Question

$\int_{}^{}{\sqrt{\tan x}dx =}$

• A. $\frac{1}{\sqrt{2}}\tan^{- 1}\left( \frac{\tan x - 1}{\sqrt{2\tan x}} \right) + \frac{1}{2\sqrt{2}}\log\left| \frac{\tan x - \sqrt{2\tan x} + 1}{\tan x + \sqrt{2\tan x} + 1} \right| + c$
• B. $\frac{1}{\sqrt{2}}\tan^{- 1}\left( \frac{\tan x - 1}{\sqrt{2\tan x}} \right) - \frac{1}{2\sqrt{2}}\log\left| \frac{\tan x - \sqrt{2\tan x} + 1}{\tan x + \sqrt{2\tan x} + 1} \right| + c$
• C. $\log\left( \frac{\tan x - 1}{\sqrt{2\tan x}} \right) + c$
• D. None of these

Answer 1

Answer: (A)

$\frac{1}{\sqrt{2}}\tan^{- 1}\left( \frac{\tan x - 1}{\sqrt{2\tan x}} \right) + \frac{1}{2\sqrt{2}}\log\left| \frac{\tan x - \sqrt{2\tan x} + 1}{\tan x + \sqrt{2\tan x} + 1} \right| + c$

Answer 2

Let $I = \int_{}^{}{\sqrt{\tan x}dx}$

Putting $\sqrt{\tan x} = t \Rightarrow \tan x = t^{2},$ we have

$\sec^{2}xdx = 2tdt \Rightarrow (1 + \tan^{2}x)dx = 2tdt$i.e., $(1 + t^{4})dx = 2tdt \Rightarrow dx = \frac{2t}{1 + t^{4}}dt\therefore I = \int_{}^{}{\sqrt{\tan x}dx = \int_{}^{}{t.\frac{2t}{1 + t^{4}}dt = \int_{}^{}{\frac{2t^{2}}{1 + t^{4}}dt}}} = \int_{}^{}\frac{(t^{2} + 1) + (t^{2} - 1)}{t^{4} + 1}dt = \int_{}^{}{\frac{t^{2} + 1}{t^{4} + 1}dt + \int_{}^{}{\frac{t^{2} - 1}{t^{4} + 1}dt}}$ $= \int_{}^{}{\frac{1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}}dt + \int_{}^{}{\frac{1 - \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}}dt}}$ $= \int_{}^{}{\frac{1 + \frac{1}{t^{2}}}{\left( t - \frac{1}{t} \right)^{2} + 2}dt + \int_{}^{}{\frac{1 - \frac{1}{t^{2}}}{\left( t + \frac{1}{t} \right)^{2} - 2}dt}}$