Q66 / 150 · MHT CET 2012 · Mathematics

The value of $\int \sqrt{1+\sec x} d x$ is

  • A
    $\sin ^{-1}(\sqrt{2} \sin x)+C$
  • Correct answer
    B
    $2 \sin ^{-1}(\sqrt{2} \sin x / 2)+C$
  • C
    $2 \sin ^{-1}(\sqrt{2} \sin x)+C$
  • D
    $2 \sin ^{-1}(\sqrt{2} x / 2)+C$

Solution

Let $I=\int \sqrt{1+\sec x} d x$ $=\int \frac{\sqrt{1+\cos x}}{\sqrt{\cos x}} d x$ $=\int \frac{\sqrt{2} \cdot \cos \frac{x}{2}}{\sqrt{1-2 \sin ^{2} \frac{x}{2}}} d x$ (let $\left.t=\sqrt{2} \sin \frac{x}{2} \Rightarrow d t=\frac{\sqrt{2}}{2} \cdot \cos \frac{x}{2} \cdot d x\right)$ $=\int \frac{2 d t}{\sqrt{1-t^{2}}}$ $=2 \sin ^{-1}(t)+C$ $=2 \sin ^{-1}\left(\sqrt{2} \cdot \sin \frac{x}{2}\right)+C$

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

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