Q69 / 150 · MHT CET 2022 (05 Aug Shift 1) · Mathematics

For $-\frac{\pi}{2}<x<\frac{\pi}{2}, \int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) d x=$ (Where $\mathrm{C}$ is a constant of integration)

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

Solution

$\begin{aligned} & \int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) d x=\int \tan ^{-1} \sqrt{\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}} d x \\ & =\int \tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) d x=\int \tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right) d x \\ & =\int \tan ^{-1} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right) d x=\int\left(\frac{\pi}{4}-\frac{x}{2}\right) d x \\ & =\frac{\pi}{2} x-\frac{x^2}{4}+C\end{aligned}$

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

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