Q116 / 200 · Manipal (MET) 2013 · Mathematics

$\int \sqrt{1+\operatorname{cosec} x} d x$ is equal to

  • A
    $\pm \sin ^{-1}(\tan x-\sec x)+c$
  • B
    $2 \sin ^{-1}(\cos x)+c$
  • C
    $\sin ^{-1}\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)+c$
  • Correct answer
    D
    $\pm 2 \sin ^{-1}\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)+c$

Solution

Let $I=\int \sqrt{1+\operatorname{cosec} x} d x=\int \frac{\sqrt{1+\sin x}}{\sqrt{\sin x}} d x$ $=\int \pm \frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\sqrt{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}} \cdot d x$ $= \pm \int \frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\sqrt{1-\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^2}} \cdot d x$ Put, $\quad \sin \frac{x}{2}-\cos \frac{x}{2}=t$ $\Rightarrow \quad\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right) d x=2 d t$ $\therefore \quad I= \pm \int \frac{2 d t}{\sqrt{1-t^2}}= \pm 2 \sin ^{-1} t+c$ $= \pm 2 \sin ^{-1}\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)+c$

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

Relevant Questions