Q83 / 150 · MHT CET 2011 · Mathematics

$\int(\sqrt{\tan x}+\sqrt{\cot } x) d x$

  • A
    $\sqrt{2} \tan ^{-1}\left(\frac{\tan x}{\sqrt{2 \tan x}}\right)+C$
  • Correct answer
    B
    $\sqrt{2} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+C$
  • C
    $\frac{\tan x}{\sqrt{2}} \cdot \tan ^{-1}\left(\frac{\cot x+1}{\sqrt{2 \tan x}}\right)+C$
  • D
    $\frac{\tan x}{\sqrt{2}} \cdot \tan ^{-1}\left(\frac{\cot x+1}{\sqrt{2 \tan x}}\right)+C$

Solution

Let $I=\int \frac{(\sin x+\cos x)}{\sqrt{\sin x \cdot \cos x}} d x$ $=\int \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2 \sin x \cdot \cos x}} d x$ $=\sqrt{2} \int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$ Put $\quad \sin x-\cos x=t$ Also, $\quad \sin 2 x=\left(1-t^{2}\right)$ $\therefore \quad I=\sqrt{2} \int \frac{d t}{\sqrt{1-t^{2}}}$ $\quad=\sqrt{2} \sin ^{-1} t+C$ $\quad=\sqrt{2} \sin ^{-1}(\sin x-\cos x)+C$ $\quad=\sqrt{2} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+C$

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

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