\theta) \int 2a\tan\theta\sec^2\theta\, \text{d}\theta \right) \text{d}\theta$
Since $\int 2a\tan\theta\sec^2\theta\, \text{d}\theta = a\tan^2\theta$, we get:
$I = \theta(a\tan^2\theta) - \int 1 \cdot (a\tan^2\theta)\, \text{d}\theta$
$I = a\theta\tan^2\theta - a\int (\sec^2\theta - 1)\, \text{d}\theta$
$I = a\theta\tan^2\theta - a(\tan\theta - \theta) + c$
$I = a\theta\tan^2\theta - a\tan\theta + a\theta + c$
Answer: $a\theta\tan^2\theta - a\tan\theta + a\theta + c$ $\left(\text{where } \theta = \tan^{-1}\left(\sqrt{\dfrac{x}{a}}\right)\right)$ and c is the constant of integration
Hence, the correct answer is option: 'B'.