Q61 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics

$\int \tan^{-1}\left(\sqrt{\dfrac{1 - \sin x}{1 + \sin x}}\right)\, dx =$

  • A
    $\dfrac{\pi}{2}x - \dfrac{x^2}{4} + C$
  • B
    $\dfrac{\pi}{4} - \dfrac{x}{4} + C$
  • C
    $\dfrac{\pi}{4}x - \dfrac{x^2}{2} + C$
  • Correct answer
    D
    $\dfrac{\pi}{4}x - \dfrac{x^2}{4} + C$

Solution

$\int \tan^{-1}\left(\sqrt{\dfrac{1 - \sin x}{1 + \sin x}}\right)\, dx$

$\sqrt{\dfrac{1 - \sin x}{1 + \sin x}} = \sqrt{\dfrac{1 - \cos(\pi/2 - x)}{1 + \cos(\pi/2 - x)}}$

Using the half-angle formulas $1 - \cos \theta = 2 \sin^2(\theta/2)$ and $1 + \cos \theta = 2 \cos^2(\theta/2)$:

$\sqrt{\dfrac{2 \sin^2(\pi/4 - x/2)}{2 \cos^2(\pi/4 - x/2)}} = \sqrt{\tan^2(\pi/4 - x/2)} = \tan(\pi/4 - x/2)$

Substituting this back into the integral:

$\int \tan^{-1}\left(\tan(\pi/4 - x/2)\right)\, dx = \int \left(\dfrac{\pi}{4} - \dfrac{x}{2}\right)\, dx$

Integrating with respect to $x$:

$\dfrac{\pi}{4}x - \dfrac{x^2}{4} + C$

Answer: $\dfrac{\pi}{4}x - \dfrac{x^2}{4} + C$

Hence, the correct answer is option: 'D'.

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